If 3 circular bars made up of glass, steel, aluminium having equal cross section applied with muiltiaxial load at same point in same direction , The bar will produce more stress: glass
If 3 circular bars made up of glass, steel, aluminium having equal cross section applied with muiltiaxial load at same point in same direction, the bar made of glass will produce more stress.
This is because the modulus of elasticity of glass is lower than that of steel and aluminium. The modulus of elasticity is a measure of a material's resistance to deformation under load. A lower modulus of elasticity means that the material is more prone to deformation and therefore, will produce more stress under the same load.
In general, the stress produced in a material under load is given by the equation:
stress = load / cross-sectional area
Since the cross-sectional area of the bars is equal and the load applied is the same, the stress produced will depend on the modulus of elasticity of the material. Therefore, the bar made of glass will produce more stress under the same load as compared to the bars made of steel and aluminium.
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The steel bar will produce the most stress when applied with a multi-axial load at the same point in the same direction, as steel is the strongest of the three materials (glass, steel, and aluminium).
The bar made of steel will produce more stress under a multi-axial load at the same point in the same direction. This is because steel has a higher Young's modulus, or the measure of the ability of a material to withstand changes in length when under tension or compression, than glass or aluminum. Therefore, it will experience greater stress under the same loading conditions. Steel has a much higher Young's modulus than glass or aluminum, meaning it will experience greater stress under the same loading conditions.
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A pan is used to boil water by placing it on a stove, from which heat is transferred at a fixed rate q0. There are two stages to the process. In Stage I, the water is taken from its initial (room) temperature Ti to the boiling point, as heat is transferred from the pan by natural convection. During this stage, a constant value of the convection coefficient h may be assumed, while the bulk temperature of the water increases with time, T infinity = T infinity(t). In Stage 2, the water has come to a boil, and its temperature remains at a fixed value, T infinity = Tb, as heating continues. Consider a pan bottom of thickness L and diameter D, with a coordinate system corresponding to x = 0 and x = L for the surfaces in contact with the stove and water, respectively. Write the form of the heat equation and the boundary/ initial conditions that determine the variation of temperature with position and time, T(x, t), in the pan bottom during Stage 1. Express your result in terms of the parameters qo, D, L, h, and T infinity, as well as appropriate properties of the pan material.
During Stage 1, the heat equation that governs the temperature distribution T(x, t) in the pan bottom is given by:
ρc_p∂T/∂t = k∂^2T/∂x^2 + q_0/V
How to use heat equation ?where ρ is the density of the pan material, c_p is the specific heat capacity of the pan material, k is the thermal conductivity of the pan material, and V is the volume of the pan bottom.The first term on the right-hand side of the equation represents heat conduction within the pan material, while the second term represents heat transfer from the pan to the water by natural convection. The term q_0/V represents the heat input rate from the stove.The boundary conditions are:
At x = 0 (the surface in contact with the stove), the temperature is fixed at Ti, and there is no heat flux:
T(x=0,t) = Ti, ∂T/∂x = 0
At x = L (the surface in contact with the water), the heat flux is given by:
-k(∂T/∂x) = h(Tb - T)
where h is the convection coefficient, Tb is the boiling point of water, and T is the bulk temperature of the water, which varies with time.The initial condition is:
T(x, t=0) = Ti
The solution to this heat equation with the above boundary and initial conditions will give the variation of temperature with position and time, T(x, t), in the pan bottom during Stage 1.Note that the properties of the pan material, including ρ, c_p, and k, need to be specified in order to solve the heat equation.
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Question
What is the angular speed of the second hand of a clock? If the second hand is 10cm long, then find the linear speed of its tip. (in rad/s and m/s)
A
0.1047,0.01047
B
1047,0.01047
C
0.1047,1047
D
0.0047,0.01047
Easy
The correct answer of the angular speed and linear speed respectively is option A) 0.1047,0.01047
To find the angular speed of the second hand of a clock, we use the formula:
Angular speed = 2π / time
Since the second hand of a clock completes one full rotation every 60 seconds, the time is 60 seconds. Therefore:
Angular speed = 2π / 60
Angular speed = 0.1047 rad/s
To find the linear speed of the tip of the second hand, we use the formula:
Linear speed = angular speed x radius (v = ωr)
Since the radius of the second hand is 10cm or 0.1m, we can plug in the values:
Linear speed = 0.1047 x 0.1
Linear speed = 0.01047 m/s
Therefore, the angular speed of the second hand of a clock is 0.1047 rad/s and the linear speed of its tip is 0.01047 m/s.
The correct answer to the angular speed and linear speed is therefore option A) 0.1047, 0.01047.
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Which of the quantities are zero throughout the flight?
The horizontal component of acceleration is always zero is the quantities are zero throughout the flight.
What is acceleration?
Acceleration was the representation rate In a change of velocity because the acceleration always depends on the object's speed. Acceleration determines the rate of the particles. Acceleration is the vector quantity. It is a vector quantity, but it has both extent and movement. Newton's law also has the acceleration of the magnitude described. The m.s-2 is the standard unit for acceleration.
What is velocity ?
The most important metric for determining an object's position and rate of movement is its velocity. The distance that an object travels in a certain amount of time might be used to define it. The object's displacement in a unit of time is referred to as velocity.
Therefore, The horizontal component of acceleration is always zero is the quantities are zero throughout the flight.
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circulation of heat in the oceans and atmosphere is an example of energy movement through _________________.
