The transient heat conduction equation for a plane wall with constant thermal conductivity and no heat generation can be derived from the energy balance equation for a rectangular volume element.
Consider a plane wall with constant thermal conductivity, k, and no heat generation. Let's assume that the wall is at temperature T at time t and T + ΔT at time t + Δt. The energy balance equation for a rectangular volume element of the wall can be expressed as:
ΔQ/Δt = -kA(ΔT/Δx), where
ΔQ/Δt is the rate of heat transfer into the volume element k is the thermal conductivityA is the cross-sectional area of the wallΔT/Δx is the temperature gradient across the wall.Rearranging the equation:
ΔT/Δt = -(k/A)(ΔT/Δx)
The above equation represents the one-dimensional transient heat conduction equation for a plane wall with constant thermal conductivity and no heat generation.
This equation can be further simplified by using the thermal diffusivity, α, which is defined as:
α = k / (ρCp) where,
ρ is the density of the material Cp is the specific heat capacity.Substituting α into the equation:
ΔT/Δt = -α(ΔT/Δx^2)
This is the final form of the one-dimensional transient heat conduction equation for a plane wall with constant thermal conductivity and no heat generation.
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what is the radius of the first null in the diffraction pattern at the moon? recall the formula for diffraction from a circular aperture from chapter q3.
The radius of the first null in the diffraction pattern at the moon would be approximately 9.64 mm
The radius of the first null in the diffraction pattern at the moon can be calculated using the formula for diffraction from a circular aperture. The formula for the radius of the first null, or the first dark fringe, is given by:
r = λ × D / a,
where λ is the wavelength of light, D is the distance from the aperture to the screen, and a is the radius of the circular aperture. In the case of diffraction at the moon, the aperture is the diameter of the moon, and the screen is the observer's eye or a detector. The diameter of the moon is approximately 3,476 kilometers, and the distance from the moon to the Earth is approximately 384,400 kilometers. The wavelength of light used for observation can vary, but for visible light, it is typically in the range of 400 to 700 nanometers.
By substituting these values into the formula, we can calculate the radius of the first null in the diffraction pattern at the moon.
Let's assume a wavelength of 550 nm (visible light) and use the average value for the diameter of the moon (3,476 km) as the radius (1,738 km) in the calculation.
r = λ × D / a
= (550 × [tex]10^{-9}[/tex] m) × (384,400 × [tex]10^{3}[/tex] m) / (1,738 × [tex]10^{3}[/tex] m)
= 0.000964 m = 9.64 mm
So, the radius of the first null in the diffraction pattern at the moon would be approximately 9.64 mm when observed with a wavelength of 550 nm and the average diameter of the moon as the radius of the circular aperture.
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Find the distance traveled by a climber if she climbs 10 meters up and then repels 10 meters back down. Include both magnitude and direction.
Answer:
Explanation:
Let up be the + direction.
Distance is the total length traveled.
Distance = 10 + 10 = 20 m
Displacement = 10 - 10 = 0
A point is located in a polar coordinate system by the coordinates r = 4.6 m and = 10°. Find the x- and y-coordinates of this point, assuming that the two coordinate systems have the same origin
Point on Cartesian plane,
x-coordinate = 4.53
y - coordinate = 0.8
What are Polar and Cartesian coordinates?In the Cartesian system the coordinates are perpendicular to one another with the same unit length on both axes. A Polar coordinate system is determined by a fixed point, a origin or pole, and a zero direction or axis. Each point is determined by an angle and a distance relative to the zero axis and the origin.
Given,
Polar coordinates of point (r , θ)
r = 4.6
θ = 10°
Cartesian coordinates of point (x , y)
x = r cosθ
⇒ x = 4.6 × cos 10°
⇒ x = 4.6 × 0.9848
⇒ x = 4.53
y = r sinθ
⇒ y = 4.6 × sin 10°
⇒ y = 4.6 × 0.174
⇒ y = 0.8
Hence, Cartesian coordinates of the point (4.53 , 0.8)
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how much work did the movers do horizontally pushing a 46 kg crate 10.3 across a rough floor without acceleration, if the effective coefficient of friction was 0.5
The work done is 2321 J.
