Answer: poop
Explanation:
How do you determine the direction of the magnetic field in a magnet?
The direction of the magnetic field in a magnet is determined by using use a compass. Place the compass near the magnet and the needle will point in the direction of the magnetic field.
There are two ways to determine the direction of the magnetic field in a magnet. The magnetic field of a magnet can be determined by two methods:
The compass method: The north end of a compass always points in the direction of the magnetic field line, and the south end points in the opposite direction. Therefore, the magnetic field direction of a magnet may be determined by positioning a compass near it.
The right-hand rule method: Consider a current-carrying wire. If the right-hand thumb points in the direction of the current, the magnetic field lines follow the direction of the curled fingers. This is true only for a straight wire, and if the current is changing or there is a gap in the wire, the magnetic field lines are different.
Therefore, if you have a magnet and you know the direction of the current or movement, you may use the right-hand rule to determine the magnetic field direction.
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snow is falling vertically at a constant speed of 3.0 m/s. at what angle from the vertical do the snowflakes appear to be falling as viewed by the driver of a car traveling on a straight, level road with a speed of 50 km/h?
The snowflakes appear to be falling at an angle of 5.71° from the vertical as viewed by the driver of a car traveling on a straight, level road with a speed of 50 km/h.
What is the apparent speed of snow as viewed by the driver of a car?From the given data:
Speed of snow, u = 3.0 m/s
Speed of car, v = 50 km/h
= 50 × 5/18 m/s
= 13.89 m/s
As the car is moving with a velocity perpendicular to the snow velocity, the relative velocity of snow with respect to the car is also 3.0 m/s.
Now we can use the concept of relative motion to calculate the angle at which the snow appears to be falling.
The formula for the angle is:
θ = sin-1(u/v)θ
= sin-1(3.0/13.89)
θ = 5.71°
Therefore, the snowflakes appear to be falling at an angle of 5.71° from the vertical as viewed by the driver of a car traveling on a straight, level road with a speed of 50 km/h.
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determine the greatest load p that can be applied to the truss in (figure 1) so that none of the members are subjected to a force exceeding either 2.2 kn in tension or 1.9 kn in compression.
Hence, we can write:1.9 kN > RCD⇒ 1.9 kN > 4.1 kN - P⇒ P < 4.1 kN - 1.9 kN⇒ P < 2.2 kN
Therefore, the maximum load P that can be applied to the truss is P < 2.2 kN.So, the greatest load that can be applied to the truss in Figure 1 is 2.2 kN.
Given Data:Maximum tension load that each bar can bear = 2.2 kN Maximum compressive load that each bar can bear = 1.9 kNTo find:The greatest load P that can be applied to the truss in Figure 1
Approach:First of all, we need to find the bar which is subjected to maximum tension or compression force.Then, we can calculate the maximum load that the bar can bear.Using the method of joints, we can find the load in each bar of the truss.Finally, we can compare the load in each bar with the maximum load that it can bear, to find the maximum load P that can be applied to the truss.
Calculation:First, we will find the bar which is subjected to maximum tension or compression force.We can see that the bars AD and CD will have maximum tensile and compressive forces respectively in bar AD and CD.Using the method of joints,
we can calculate the force in each bar:∑Fy = 0⇒ RCD + RDA = P⇒ RDA = P - RCD∑FCD = 0⇒ RCD - 2.2 - 1.9 = 0⇒ RCD = 4.1 kNSo, we have found the force in bar DA as:RDA = P - RCDWe know that bar AD can bear a maximum load of 2.2 kN in tension. Hence, we can write:2.2 kN > RDA⇒ 2.2 kN > P - RCD⇒ P < RCD + 2.2 kN⇒ P < 4.1 + 2.2 kN = 6.3 kNWe also know that bar CD can bear a maximum load of 1.9 kN in compression.
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A 10.0 g piece of metal at 100 C is transferred to a calorimeter containing 50.0 mL of water initially at 23.0 C. Calculate the specific heat capacity of the metal if the heat capacity of the calorimeter, C cal, is 25.0 J/K. The final temperature, T final is 25.6 C.
The specific heat capacity of the metal is 0.73 J/g°C.
The specific heat capacity of the metal can be calculated from the equation
q = (m × c × ΔT)metal + (Ccal × ΔT)calorimeter,
where q is the heat absorbed by the calorimeter, m is the mass of the metal, c is the specific heat capacity of the metal, ΔT is the change in temperature, and Ccal is the heat capacity of the calorimeter.
The final temperature, Tfinal, is 25.6°C.