Circulation of heat in the oceans and atmosphere is an example of energy movement through convection.
What is convection?The movement of heat energy through a fluid is known as convection. This kind of heating is most frequently seen in the kitchen, typically with a liquid that is brought to a boil. The air that makes up the atmosphere behaves like a fluid. The rocks are warmed up as a result of the sun's radiation penetrating the ground.
As a result of conduction, the temperature of the rock will increase, which will cause heat energy to be released into the environment. This will result in the formation of a bubble of air that is warmer than the air around it. This pocket of air climbs into the atmosphere and continues its journey. The heat that was held within the bubble dissipates into the atmosphere as it rises, causing the bubble to gradually become cooler.
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Continuing the preparation of the proposal for the Eau Gaullie treatment plant (Problem 6-38), design the flocculation tank by providing the follow- ing for the first two compartments only: 1. Water power input in kW 2. Tank dimensions in m 3. Diameter of the impeller in m 4. Rotational speed of impeller in rpm Use the following assumptions: 1. Number of tanks = two 2. Tapered G in three compartments: 90, 60, and 30 s 3. GO= 120,000 4. Compartment length = width = depth 5. Impeller type: axial-flow impeller, three blades, Np = 0.31 6. Available impeller diameters: 1.0, 1.5, and 2.0 m 1 7. Assume B = 1/3 H Answers for first compartment only: P = 295.31, or 295 W or 0.295 kW L=W=D = 3.3 m Impeller diameter = 1.5 m Rotational speed = 30 rpm
Eau Gaullie treatment a) expected flow rate is 1852 gpm. b) Tank diameter is 41.7 ft. c) input power is 21.1 hP
From flow in tank coefficient equation C_Q = Q/(W×D³)
Q ∝ d³
Qd₁/Qd₂ = d₁³/d₂³
where Eau Gaullie treatment Qd₁ and Qd₂ are different discharges with respect to different diameter, (Qd₁ = 3200 gpm) and d₁ is diameter one ( 12 in) while d₂ is diameter 2 ( 10 in),
now we substitute
Qd₂ = Qd₁(d₂³/d₁³)
Qd₂ = 3200( 10/12)³
Qd₂ = 1852 gpm
∴ expected flow rate is 1852 gpm
next we calculate the head
h ∝ d²
hd₁/hd₂ = d₁²/d₂²
where hd₁ and hd₂ are different heads with respect to different diameter, (hd₁ = 60 ft) and d₁ is diameter one ( 12 in) while d₂ is diameter 2 ( 10 in),
now we substitute
hd₂ = hd₁(d₂²/d₁²)
hd₂ = 60 ( 10/12 )²
hd₂ = 41.7 ft
∴ head is 41.7 ft
Now calculate the input power
W ∝ d⁵
Wd₁/Wd₂ = d₁⁵/d₂⁵
where Wd₁ and Wd₂ are different power with respect to different diameter, (Wd₁ = 60 hp) and d₁ is diameter one ( 12 in) while d₂ is diameter 2 ( 10 in),
now we substitute
Wd₂ = Wd₁(d₂⁵/d₁⁵)
Wd₂ = 60 ( 10/12 )⁵
Wd₂ = 21.1 hP
∴ input power is 21.1 hP
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which of the following changes will result in a larger frequency for an oscillating spring-mass system?
An increase in the spring constant changes will result in a larger frequency for an oscillating spring-mass system.
About mass -spring systemThe mass spring system is a system composed of objects that have mass and are connected to springs. The spring circuit can be composed of several springs mounted in series or parallel as needed. Springs connected in series will decrease the value of the spring constant, while the installation of springs in parallel will increase the value of the spring constant.
Spring system formation is a simple spring system formation arranged in series or parallel. The stages of the research carried out included determining the mathematical equation for mass movement determined from the equilibrium position in the formation of a spring system, including a spring system of a mass connected to two springs arranged in series, a spring system of a mass connected to two springs arranged in parallel and a spring system of two masses. connected by two springs arranged in series, then create a spring system simulation program.
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newtons ring experiment for engineering
Newton's ring is a phenomenon of interference pattern of the light rays which is created by reflection of light rays.
What is Newton's ring experiment?Newton's rings is a phenomenon in which an interference pattern of the light is generally created by the reflection of light rays between the two surfaces, typically a spherical surface and an adjacent touching flat surface in space.
Newton's rings, in optics, is a series of concentric light- and dark-colored bands which are observed between any two pieces of glass when one is convex and it rests on its convex side on another piece which is having a flat surface. Thus, a layer of air exists between the two.
Newtons ring experiment is used for the determination of wavelength of monochromatic lights. It is also used for the determination of refractive index of transparent liquid.
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An object that can move in either direction along a horizontal line ( the positive axis). Assume that friction is so small that it can be neglected. Sketch the shape of the graph of the force applied to the object that would produce the motion described.
A) The object moves away from the origin with a constant velocity.
B)The object moves toward the origin with a constant velocity.
C) The object moves away from the origin with a steadily increasing velocity (a constant acceleration)
A) and B): No force whenever, F(t) = 0, ∀t. C): A constant force,
F(t) = F0,∀t.
Newton's Laws
In the year 1687, Isaac Newton distributed his fundamental work "Philosophiae Naturalis Principia Mathematica", in which he included three essential standards of movement. These standards are referred to now as Newton's maxims or Newton's laws.
A) and B): No force whenever,
F(t) = 0, ∀t.
As indicated by Newton's most memorable regulation, an article moves with constant velocity when no net force is following up on it.