What is the work that is done?We have to note that the acceleration has to do with the change in the velocity of the object with time. We are told that there is a movement of the crate as we can see. Let us note that the work that is done by the object subtracted from the work done by the friction. Thus we have that;
Work = (46 * 9.8 * 10.3) - (0.5 * 46 *9.8 * 10.3)
Work = 4643 - 2321
= 2321 J
The work that has been done is 2321 J. This is done in moving the object
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What is the value of the electric potential at the center of the triangle if q1 = 4.1 μC , q2 = 3.4 μC , q3 = -3.5 μC ?
Answer:
triangle if q1 = 4.1 μC , q2 = 3.4 μC , q3 = -3.5 μC ?
A cello string vibrates with a frequency of 64 Hz. If the wavelength of t waves in the string is 180 cm, calculate the wave speed.
Answer:
V=115.2m/s
Explanation:
V=fλ
where v=wave speed,f=frequency,λ=wavelength
convert 180cm to m
1.8m
V=64×1.8
V=115.2m/s
A thin circular disk of radius R = 30 cm is oriented in the yz-plane with its center as the origin. The disk carries a total charge Q = +3 mu C distributed uniformly over its surface. Calculate the magnitude of the electric field due to the disk at the point x = 15 cm along the x-axis. 4.98 times 10^5 N/C 2.49 times 10^5 N/C 3.32 times 10^5 N/C 9.95 times 10^5 N/C 1.99 times 10^5 N/C
The magnitude of the electric field due to the disk at the point x = 15 cm along the x-axis is approximately 2.99 x 10⁶ N/C. Rounded to two significant figures, the answer is 3.0 x 10⁶ N/C.
What is the magnitude?The magnitude of a physical quantity is a measure of the size of that quantity, and is usually expressed using an appropriate unit of measurement. Magnitude is a way of expressing the size of something without specifying its exact size. For example, when referring to the magnitude of an earthquake, one might say "it was a magnitude 7 earthquake".
To calculate the electric field at a point on the x-axis due to a uniformly charged disk, we can use the formula:
[tex]\mathrm{E = (1/4\pi \p \epsilon_0) \times Q / r^2 \times sin(\theta)}[/tex]
where ε₀ is the permittivity of free space,
Q is the total charge on the disk, r is the distance from the center of the disk to the point where we want to calculate the electric field,
and θ is the angle between the normal to the disk and the line connecting the center of the disk to the point where we want to calculate the electric field.
In this problem, the disk is oriented in the yz-plane, so the normal to the disk is in the x-direction. The distance from the center of the disk to the point where we want to calculate the electric field is 15 cm.
The angle θ is the angle between the normal to the disk and the line connecting the center of the disk to the point where we want to calculate the electric field.
Since the disk is centered at the origin, the line connecting the center of the disk to the point on the x-axis is also in the x-direction, so θ = 0.
We can therefore simplify the formula to:
[tex]\mathrm{E = (1/4\pi \p \epsilon_0) \times Q / r^2 }[/tex]
Substituting the given values, we get:
[tex]\mathrm{E = (1/4\pi \p \epsilon_0) \times (3\mu C) / (0.15)^2 }[/tex]
We can calculate the value of 1/4πε₀ as 9 so:
[tex]\mathrm{E = (9.0 \times 10^9 \ N \cdot m^2/C^2) \times (3\mu C) / (0.15)^2 }[/tex]
E = 2.99 x 10⁶ N/C
Therefore, the magnitude of the electric field due to the disk at the point x = 15 cm along the x-axis is approximately 2.99 x 10⁶ N/C. Rounded to two significant figures, the answer is 3.0 x 10⁶ N/C.
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A 1825 kg plane in a shallow dive accelerates at 7.6 m/s^2. What is the drag force on the plane if the forward force of gravity and thrust add up to 15,000 Newtons?
Answer:
5996667544211123345678
Help please how do you do this
The impulse is given as -75 n.s
How to solve for impulseWe would first have to solve for the initial momentum and then the final momentum
The initial momentum is given as 6 x 5 = 30
The final momentum = -9 x 5 = -45
Then the impulse would be the change in momentum which is gotten as
I = Pf - Pi
I = -45 - 30
I = -75
The impulse is given as I = -75
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When the bottom string is pulled slowly in the heavy-ball-and-strings demonstration (Figure 4.8 in your book), tensiona. is about the same in both strings.b. in the bottom string is zero.c. in the top string is due to your pull plus the weight of the ball.
In the heavy ball and string demonstration, if you pull the bottom string slowly, the tension c) The top string is due to your pull and the weight of the ball.