The heat absorbed by the calorimeter, q, can be calculated from the equation
q = mcΔT,
where m is the mass of the water and c is the specific heat capacity of water.
Since the calorimeter contains 50.0 mL of water, which has a density of 1.00 g/mL, the mass of the water is 50.0 g.
Therefore, the heat absorbed by the calorimeter is
q = (50.0 g) × (4.18 J/g°C) × (25.6°C − 23.0°C) = 544 J.
The heat absorbed by the metal can be calculated from the equation
qmetal = −qcalorimeter = −544 J.
Since the metal is transferred to the calorimeter at 100°C, the initial temperature of the metal, Ti, is 100°C.
Therefore, ΔTmetal = Tfinal − Ti = 25.6°C − 100°C = −74.4°C.
Since the metal has a mass of 10.0 g, the specific heat capacity of the metal can be calculated from the equation cmetal = qmetal ÷ (m × ΔTmetal) = −544 J ÷ (10.0 g × −74.4°C) = 0.73 J/g°C.
Therefore, the specific heat capacity of the metal is 0.73 J/g°C.
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The temperature of a gas stream is to be measured by a thermocouple whose junction can beapproximated as a 1.2-mm-diameter sphere. The properties of the junction are k =35 W/m °C, p=8500kg/m3, and Cp = 320 J/kg °C, and the heat transfer coefficient between the junction and the gas is h=65W/m2 °C. Determine how long it will take for the thermocouple to read 99 percent of the initialtemperature difference. (∅/∅i= 0.01)
it will take 30.65 minutes for the thermocouple to read 99 percent of the initial temperature difference. (∅/∅i = 0.01).
The temperature of a gas stream is to be measured by a thermocouple whose junction can be approximated as a 1.2-mm-diameter sphere. So, the radius, r = 0.6 mm = 0.0006 m, the volume of the sphere, V = (4/3)πr³, and the area of the sphere, A = 4πr².
The properties of the junction are k = 35 W/m °C, p = 8500 kg/m³, and Cp = 320 J/kg °C, and the heat transfer coefficient between the junction and the gas is h = 65 W/m² °C.
We have, thermal conductivity of the sphere = k = 35 W/m °C, density of the sphere = p = 8500 kg/m³, specific heat of the sphere = Cp = 320 J/kg °C, and heat transfer coefficient between the sphere and the gas, h = 65 W/m² °C.
The initial temperature difference is given by, ΔT₀ = 1°C = 1 K. Let, the time taken for the thermocouple to read 99% of the initial temperature difference, ΔT99 = 0.99 K.
Let, the thermal diffusivity of the sphere be,
α = k / (pCp) = (35 W/m °C) / (8500 kg/m³ x 320 J/kg °C) = 0.000012868 m²/s.
And, the Biot number is given by, Bi = (h x A) / k = [(65 W/m² °C) x 4π(0.0006 m)²] / (35 W/m °C) = 0.0492.
The equation for the unsteady-state temperature profile of a sphere is, θ(r,t) = Σ [(-1)n+1 / n] exp(-n²π²αt / r²) sin (nπr / R), where R is the radius of the sphere. We can estimate the time taken for the thermocouple to read 99% of the initial temperature difference using a semi-log plot of θ/ΔT vs. t/ti.
This plot is linear and of the form, θ/ΔT = 1 - A exp (-Bt/ti), where A = 0.01 and B = (nπ/R)².So, θ/ΔT = 0.99 = 1 - A exp (-Bt/ti), or 0.01 exp (-Bt/ti) = 0.01/0.99, or exp (-Bt/ti) = 1/99, or -Bt/ti = ln (1/99), or t/ti = ln (99).
Therefore, the time taken for the thermocouple to read 99% of the initial temperature difference is, ti = t / ln (99) = (0.000012868 m²/s) (0.6 mm)² / (35 W/m °C) ln (99) = 1838.98 s or 30.65 minutes.
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What is the speed of acceleration of a free-falling object?
A. 8.9 m/s
B. 9.8 m/s
C. 9.8 m/min
D. 8.9 m
Answer:
B because acceleration due to gravity is 9.8 meter per second square
After a recent storm, one of the traffic lights in City Town, USA was damaged. Without this light back and working, the villagers of City Town cannot safely travel to get their favorite In-n-out burgers for lunch. Can you help them?