C): A constant force,
F(t) = F0,∀t.
As indicated by Newton's subsequent regulation, the speed increase is corresponding to the acting force. Assuming that the speed increase is constant, the force is constant.
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what a geologist sees. a geology student sits beside an outcrop on mont royal, the namesake of montreal. this exposure shows the intrusion of molten basalt into preexisting limestone.
The geologist sees depositional contact. Therefore, option A is correct.
What is depositional contact?A sedimentary or volcanic rock is said to have been deposited on an older rock at a depositional contact (of any type). Where volcanic rocks encroach on older rock, this is known as an intrusive contact (of any type).
The ten different types of contacts are as follows: (1) bedding planes, (2) diastems, (3) angular irregularities, (4) disconformities, (5) para-conformities, (6) nonconformities, (7) pedologic contacts, (8) faults, 9) intrusive contacts, and 10) extrusive contacts.
Hence, option A is correct.
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your question is incomplete, most probably the full question is this:
what a geologist sees. a geology student sits beside an outcrop on mont royal, the namesake of montreal. this exposure shows the intrusion of molten basalt into preexisting limestone.
A parallel plate capacitor with plates of area A and plate separation d is charged so that the potential difference between its plates is V. If the capacitor is then isolated and its plate separation is decreased to d/2, what happens to the potential difference between the plates? A) The final potential difference is 4V B) The final potential difference is 2V. C) The final potential difference is 0.51 D) The final potential difference is 0.25V E) The final potential difference is V.
The new potential difference when the distance between the plates vary is calculated to be 0.5 V. Correct option is C.
The parallel plate capacitor's area is denoted by the letter A.
Distance between the plates is d.
Potential difference is V.
If the distance is reduced to d/2, the potential difference is to be found out.
We know that, Q = C V
To find out potential difference, make it as subject,
V = Q/C
From the above equation, it is seen that, capacitance is inversely proportional to potential difference.
So, V C = k
where,
k is contant
The parallel plate capacitor's capacitance is stated as,
C = εA/d
When the distance apart is d. Then, C₁ = εA/d
When the distance is half d/2
C₂ = εA/(d/2)
C₂ = 2εA/d
Then, applying V C = k
V₁ is voltage of the full capacitor V1 = V
V₂ is the required voltage let say V'
Then,
V₁ C₁ = V₂ C₂
V × εA/d=V' × 2εA/d
VεA/d = 2V'εA/d
Then the εA/d cancels on both sides and remains
V = 2V'
Then, V' = V/2
Thus, the potential difference is half when the distance between the parallel plate capacitor was reduce to d/2.
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A pool ball moving 1.83 m/s
strikes an identical ball at rest.
Afterward, the first ball moves
1.15 m/s at a 23.3° angle. What is
the x-component of the velocity of
the second ball?
Answer:n to the right to the right
Explanation:
A velocity vector 39∘ above the positive x -axis has a y -component of 10m/s .What is the value of its x -component?
The y component of the velocity vector is 10 m/s. Then the v is 12.86 m/s. Now the x component of this velocity is v sin θ.That is , x component is 8 m/s.
What is velocity ?Velocity of an object is the measure of its distance travelled per unit time. Velocity is a vector quantity thus having both magnitude and acceleration. The magnitude is called speed.
For a velocity vector there are three translation components possible for three different axes.
here, x component = v sinθ
y component = v cos θ
Given that v cos θ = 10 m/s
θ = 39
Vy = v cos 39 = 10 m/s
then v = 12.8 m/s
Now, the x component is calculated as:
vx = 12.8 m/s sin 39 = 8 m/s.
Therefore, the x component of velocity here is 8 m/s.
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A student in a science lab is investigating heat transfer and thermal energy conservation as she mixes hot and cold water. She first measures out her desired amount of cold water into a Styrofoam cup. She then measures out hot water from the faucet or from a pot hot water on her stove. After she measures the temperature of the hot and cold waters, she pours the hot water into the cold water. She monitors the temperature of the mixed waters and records the final temperature. She uses a standard thermometer to record the temperatures. She does three trials, which are shown below:
Trial 1:
For her first trial, the student decided to mix 250 mL of water at 20 °C with 250 mL of water at 98 °C. After waiting some time, she recorded the temperature of the mix to be 56 °C.
Trial 2:
For her second trial, the student decided to mix 200 mL of water at 20 °C with 400 mL of water at 98 °C. After waiting some time, she recorded the temperature of the mix to be 72 °C.
Trial 3:
For her third trial, the student decided to mix 300 mL of water at 15 °C with 150 mL of water at 90 °C. After waiting some time, she recorded the temperature of the mix to be 41 °C.
Include a data table that organizes the data collected from the three trials.
Make another table, or add to your table, to show data calculations. You will calculate the change in temperatures of the cold and hot water, as well as the mass of the cold and hot waters.
Use the beginning temperature of the hot and cold water and the final temperature of the mixture to calculate the change in temperature of the cold water and the change in temperature of the hot water. For example, the temperature of the cold water was raised from its beginning temperature to the final temperature of the mixture.
Since one milliliter (mL) of water has a mass of one gram (g), it is very easy to determine the mass of the cold and hot water. For example: If you have 100 mL of water, then the mass of the water is 100 g. Remember, 1 kg = 1000 g. Convert the mass of the hot and cold water to kilograms.
Use the equation Q = (m)(c)(Δ T) to calculate the heat gained by the cold water for each trial. Show your work using the problem-solving method shown in previous rubrics. The specific heat for water (c) is 4186 J/(kg * C°).