About stress/tensionWhen a force is applied to an elastic object, the object will stretch to a certain size. The amount of stress is the ratio of the tensile force acting on the cross-sectional area of the body. If the area of a bar is A, then both ends of the bar are subjected to a tensile force of equal magnitude and opposite direction F. The rod should be under tension.
Looking at a cross-section perpendicular to the length of the rod, the tensile force F is evenly distributed over the cross-sectional area A. Stress is therefore defined as the ratio of the magnitude of the force F to the cross-sectional area. . Cross-sectional area of A. The stress is expressed as σ (sigma) and its unit is Nm-².
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A 193.4 g sample of water at 13.5°C is heated to steam at 131.1°C. How much heat was absorbed?
A 193.4 g sample of water at 13.5°C is heated to steam at 131.1°C. The amount of heat absorbed will be 96000 KJ.
What is Heat energy?Thermal energy refers to the energy which is contained within a system that is responsible for the temperature. Heat is the flow of thermal energy in the system.
Given that,
Mass, m = 193.4 g
We know that,
Q₁ = mcΔt
Q₁ = 193.4 × 4200 × (131.1 - 13.5)
Q₁ = 193.4 × 4200 × 117.6
Q₁ = 95,524,128 Joules
Now, the latent heat
Q₂ = mL
Q₂ = 193.4 × 2260
Q₂ = 437,084 Joules
Now, the total heat energy required,
Q = Q₁ + Q₂
Q = 95,524,128 + 437,084
Q = 95,961,212 Joules
Therefore, the heat energy is approximately 96000 KJ.
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Your question is incomplete, most probably the complete question is:
Calculate the heat energy required to convert completely a 193.4 g sample of water at 13.5°C is heated to steam at 131.1°C, given that the specific heat capacity of water is 4200 J/(kg°C) and the specific latent heat of vaporization of water is 2260 kJ/kg.
12. Consider the falling and rolling motion of the 2.0 kg ball in the following two resistance-free
situations. In one situation, the ball falls off the top of the platform to the floor. In the other situation,
the ball rolls from the top of the platform along the staircase-like pathway to the floor. Fill in answer
for each the blanks lettered A-K.
PE=100J
KE=0J
PE= F J
KE- GJ
v=__H_m/s
PE=50J
KE-A J
V= B_m/s
PE=_C_J
KE=D J
v=__E__m/s
Answer the questions using logic and a little bit of arithmetic.
Top & Bottom:
1. What is the PE of the ball at the very top of the diagram above?
2.
What is the KE of the ball at the very top of the diagram above?
3. What is the PE of the ball the location shown as I on the diagram above?
4. What is the KE of the ball the location shown as J on the diagram above?
5.
What is the velocity of the ball the location shown as K on the diagram above?
requires a little mental math - but you can use a calculator if you need it.
6. What is the PE of the ball the location shown as F on the diagram above as the ball hits the ground?
10
PE=IJ
KE J J
v=K_m/s
8. What is the velocity of the ball the location shown as H on the diagram above?
your previous answers-don't solve the velocity equation.
7. What is the KE of the ball the location shown as G on the diagram above as the ball hits the ground?
Rolling down the ramp:
9. What is the KE of the ball the location shown as A on the diagram above?
This one
Use logic &
10. What is the velocity of the ball the location shown as B on the diagram above? Use logic & your previous
answers-don't solve the velocity equation.
11. What is the PE of the ball the location shown as C on the diagram above?
12. What is the KE of the ball the location shown as D on the diagram above?
13. What is the velocity of the ball the location shown as E on the diagram above? Use logic & your previo
answers-don't solve the velocity equation.
Gravity is the only force that is active. The overall mechanical energy is preserved since it is an inner or conservative force. As a result, each location has the 100 J of initial mechanical energy. So, 50 J is the KE for A.
The KE is 50 J because the PE at the exact same stairstep is Fifty J (C) (D).
The PE is 0 J at zero height (F and I). As a result, there is 100 J of velocity just at bottom of the hill (G and J).
The velocity may be calculated as 7.07 m/s for B and E as well as 10 m/s for H and K using the formula KE = 0.5*m*v2.
The answers provided here for the speed numbers are based on the assumption that 100% of the ball's kinetic energy is present as translational kinetic energy. The kinetic energy really would take some of the form of rotating kinetic energy. The real speed figures would thus be a little bit lower than those reported. (Torque and angular velocity is not covered in this tutorial for the physics classroom.)