Use the equation Q = (m)(c)(Δ T) to calculate the heat "lost" by the hot water for each trial. Show your work using the problem-solving method shown in previous rubrics. The specific heat for water (c) is 4186 J/(kg * C°).
Compare the values for heat gain and heat loss in questions 3 and 4.
In an isolated system, the total heat given off by warmer substances equals the total heat energy gained by cooler substances. Now look at your answer to Question 5. What might have caused the difference you have reported? Even though this data was provided to you, think of the errors the student could have encountered when collecting the data.
Write a complete conclusion for this activity.
The heats gained by cold water in trials 1, 2 and 3 are 3048.6 J, 4311.52 J and 3048.6 J respectively and the heat lost are -6858 J, -4307.84 J and -3593.9 J respectively.
How to find heat gained and lost?Calculation of Heat Gained by Cold Water:
Trial 1:
Q = (m)(c)(Δ T)
Q = (0.25 kg)(4186 J/(kg x C°))(36°C)
Q = 3048.6 J
Trial 2:
Q = (m)(c)(Δ T)
Q = (0.2 kg)(4186 J/(kg x C°))(52°C)
Q = 4311.52 J
Trial 3:
Q = (m)(c)(Δ T)
Q = (0.3 kg)(4186 J/(kg x C°))(26°C)
Q = 3048.6 J
Calculation of Heat Lost by Hot Water:
Trial 1:
Q = (m)(c)(Δ T)
Q = (0.25 kg)(4186 J/(kg * C°))(-52°C)
Q = -6858 J
Trial 2:
Q = (m)(c)(Δ T)
Q = (0.4 kg)(4186 J/(kg x C°))(-26°C)
Q = -4307.84 J
Trial 3:
Q = (m)(c)(Δ T)
Q = (0.15 kg)(4186 J/(kg x C°))(-75°C)
Q = -3593.9 J
Comparison of Heat Gain and Heat Loss:
In all three trials, the heat lost by the hot water is not equal to the heat gained by the cold water. This discrepancy is likely due to errors in the measurement of the temperature and volume of the water, or due to heat loss to the environment.
Conclusion:
This activity allowed the student to investigate heat transfer and thermal energy conservation. Through the measurement of the temperature of hot and cold water and the calculation of heat gained and lost by each, the student was able to gain a better understanding of these concepts. However, it is important to note that the results may have been affected by errors in measurement, so further experimentation and refinement of techniques may be necessary to obtain more accurate results.
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If a car is pushed with a force of 18N for 8m, how much work has been done?
According to the question, the work done by a car is calculated as 144Nm.
What is force?Force may be defined as a process of pushing or pulling on an object that significantly produces acceleration in the body on which it acts. It is an external agent capable of changing a body's state of rest or motion. It has a magnitude and a direction.
According to the question,
The force applied on a car = 18 N
The displacement made by a car = 8m.
Now, the work done is calculated with the help of the given formula:
Work done = Force × Displacement.= 18 N × 8m = 144Nm.
Therefore, the work done by a car is calculated as 144Nm.
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An object is launched from the origin with a velocity of 10.0 m/s at an angle of 30.0 degrees above the horizontal. What is the velocity of the object 2.00 seconds later? A. Vx=8.7 m/s, Vy=-14.6 m/s B. Vx - 14.6 m/s, Vy=-5.6 m/s C. Vx = -8.7 m/s, Vy=-14.6 m/s D. Vx-14.6 m/s. Vy - 8.7 m/s E. Vx = 5.6 m/s, Vy=-14.6 m/s
The velocity of the object both horizontally and vertically at 2.00 seconds is (A) 8.7 m/s and - 14.6 m/s. The result is obtained by us using formula for projectile motion.
How to find horizontal and vertical velocity of projectile motion?In horizontal motion, the velocity is not affected by the gravitational acceleration. It can be calculated by
vx = v₀ cos θ
In vertical motion, the velocity is affected by the gravitational acceleration. It ca be expressed as
vy = v₀ sin θ - gt
An object is moving in projectile motion.
we have:
initial velocity, v₀ = 10.0 m/sAngle above the horizontal, θ = 30°Time, t = 2.00 sThe horizontal velocity of the object is
vx = 10.0 × Cos 30°
vx = 10.0 × ½ √3
vx = 5√3
vx = 8.7 m/s
The vertical velocity of the object is
vy = v₀ sin 30 - gt
vy = 10.0 (½) - 9.8 (2.00)
vy = 5 - 19.6
vy = - 14.6 m/s
Hence, the velocity of the object is vx = 87 m/s and vy = - 14. m/s.
The correct option is (A).
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The position of a particle moving in the x-y plane at any time t is given by : x=(3t^3−6t) metres; y=(t^2−2t) metres. Select the correct statement.
a. acceleration is zero at t=0
b. Valocity is zero at t=0
c. Valocity is zero at t=ts
d. Valocity and acceleration of the particle are never zero
The required velocity and acceleration of this particle when position of the particle is given are never zero. Correct option is D.
The position of the particle is given as x = 3t³ - 6t, y = t² - 2t
The particle's velocity is determined by,
v = dx/dt i + dy/dt j = (9t² - 6) i + (2t - 2) j
The velocity at the point t = 0 is,
v = (9t² - 6) i + (2t - 2) j = - 6 i - 2 j
The velocity at the point t = 1 is,
v = (9t² - 6) i + (2t - 2) j = 3 i
The acceleration of the particle is,
a = d²x/dt² i + d²y/dt² j = 18t i + 2 j
The particle's acceleration at time zero is,
a = 2 j
Thus, the velocity and acceleration of this particle are never zero.