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Internal hard disk is Select one: a. Removable but not fixed b. Not fixed c. Removable d. Fixed
Internal hard disk is Removable but not fixed. Thus, option A is the correct answer.
What is Hard disk?A hard disk drive (HDD) is a type of non-volatile storage device that stores digital data on rapidly rotating disks (platters) coated with magnetic material. Hard disk drives are commonly used in computers as the primary storage for the operating system and for user data, as well as in other devices such as external hard drives, network-attached storage (NAS) devices, and digital video recorders (DVRs). An HDD consists of one or more disks that spin at high speeds (usually 5,400 or 7,200 RPM) and are read from and written to by magnetic read/write heads that move across the disk surfaces. The data is stored on the disk in magnetic patterns, and the read/write heads can detect and change these patterns to read and write data.
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please help with this problem
Answer:
option d
it is possible in the case of roller only
Explanation:
Two spacecraft
and
are traveling directly towards each other, intending to meet to dock together. Mission control on Earth initially sees spacecraft
moving with a speed of 154.5 m/s and spacecraft
moving with a speed of 133.0 m/s. Each spacecraft will need to come to a complete stop when they meet, and each will slow down using its rockets. The rockets on spacecraft
cause it to slow down at a constant rate of 17.1 m/s2, and the rockets on spacecraft
cause it to slow down at a constant rate of 22.2 m/s2. Spacecraft
turns on its rockets first.
At what distance
from spacecraft
should spacecraft
turn on its rockets in order for the two spacecraft to meet, have zero velocity relative to the Earth, and arrive at the same time?
A spaceship is a machine or vehicle built to travel through space.
What is Spacecraft?Spacecraft are a form of artificial satellite used for a wide range of tasks, such as communications, Earth observation, meteorology, navigation, space colonization, planetary exploration, and cargo and person transfer.
A spacecraft that is performing a sub-orbital spaceflight launches into space and then lands without gaining enough momentum or energy to complete a full orbit of the Earth. Spacecraft enter confined orbits around the Earth or other celestial bodies for orbital spaceflights.
Only humans are carried on human spacecraft either as crew or guests from launch or while in orbit (space stations).
Therefore, A spaceship is a machine or vehicle built to travel through space.
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Vector A⃗ =2i^+3j^ and vector B⃗ =2i^−2j^+4k^ .
The summation of the two vector quantities is [tex]4 \hat{i} + \hat{j} + 4 \hat{k}[/tex].
What is vector quantity?In physics, a vector is a quantity with both a magnitude and a direction. It is often represented by an arrow whose length is proportional to the magnitude of the quantity and whose direction is the same as that of the quantity. A vector does not have position, while having magnitude and direction.
Ordinary quantities with a magnitude but no direction are referred to as scalars in contrast to vectors. For instance, while speed, time, and mass are scalars, displacement, velocity, and acceleration are vector quantities.
Given [tex]\vec{A} =2\hat{i} +3\hat{j}[/tex] and [tex]\vec{B} = 2 \hat{i} -2\hat{j} + 4 \hat{k}[/tex]
Hence,
[tex]\vec{A}+\vec{B} = 4 \hat{i} + \hat{j} + 4 \hat{k}[/tex]
Hence, sum of the two given vector quantity is [tex]4 \hat{i} + \hat{j} + 4 \hat{k}[/tex].
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Two balls are rolling directly toward each other. One, of mass 0.25 kg, has a speed of 1.7 m/s; the other has a mass of 0.18 kg and a speed of 2.5 m/s. After the collision, the 0.25 kg ball has reversed its direction and has a speed of 0.10 m/s. What is the velocity (magnitude and direction) of the other ball?
m/s direction:
(same/opposite) direction of initial motion for ball
The velocity of the balls that rolling directly toward each other. One, of mass 0.25 kg, has a speed of 1.7 m/s; the other has a mass of 0.18 kg and a speed of 2.5 m/s. After the collision, the 0.25 kg ball has reversed its direction and has a speed of 0.10 m/s = 0 (zero)
The linear momentum of a system of interacting objects equals the linear momentum of the system after the interaction if the net force acting on the system of interacting objects is zero.
The equation:
p = mv
Where,
p = momentum
m = mass of the object
v = the velocity of the object.
p₁ = p₂
(mv + mv)₁ = (mv + mv)₂
(0.25 x 1.7) = (0.25 x (-0.10)
v₂ = 0
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a position-type motion control system is in an instability condition when overshoot and oscillations occur. (true or false)
True, a position-type motion control system is in an instability condition when overshoot and oscillations occur.
What kinds of motion control systems are there?
Motion control systems can be classified as either open loop or closed loop. The fundamental distinction between the two is that closed systems use feedback whereas open loop systems do not. The distinction between open and closed loop systems is discussed in more detail here.
Any transient value of a parametric measurement that exceeds its steady-state or final value while transitioning from one value to another is referred to as overshoot. When a parameter transitions from one value to another, overshoot refers to the temporary values that are greater than their final (steady state) value in electronics.
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aball of mass 2kg has a diameter of 50cm falls in the pool. calculate its buoyant force and volume of water displaced?
The buoyant force and volume of water displaced are 19.6 N and 0.065 [tex]m^{3}[/tex] respectively.
Explanation:
According to Archimedes' principal, When a body is immersed fully or partially in a fluid, the the body experiences an upward force called buoyant force which is equal to the weight of fluid dispersed.
Buoyant force ([tex]F_{b}[/tex]) = mg
where, m is the mass of ball
g is acceleration due to gravity i.e. 9.8 [tex]ms^{-2}[/tex]
From the question putting the values we get,
[tex]F_{b} =[/tex] 2×9.8 = 19.6 N
Now,
Water displaced = Volume occupied by the ball
[tex]= \frac{4}{3} \pi r^{3}[/tex] ( as ball is sphere)
{diameter of the ball = 50cm or, 0.5m So radius =[tex]\frac{0.5}{2}[/tex]}
= [tex]\frac{4}{3}*\frac{22}{7}*(\frac{0.5}{2}) ^{3}[/tex]
[tex]=\frac{4}{3}[/tex] × [tex]\frac{22}{7}[/tex]×[tex]\frac{0.125}{8}[/tex]
[tex]= \frac{4}{3}[/tex] × [tex]\frac{22}{7}[/tex] × [tex]\frac{125}{8000}[/tex]
[tex]= 0.065 m^{3}[/tex]
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In which one of the following situations does the car have an acceleration that is directed due north? a) A car travels northward with a constant speed of 12 ms. b) A car is traveling southward as its speed increases from 12 ms to 23 m/s c) A car is traveling southward as its speed decreases from 18 ms to 12 m/s d) A car is traveling northward as its speed decreases from 24 ms to 12 m/s e) A car travels southward with a constant speed of 24 mis.
A car is traveling southward as its speed decreases from 18 m/s to 12 m/s have an acceleration that is directed due north.
What is acceleration?
If an object's velocity changes, it is said to have been accelerated. An object's velocity can alter depending on whether it moves faster or slower or in a different direction. A falling apple, the moon orbiting the earth, and a car stopped at a stop sign are a few instances of acceleration. Through these illustrations, we can see that acceleration happens whenever a moving object changes its direction or speed, or both.
Since the speed is decreasing, the acceleration will be in the opposite direction hence option c.
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the electric field everywhere on the surface of a thin, spherical shell of radius 0.730 m is of magnitude 875 n/c and points radially toward the center of the sphere.(a) What is the net charge within the sphere's surface? (b) What can you conclude about the nature and distribution of the charge inside the spherical shell?
The net charge within the sphere's surface is 2.36 x 10^-7 C and it can be deduced that the charge within the spherical shell is positively charged and uniformly distributed on its surface.
(a) To ascertain the net charge enclosed within the boundaries of the spherical shell of radius 0.730 meters, we must delve into the intricacies of the electric field and its correlation with the charge distribution on a conductor.
The electric field on the surface of a conductor can be mathematically represented as E = kσ, where k stands for the Coulomb constant and σ symbolizes the surface charge density.
The surface area of a spherical shell can be calculated as 4π[tex]r^2[/tex], where r signifies the radius of the sphere. By fusing these two equations, we can determine the surface charge density as follows:
σ = E / (k) = (875 n/C) / (9 x 10^9 Nm^2/C^2) = 9.72 x 10^-10 C/m^2
The charge contained within the sphere can then be computed as:
Q = σ * 4πr^2 = (9.72 x 10^-10 C/m^2) * 4π * (0.730 m)^2 = 2.36 x 10^-7 C
(b) The electric field, exhibiting a radially inward direction, implies that the charge within the spherical shell is positively charged.