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A 5-kg block is suspended by a rope from the ceiling of an elevator that accelerates downward at 3.0 m/s2 . The tension force of the rope on the block is: (specify magnitude and direction)
The tension force is 15 N.
What is the tension force?Tension force is a pulling force transmitted through a flexible medium, such as a rope or cable, when it is pulled at both ends. It is directed along the length of the medium and is equal in magnitude at both ends.
In this case we have been told that A 5-kg block is suspended by a rope from the ceiling of an elevator that accelerates downward at 3.0 m/s2 .
We know that the tension force is the force that acts along the rope hence;
Tension = ma
= 5 Kg * 3.0 m/s2
= 15 N
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Sound is emitted by a point source. You wish to compare the sound intensity and sound intensity level from this source at two different sites. If the distance to the second site is a factor of 8 greater than the distance to the first, determine the following.
(a) Determine the multiplicative factor by which the sound intensity decreases as you go from the first to the second site. (Assume the intensities at the first and second sites are I1 and I2, respectively.)
I1
I2
= The answer to this one is 64.
(b) Determine the additive amount by which the sound level intensity decreases as you go from the first to the second site. (Assume the sound level intensities at the first and second sites are ?1 and ?2, respectively.) How is sound intensity level related to sound intensity? Recall that
log10(ab) = log10(a) + log10(b).
?1 ? ?2 = dB
a) The multiplicative factor by which the sound intensity decreases as you go from the first to the second site is 64.
b) The additive amount by which the sound level intensity decreases as you go from the first to the second site is -15.56 dB.
a) The intensity of the sound is measured in decibels and the sound intensity level with the intensity I is calculated by the expression,
β(dB) = 10 log(I/I₀)
where,
β/dB is the intensity level
I is sound intensity
The intensity is given by the relation,
Intensity = Power/Area
I = (P/πr²)
Intensity is inversely related to the square of radius.
I₁/I₂ = (r₂/r₁)² = (8/1)² = 64
I₁ = 64 I₂
So, the multiplicative factor is 64.
b) The formula for the intensity level is given by,
β(dB) = 10 log(I/I₀)
β₁ = 10 dB log(I₁/I₀)
β₂ = 10 dB log(I₂/I₀)
β₂ - β₁ = 10 dB[log(I₂/I₀) - log(I₁/I₀)]
β₂ - β₁ = 10 dB log(I₂/I₁)
β₂ - β₁ = 10 dB log(1/36) = -15.56 dB
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Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150 MW by the radioactive decay of fission products. This heat transfer causes a rapid increase in temperature if the cooling system fails (1 watt = 1 joule/second or 1 W = 1 J/s and 1 MW = 1 megawatt).
(a) Calculate the rate of temperature increase in degrees Celsius per second (
∘
C/s) if the mass of the reactor core is 1.60
×
10
5
kg and it has an average specific heat of 0.3349 kJ/kg
∘
⋅
C.
(b) How long would it take to obtain a temperature increase of 2000
∘
C, which could cause some metals holding the radioactive materials to melt? (The initial rate of temperature increase would be greater than that calculated here because the heat transfer is concentrated in a smaller mass. Later, however, the temperature increase would slow down because the 5
×
10
5
-kg steel containment vessel would also begin to heat up.)
(a) The rate of temperature increase is about 1.40 degrees Celsius per second. (b) It would take about 1428.57 seconds or 23.81 minutes to obtain a temperature increase of 2000 degrees Celsius.
(a) The rate of temperature increase can be calculated by first finding the total energy being transferred per second by the decay of fission products, which is 150 MW. This is equivalent to 150 x 10⁶ J/s. Then, we can use the formula:
rate of temperature increase = (energy transferred per second) / (mass x specific heat)
Plugging in the values, we get:
rate of temperature increase = (150 x 10⁶ J/s) / (1.60 x 10⁵ kg x 0.3349 kJ/kg∘C)
rate of temperature increase ≈ 1.40∘C/s
Therefore, the rate of temperature increase is about 1.40 degrees Celsius per second.
(b) To find the time it would take to obtain a temperature increase of 2000 degrees Celsius, we can use the formula:
time = (change in temperature) / (rate of temperature increase)
Plugging in the values, we get:
time = (2000∘C) / (1.40∘C/s)
time ≈ 1428.57 s
Therefore, it would take about 1428.57 seconds or 23.81 minutes to obtain a temperature increase of 2000 degrees Celsius.
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Complete question:
Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150 MW by the radioactive decay of fission products. This heat transfer will cause a rapid increase in temperature if the cooling system fails.
(a) Calculate the rate of temperature increase, in degrees Celsius per second (°C/s), if the mass of the reactor core is 1.4 × 105 kg and it has an average specific heat of 0.3349 kJ/(kg.°C).
(b) How long, in minutes, would it take for the temperature to increase by 2000°C, which could cause some metals holding the radioactive materials to melt? (The initial rate of temperature increase would be greater than that calculated here because the heat transfer is concentrated in a smaller mass. Later, however, the temperature increase would slow down because the 5 x 105-kg steel containment vessel would also begin to heat up.)
A)How much heat does it take to increase the temperature of 2.10 molmol of an ideal gas by 60.0 KK near room temperature if the gas is held at constant volume and is diatomic?
B)What is the answer to the question in part A if the gas is monatomic?
A) It would take 2079 J of heat to increase the temperature of 2.10 mol of a diatomic gas by 60.0 K at a constant volume.
B) It would take 1244 J of heat to increase the temperature of 2.10 mol of a monatomic gas by 60.0 K at a constant volume.
What is the difference between diatomic and monoatomic?The terms "diatomic" and "monoatomic" describe how many atoms make up a molecule or an ion. Diatomic molecules, like O2 or HCl, are made up of two covalently connected atoms of the same element. On the other hand, monoatomic species are made of a single atom, which could be neutral like helium or argon or charged like cations and anions. Diatomic molecules have unique chemical properties and are frequently involved in chemical processes, whereas monoatomic species normally exist as gases under normal conditions and are relatively inert. In the study of chemistry and physics, the contrast between diatomic and monoatomic particles is significant, particularly in understanding the behavior of various elements and their interactions with other substances.
(A) To calculate the amount of heat required to increase the temperature of 2.10 mol of a diatomic gas by 60.0 K at constant volume, we can use the formula: Q = nCvΔT
where Q is the amount of heat, n is the number of moles of the gas, Cv is the molar heat capacity at constant volume, and ΔT is the change in temperature. For a diatomic gas, Cv = (5/2)R, where R is the gas constant.
So, substituting the given values, we get:
Q = (2.10 mol)(5/2)(8.31 J/mol·K)(60.0 K)
Q = 2079 J
Therefore, it would take 2079 J of heat to increase the temperature of 2.10 mol of a diatomic gas by 60.0 K at a constant volume.
B) For a monatomic gas, Cv = (3/2)R. So, using the same formula as above, we get:
Q = (2.10 mol)(3/2)(8.31 J/mol·K)(60.0 K)
Q = 1244 J
Therefore, it would take 1244 J of heat to increase the temperature of 2.10 mol of a monatomic gas by 60.0 K at a constant volume.
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) To calculate the amount of heat required to increase the temperature of 2.10 mol of a diatomic gas by 60.0 K at constant volume, we can use the formula:
Q = nCvΔT
where Q is the amount of heat, n is the number of moles of the gas, Cv is the molar heat capacity at constant volume, and ΔT is the change in temperature. For a diatomic gas, Cv = (5/2)R, where R is the gas constant.
So, substituting the given values, we get:
Q = (2.10 mol)(5/2)(8.31 J/mol·K)(60.0 K)
Q = 2079 J
Therefore, it would take 2079 J of heat to increase the temperature of 2.10 mol of a diatomic gas by 60.0 K at constant volume.
B) For a monatomic gas, Cv = (3/2)R. So, using the same formula as above, we get:
Q = (2.10 mol)(3/2)(8.31 J/mol·K)(60.0 K)
Q = 1244 J
Therefore, it would take 1244 J of heat to increase the temperature of 2.10 mol of a monatomic gas by 60.0 K at constant volume.
a car comes to a bridge during a storm and finds the bridge washed out. the driver must get to the other side, so he decides to try leaping it with his car. the side the car is on is 18.1 m above the river, whereas the opposite side is a mere 2.0 m above the river. the river itself is a raging torrent 68.0 m wide. for help with math skills, you may want to review: vector magnitudes for general problem-solving tips and strategies for this topic, you may want to view a video tutor solution of different initial and final heights.
Equation to solve for v:v = sqrt(u² + 2as)Substituting the given values, we get:v = sqrt((25.7 m/s)² + 2(-9.8 m/s²)(16.1 m))v = 14.9 m/s
What is speed?It is a scalar quantity that represents the speed at which an item moves or changes position with respect to time.
A) To determine the initial velocity required to clear the river and land safely on the other side, we need to use the following equation d = vt + (1/2)at²
Where d is the distance the car needs to travel horizontally, which is the width of the river (68 m).v is the initial velocity of the car. An is the acceleration due to gravity (-9.8 m/s²).
T is the time the car spends in the air. We can rearrange this equation to solve for v:v = (d – (1/2)at²) / substituting the given values, we get:v = (68 m – (1/2)(-9.8 m/s²)(18.1 m)²) / (2(18.1 m)/v)v = 25.7 m/s
B) To determine the speed of the car just before it lands on the other side, we can use the following equation: where:u is the initial velocity of the car. An is the acceleration due to gravity (-9.8 m/s²).s [tex]v^² = u^² + 2as[/tex]is the vertical distance the car travels from the cliff to the other side (18.1 m – 2 m = 16.1 m).
Therefore, (A) To clear the river and safely land on the other side, the car must be moving at a speed of 25.7 m/s (or approximately 92.5 km/h) immediately as it exits the cliff.
(B) The car is travelling at a speed of 14.9 m/s, or approximately 53.6 km/h, shortly before it safely falls on the opposite side.
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A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 18.1 m above the river, whereas the opposite side is a mere 2 m above the river. The river itself is a raging torrent 68 m wide.A) How fast should the car be travelling just as it leaves the cliff in order to just clear the river and land safely on the opposite side? B) What is the speed of the car just before it lands safely on the other side?
4) Collision in which K.E and momentum of system remain same is called__________
(*)Elastic Collision
(*) Inelastic collision
(*) Conserved collision
(*) Linear collision
A collision that is elastic occurs when there is no net loss of kinetic energy in the system as a result of the collision. Kinetic energy and momentum are both conserved in elastic collisions.