The proportionality between the electric field's magnitude and the net charge within the spherical shell suggests that the charge is distributed uniformly on the surface of the sphere.
In conclusion, it can be deduced that the charge within the spherical shell is positively charged and uniformly distributed on its surface.
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If the magnitude of the resultant force acting on the bracket is to be 430 N directed along the positive u axis, determine the magnitude of F. (Figure 1) Express your answer with the appropriate units. - LHẢ * * 0 Ea ? F1 - 480.5 N Submit Previous Answers Request Answer X Incorrect; Try Again; One attempt remaining Review your calculations, you may have made a rounding error or used the wrong number of significant figures. Part B Determine the angle o Express your answer with the appropriate units. < 1 of 1 > CHAO ? $ = 8.619 30° F = 200 N Submit Previous Answers Request Answer X Incorrect; Try Again; 5 attempts remaining Review your calculations, you may have made a rounding error or used the wrong number of significant figures. F; 260 N
If the magnitude of the resultant force acting on the bracket is to be 430 N directed along the positive u-axis, the magnitude of F is as follows:
The vector components should be drawn which is acting on the bracket as follows.
The x and y components of each force are represent by the dashed arrows. Force, (in blue) is along the x-axis, thus it has no y-component.
Now, the resultant force is drawn (as stated in the question) and its components also.
After this, can now write the x and y component values of each force.
(F1)x = F1 sinϕ
(F1)y = F1 cosϕ
(F2)x =200N
(F2)y = 0
(F3)x = 260(5/13)=100N
(F3) y =260( 12/13 )=240N
(FR)x =450 cos 30 =389.71N
(FR) y =450 sin 30 =225N
Next is to sum the forces along the x and y axes. To do this, we will establish the positive sides. Pick the forces acting up, and forces acting to the right to be positive.
+→∑(FR) x =∑(Fx)
389.71= F1 sinϕ+200+100
F1 sinϕ=89.71———————————(1)
+↑∑(FR)y=∑(Fy)
225 = F1 cosϕ−240
F1 cosϕ=465———————————(2)
The last step is to solve equations (1) and (2) simultaneously. To do so, remember the identity, sinθ/cosθ = tanθ
Solving gives us:
ϕ=10.9 and F1 = 474N
Therefore, the ϕ=10.9 and F1 = 474N
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Two pans of a balance are 33.6 cm apart. The fulcrum of the balance has been shifted 1.35 cm away from the center by a dishonest shopkeeper. By what percentage is the true weight of the goods being marked up by the shopkeeper? Assume the balance has negligible mass. Answer in units of %.
The true weight of the goods being marked up by 8.3% by the shopkeeper.
What is bending moment?A bending moment is the reaction that occurs in a structural element as a result of an external force or moment that causes the element to bend, as defined by solid mechanics. The beam is the most prevalent or most basic structural component that experiences bending moments.
Calculated the bending moment about the balance:
P₁(33.6 + 1.35) = P₂ ( 33.6 - 1.35)
P₂ = 1.083 P₁
Hence, percentage change = (P₂-P₁)/P₁ × 100% = 8.3 %
So, the true weight of the goods being marked up by 8.3% by the shopkeeper.
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A rod of length 2a with total charge Q uniformly distributed along its length is placed along the zaxis with its center at the origin. Find the electric field everywhere?
The electric field due to a charged rod of length 2a with total charge q uniformly distributed along its length, placed along the z axis with its center at the origin.
Through the application of Coulomb's equation and the concept of superposition, we can determine the electric field everywhere caused by the charged rod. The electric field due to a tiny piece of the rod at a point P in space is given by:
dE = k dq / r²
where
k = Coulomb's constant,
dq i= the segment's charge,
r = the segment's distance from point P, and
dE = the segment's contribution to the electric field.
We integrate this formula along the entire length of the rod, from -a to +a, to determine the total electric field at P as follows:
E = ∫ dE = k ∫ dq / r²
where the rod's length is used as the integral. The charge per unit length of the rod, or dq, can be expressed in terms of the charge density as dq = ds, where ds is an element along the rod with an infinitesimal length. r = √(s² + z²) where z is the segment's distance from the z axis, we can also describe r in terms of the distance s from the origin to the segment.