Give an example of an elastic collision.When two balls collide at a pool table, that is an instance of an elastic collision. When you throw a ball on the ground and it bounces back into your hand, there is no net change in the kinetic energy, making it an elastic collision.
Give an illustration of what an elastic collision is.Two balls colliding at a pool table is an example of an elastic collision. When a ball is tossed to the ground and subsequently returns to your hand, there is no net change in the kinetic energy, making it an elastic collision.
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For each of the regions, use the midpoint method to identify whether the supply of this good is elastic or inelastic.Region Elastic or InelasticBetween Y and Z Between W and X
(a). Elasticity of supply between W and X is elastic.
(a)-1.The supply elasticity relationship Y and Z becomes inelastic.
b. The assertion is accurate.
What kind of products are elastic?Like a dishwasher or a car, these are infrequently acquired things that can be put off if the price increases. Elasticity is a crucial economic metric because it shows how much of an item or service consumers consumes when the price varies, which is especially essential for businesses that sell goods or services. That whenever a commodity is elastic, a price fluctuation prompts a shift in demand for it rapidly.
What three forms of elasticity are there?Economists use elasticity to determine how different factors interact. Pricing, cross-price quantity demanded, and quantity demanded are the three main types of elasticity. Goods' elasticity gauges how sensitive they are to price changes.
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HELPPPPP MEE
LATE SCIENCE WORK
The true statement about the motion of the helicopter is " the helicopter travels at a greater speed between D and E than it did between A and B.
option D.
What is motion?
Motion is the process of an object changing its position with respect to a reference point over time. It involves the movement of an object from one point in space to another.
Motion can be described in terms of its speed, direction, and acceleration. Speed refers to the rate at which an object is moving, while direction refers to the path that the object is following. Acceleration refers to the rate at which an object's speed or direction changes.
v = Δx / Δt
where;
Δx is change in positionΔt is change in timeThe change in position of the helicopter between A and B = 0
because, A = 100 m and B = 100 m
so the speed of the helicopter between A and B = 0 m/s.
Thus, the last statement is the only correct option,
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a marathon runner starts her race at a speed of 3 m/s. by the end of the race, 5 hours later, her speed is 2.8 m/s. what was her average acceleration?
Answer: Acceleration, a= -1.11*10^-5
Explanation:
Using a= (V(final speed)- U(Initial speed) )/ TIME IN SECONDS
5hr to seconds is 5*60*60=18,000s
(2.8m/s-3m/s)/ 18,000s = -1.11*10^-5
I'm sorry I wasn't able to explain well, but I hope this helps.
Suppose that a right-moving em wave overlaps with a left-moving em wave so that, in a certain region of space, the net electric field in the y direction and magnetic field in the z direction are given by Ey = E0 sin(kx-wt)+E0 sin(kx+wt) and Bz =B0 sin(kx-wt)+B0 sin(kx+wt). (a) Find the mathematical expression that represents the standing electric and magnetic waves in the y and z directions, respectively. (b) Determine the Poynting vector and find the x locations at which it is zero at all times.
The standing wave can be obtained by taking the sum of the two traveling waves with equal amplitude and opposite phase velocities. To do this, we can use the trigonometric identity:
How to use wave amplitude?sin(A) + sin(B) = 2 cos((A+B)/2) sin((A-B)/2)
Applying this identity to the electric field Ey, we get:Ey = 2 E0 cos(wt) sin(kx)
This represents a standing wave with nodes (zero amplitude) at x = nλ/2k, where n is an integer.
Similarly, for the magnetic field Bz, we have:
Bz = 2 B0 cos(wt) sin(kx)
This also represents a standing wave with nodes at the same positions as the electric field.(b) The Poynting vector represents the flow of energy of the electromagnetic wave and is given by:S = E x B
where x represents the vector cross product. Substituting the expressions for E and B, we get:
S = E0 B0 sin^2(kx) / μ0
where μ0 is the permeability of free space.
The Poynting vector is zero when sin^2(kx) = 0, which occurs at x = nπ/k for n an integer. This represents positions where the electric and magnetic fields are out of phase and the energy flow is momentarily zero.
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A capacitor consists of two metal plates each 10 cm by 20 cm; they are separated by a 2.0 mm thick insulator with dielectric constant 4.1 and dielectric strength 6.0107 V/m. What is the capacitance in pF(10 −12
F)?
The equation gives the capacitance of a parallel-plate capacitor:C = εA/d. Where C is the capacitance, ε is the permittivity of the dielectric material between the plates, A is the area of each plate, and d is the distance between the plates.
In this case, the area of each plate is 10 cm × 20 cm = 200 cm^2 = 0.02 m^2. The distance between the plates is 2.0 mm = 0.002 m. The permittivity of the dielectric material is ε = ε0εr, where ε0 is the vacuum permittivity (8.85 × 10^-12 F/m), and εr is the relative permittivity or dielectric constant (4.1).
So, substituting these values into the equation, we get:
C = εA/d
= (ε0εr)(0.02)/(0.002)
= (8.85 × 10^-12)(4.1)(0.02)/(0.002)
= 7.26 × 10^-11 F
= 72.6 pF
Therefore, the capacitance parallel-plate capacitor is 72.6 pF.
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A 10,000 kg railroad car is rolling at 8.00 m/s when a 6000 kg load of gravel is suddenly dropped in. What is the car's speed just after the gravel is loaded? Express your answer with the appropriate units.