E = k ∫ dq / r^2 = k ∫ λ ds / (s² + z²)
E = k λ z ∫_{-arctan(a/z)}^{arctan(a/z)} dθ / (1 + (a/z)² tan² θ)
E = k λ z / (2πε_0) ∫_{-a/z}^{a/z} du / (1 + u²)
E = k λ z / (2πε_0) [arctan(a/z) - arctan(-a/z)]
E = k λ a / (2πε_0 z√(a² + z²))
This is the equation for the electric field produced by a charged rod of length 2a, arranged along the z axis with its center at the origin, and with total charge q evenly distributed throughout its length. The electric field radiates outward from the rod, and as you move away from the rod, its strength diminishes as 1/r².
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in the proton-proton chain select all that apply a.no neutrinos are produced.
b.energy is released because a helium nucleus has a greater mass than a hydrogen nucleus.
c.no photons are produced.
d.carbon serves as a catalyst for the nuclear reaction.
e.energy is produced in the form of gamma rays and the velocity of the created nuclei.
The proton-proton chain is a series of nuclear reactions that take place in the cores of stars and are responsible for producing energy through fusion.
Option b and e will apply for the proton-proton chain.
In the chain, hydrogen nuclei (protons) collide and combine to form helium nuclei, releasing energy in the process.
b. Energy is released because the mass of the helium nucleus formed from the combination of two hydrogen nuclei is less than the total mass of the two individual hydrogen nuclei.
This difference in mass is converted into energy according to Einstein's equation, [tex]E=mc^2[/tex], where E is energy, m is mass and c is the speed of light.
e. Energy is produced in the form of gamma rays and the velocity of the created nuclei. The high-energy gamma rays and fast-moving helium nuclei carry away the energy released in the reaction.
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An electronic package with a surface area of 1 m2 placed in an orbiting space station is exposed to space. The electronics in this package dissipate all 1950 W of its power to space through its exposed surface. The exposed surface has an emissivity of 1.0 and an absorptivity of 0.25. Given: 0 = 5.67x10-8 W/m2K4Determine the steady state exposed surface temperature of the electronic package if the surface is not exposed to the sun. If the surface is not exposed to the sun, the surface
The electronic package's steady state temperature can be calculated using the Stefan-Boltzmann Law, which states that the total energy radiated per unit surface area per unit time is proportional to the fourth power of the temperature.
The electronic package's energy balance equation is as follows:
Power absorbed minus power radiated equals power generated.
where P = absorptivity * incoming radiation P = emissivity * Stefan-Boltzmann constant * T4 Power generated = 1950 W (constant)
Because the surface is not in direct sunlight, we can assume that the incoming radiation is 0 and the power absorbed is 0.
Thus, the energy balance equation becomes:
0 - Power radiated = 1950 W
or
Power radiated = 1950 W
We can use the Stefan-Boltzmann Law to calculate the surface temperature:
Power radiated = emissivity * Stefan-Boltzmann constant * T^4
1950 W = 1 * 5.67 x 10^-8 W/m2K^4 * T^4
T^4 = (1950 / (5.67 x 10^-8))
T = ( (1950 / (5.67 x 10^-8))^(1/4))
The electronic package's steady state temperature is the temperature at which the amount of power generated by the electronics equals the amount of power radiated to space.
Full question.
An electronic package with a surface area of 1 m2 placed in an orbiting space station is exposed to space. The electronics in this package dissipate all 1 kW of its power to the space through its exposed surface. The exposed surface has an emissivity of 1.0 and an absorptivity of 0.25.Given: 0 = 5.67x10-8 W/m2K4 Determine the steady state exposed surface temperature of the electronic package (a) if the surface is exposed to a solar flux of 750 W/m2, and (b) if the surface is not exposed to the sun.
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suppose that a ball is dropped from the upper observation deck of a building, 300 m above the ground. (a) what is the velocity of the ball after 2 seconds? (b) how fast is the ball traveling when it hits the ground? solution we will need to find the velocity both when t
The velocity of the ball is 19.6 m/s and when the ball lands, it is moving at a speed of 77.134 m/s.
What is gravity?All objects with mass are subject to gravity, which is a fundamental force of nature. It is the force that pulls everything toward the core of a planet or other entity.