The car's speed just after the gravel is loaded is 4.00 m/s.
The momentum of the system (railroad car + gravel) is conserved before and after the gravel is dropped.
Therefore, we can use the law of conservation of momentum to find the velocity of the combined system just after the gravel is loaded.
Before the gravel is dropped, the momentum of the railroad car is:
p1 = m1v1 = (10000 kg)(8.00 m/s) = 80000 kg*m/s
where m1 is the mass of the railroad car and v1 is its velocity.
When the gravel is dropped, the total mass of the system becomes:
m2 = m1 + m_gravel = 10000 kg + 6000 kg = 16000 kg
where m_gravel is the mass of the gravel.
The momentum of the system just after the gravel is dropped is:
p2 = m2v2
where v2 is the velocity of the combined system just after the gravel is loaded.
Since momentum is conserved, we can equate p1 to p2:
p1 = p2
m1v1 = m2v2
Solving for v2, we get:
v2 = (m1v1) / m2
Substituting the given values, we have:
v2 = (10000 kg)(8.00 m/s) / 16000 kg
v2 = 4.00 m/s
Therefore, the car's speed just after the gravel is loaded is 4.00 m/s.
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The motion of the kangaroo is under free-fall. We are looking for the initial velocity, and we know that the velocity in the highest position is zero.
From
v
2
=
(
v
0
)
2
+
2
ay
,
v 2
=(v 0
) 2
+2ay,
we have
v
2
=
(
v
0
)
2
+
2
ay
v
2
−
2
ay
=
(
v
0
)
2
v
0
=
v
2
−
2
ay
v 2
v 2
−2ay
v 0
=(v 0
) 2
+2ay
=(v 0
) 2
= v 2
−2ay
Substituting the known values,
v
0
=
v
2
−
2
ay
v
0
=
0
2
−
2
(
−
9.81
m/s
2
)
(
2.50
m
)
v
0
=
7.00
m/s
v 0
v 0
v 0
= v 2
−2ay
= 0 2
−2(−9.81m/s 2
)(2.50m)
=7.00 m/s
Therefore, the vertical speed of the kangaroo when it leaves the ground is 7.00 m/s.
Part B
Since the motion of the kangaroo has uniform acceleration, we can use the formula
y
=
v
�
t
+
1
2
a
t
2
y=v o
t+ 2
1
at 2
The initial and final position of the kangaroo will be the same, so �
y is equal to zero. The initial velocity is 7.00 m/s, and the acceleration is -9.81 m/s2.
y
=
v
0
t
+
1
2
a
t
2
0
=
(
7.00
m/s
)
t
+
1
2
(
−
9.81
m/s
2
)
t
2
0
=
7
t
−
4.905
t
2
7
t
−
4.905
t
2
=
0
t
(
7
−
4.905
t
)
=
0
t
=
0
or
7
−
4.905
t
=
0
y
0
0
7t−4.905t 2
t(7−4.905t)
t=0
=v 0
t+ 2
1
at 2
=(7.00 m/s)t+ 2
1
(−9.81 m/s 2
)t 2
=7t−4.905t 2
=0
=0
or7−4.905t=0
Discard the time 0 since this refers to the beginning of motion. Therefore, we have
7
−
4.905
t
=
0
4.905
t
=
7
t
=
7
4.905
t
=
1.43
s
7−4.905t
4.905t
t
t
=0
=7
= 4.905
7
=1.43 s
The kangaroo is about 1.43 seconds long in the air.
The motion of the kangaroo is under free-fall, its vertical speed when it leaves the ground 7.00m/s and it is in air for 1.43s.
The stir of the kangaroo is under free- fall. We're looking for the original haste, and we know that the haste in the loftiest position is zero.
From,
v ² = ( vo) ² 2ay,
we have,
v ² = ( vo) ² 2ay,
v ²- 2ay = vo ²
vo = √ v ²- 2ay
Vo = 7.00 m/ s
thus, the perpendicular speed of the kangaroo when it leaves the ground is 7.00 m/s.
Since the stir of the kangaroo has invariant acceleration, we can use the formula,
y = vo * t1/2 at ²
7t-4.905 t ²
t = 0 or t = 1.43
thus, kangaroo is about 1.43 seconds long in the air.
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Complete question:
A kangaroo can jump over an object 2.50 m high. (a) Calculate its vertical speed when it leaves the ground. (b) How long is it in the air?
The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy E_a = 30.0 kJ/mol. If the rate constant of this reaction is 5.0 times 10^4 M^-1 s^-1 at 201.0 degreeC, what will the rate constant be at 172.0 degreeC? Round your answer to 2 significant digits.
The rate constant be at 172.0°C is 1.11 x10⁵ m⁻¹s⁻¹
We employ the Arrhenius equation, which is: to determine the rate constant for the reaction at two different temperatures.
Ln(k 172.0°C/ k 201°C) = Ea/R (1/T₁ - 1/T₂)
where,
k 201°C = equilibrium constant at 201°C = 5x 10⁴m⁻¹s⁻¹
k 172.0°C = equilibrium constant at 172.0°C = ?
Ea = Activation energy = 30.0 kJ/mol = 30000 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
T₁ = initial temperature = 201°C = 474°K
T₂ = final temperature = 172.0°C = 445 °K
Putting the value
Ln(k 172.0°C/ k 201°C) = 30000 J/mol /8.314 J/mol K (1/ 474°K - 1/445 °K)
k 172.0°C = 1.11 x10⁵ m⁻¹s⁻¹
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