How do you determine it?(a) The equation can be used to determine the velocity of a dropped ball.
v = v0 + a * t
where a is the acceleration caused by gravity (9.8 m/s2), v0 is the beginning velocity (0 in this case because the ball is dropped from rest), and t is the amount of time that has passed.
Therefore, if t = 2 seconds, the ball's velocity at that time would be:
v = 0 + 9.8 * 2 = 19.6 m/s
(b) The ball's final velocity is identical to its velocity when it lands. The following equation can be used to determine how long it takes the ball to touch the ground:
t = √(2h / a)
where a is the acceleration brought on by gravity (9.8 m/s2) and h is the height (in this case, 300 m) from which the ball is dropped.
Thus, the duration of time required for the ball to touch the ground is:
t = √(2 * 300 / 9.8) = √(600 / 9.8) = √(61.22) = 7.83 s
The formula: can be used to determine the ball's velocity when it lands.
v=v0 + a*t = 0 + 9.8 * 7.83 = 77.134 m/s
So, when the ball lands, it is moving at a speed of 77.134 m/s.
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question which of the following affect the speed and direction of your vehicle? a. responses momentum b. momentum kinetic energy c. kinetic energy safety belts d. safety belts gravity
e. gravity potential energy f. potential energy inertia g. inertia
The correct answers format are:
a. Responses momentum
b. Momentum kinetic energy
f. Potential energy inertia
g. Inertia
How the factors that affect vehicle speed and direction?A. Momentum - A vehicle's momentum is the product of mass and velocity. The greater the momentum, the more difficult it is to change the speed or direction of the vehicle.
B. Kinetic Energy – A vehicle's kinetic energy is the energy it has as a result of its motion. The greater the kinetic energy, the faster the vehicle and the harder it is to turn.
D. Seat Belts - Seat belts are used to restrain occupants during sudden vehicle stops and collisions. Seating occupants reduces the risk of injury and can affect vehicle speed in the event of an accident.
G. Inertia - Inertia is the impulse of an object to continue changes in its state of motion. The more inertia a vehicle has, the more difficult it is to change speed or direction.
So the correct options to affect the speed and direction of the vehicle are: reaction dynamics, b. momentum kinetic energy, i. seat belt gravity and g. inertia.
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A student sees a mirror image of a tree in a pond on a very still day. The student is able to see a mirror image of the tree in the pond because light
striking the surface of the water -
bends downward.
bounces off.
becomes brighter.
is absorbed.
Answer:
B.
Explanation:
What causes a reflection in still water?
Reflection in water is caused when light hits the surface of still water and travels to our eye so we can see the complete image and light it's reflecting.
All materials reflect light in some way.
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A hydraulic jack uses hydrostatic pressure forces to leverage a small applied force to generate a large resulting force (alternatively you can think of it as a type of manometer). A force of magnitude, Fapp = 25 lb is applied to the left side of the jack (which has a square cross-section of side d=2.5 in). The right side of the jack also has a square cross-section but each of side D=10 in. The specific gravity of the oil in the jack is soil = 0.8. The weights of the bearing blocks can be neglected. Note that the surfaces of the bearing blocks on both sides of the jack are horizontala. What is the magnitude of the weight, W, that can be supported by the right side of the jack for the given difference in elevation, 2.4 ft as shown?b. The vertical right side of the right side of the jack is a total height of Htot =2.6 ft (and length into the page of D=10 in ). What is the magnitude of the force due to the hydrostatic pressure on that vertical side? Note that this vertical surface is rectangular in geometry and is of height Htot and length, D, into the page.
92.16 lb is the magnitude of the weight that can be supported by the right side of the jack.
Hydraulic jacks are devices that leverage applied force to generate a larger force. In this case, a 25 lb force is applied to the left side of the jack with a square cross-section of side d = 2.5 in.
The right side of the jack is also square with a cross-section of side D = 10 in and has a difference in elevation of 2.4 ft.
To calculate the magnitude of weight, W, that can be supported by the right side of the jack, we need to calculate the pressure due to the hydrostatic force on the vertical side of the jack.
This vertical side is a rectangular shape of height Htot = 2.6 ft and length D = 10 in. Taking into account the specific gravity of the oil in the jack as soil = 0.8, the hydrostatic pressure on the vertical side is equal to the difference in elevation multiplied by the weight of the oil.
W = 2.4 ft x 0.8 x 62.4 lb/ft³ = 92.16 lb. This is the magnitude of the weight that can be supported by the right side of the jack.
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