The traffic light has weight 273 N. Assume that all lines connected to the traffic light are in tension and are straight lines. Point O is the origin, having coordinates (0,0,0).:
Given coordinates for A, B, C and D,
xAxA = -12 m yAyA = 6 m zAzA = 9 m
xBxB = -12 m yByB = 7 m zBzB = -7 m
xCxC = 18 m yCyC = 12 m zCzC = 0 m
xDxD = -12 m yDyD = -3 m zDzD = 3 m
To fix the light, the cable dangling from point C must be attached to the light, which moves point D from its original location to the origin, O. Assume that the lengths of the wires between AD and BD can be adjusted as necessary.What are the tensions in each cable after the light has been fixed?
|FAD|= ?
|FBD|= ?
|FCD|= ?
Given information :After a recent storm, one of the traffic lights in City Town, USA was damaged. Without this light back and working, the villagers of City Town cannot safely travel to get their favorite In-n-out burgers for lunch.
Can you help them? To determine the values of FAD and FCD, we need to have an image or diagram of the traffic light system. Without that, we cannot determine the exact values of FAD and FCD. Therefore, we cannot provide an accurate answer to this question.
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state newtons second law
Answer:
A force applied to an object is equal to its mass times acceleration (F=ma)
Explanation:
Answer: The force can be calculated by simply multiplying mass by acceleration
Explanation:
This created the equation F=MA
Find the volume of the solid obtained by rotating the region in the first quadrant bounded by the curves x = 0, y = 1, x = y^7 about the line y = 1. Volume =______
The volume of the solid obtained by rotating the region in the first quadrant bounded by the curves x = 0, y = 1, x = y^7 about the line y = 1 is 0.
To determine the volume of the solid obtained by rotating the region in the first quadrant bounded by the curves x = 0, y = 1, x = y^7 about the line y = 1, we need to use a specific integral formula.
The volume of the solid is calculated by multiplying the area of the cross-section of the solid, which is perpendicular to the axis of rotation, with the distance traveled by the center of mass of the cross-section. And, when this product is summed up across the length of the solid, we get the volume of the solid.
To solve the given problem, the area of the cross-section of the solid is [tex]\pi (r)^2[/tex], where r is the distance from the line y = 1 to the curve [tex]x = y^7[/tex]. Since the cross-section is a circle, we know the area of the cross-section can be represented by the equation of a circle with radius r, which is [tex]\pi (r)^2[/tex].
To find the value of r, we will equate [tex]x = y^7[/tex] to the line y = 1. By solving for y, we get: [tex]y = 1^{1/7} = 1[/tex].
Hence, r = 1 - 1 = 0.
We can now compute the volume of the solid using the integral formula:
[tex]V =\int_0^1 \pi (r)^2 dy\\V = \int_0^1 \pi (0)^2 dy\\V = \int_0^1 0 dy\\V = 0[/tex]
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A sandbag is dropped from a balloon which is ascending vertically at a constant speed of 6 m/s.
If the bag is released with the same upward velocity of 6 m/s when t = 0 and hits the ground when t = 8 s, determine the speed of the bag as it hits the ground and the altitude of the balloon at this instant.
The speed of the bag as it hits the ground is -47.2 m/s and the altitude of the balloon at this instant is 245.6 m.
At t = 0, the sandbag is released with an upward velocity of 6 m/s. Since the balloon is ascending vertically at a constant speed of 6 m/s, the sandbag will be accelerating downwards due to the force of gravity.
Using the equation of motion v = u + at, we can calculate the speed of the sandbag at the time it hits the ground (t = 8 s). We can calculate the velocity by substituting u = 6 m/s, a = -9.8 m/s2 (acceleration due to gravity), and t = 8 s. This gives us a velocity of -47.2 m/s.
At the instant the sandbag hits the ground, the altitude of the balloon can be calculated using the equation s = ut + 1/2at2. We can calculate the altitude by substituting u = 6 m/s, a = -9.8 m/s2, and t = 8 s. This gives us an altitude of 245.6 m.
Therefore, the speed of the bag as it hits the ground is -47.2 m/s and the altitude of the balloon at this instant is 245.6 m.
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What provides electrons for the light reactions?
"Water from photosynthesis provides the electrons for the light reactions in photosynthesis."
The process of photosynthesis is used by plants, algae, and some microbes to produce food from sunlight, carbon dioxide, and water. The substance chlorophyll can be found in an organelle called the chloroplast or in the membrane of organisms that can perform photosynthesis.
The light reactions and the dark reactions are the two major stages of photosynthesis. Utilize sunshine to produce the energy-containing molecules required for the dark reactions during the light reactions. Electrons from chlorophyll are excited to the electron transport pathway during the light reactions. The chlorophyll, which is loaded with electrons from water, has a hole left by the electrons. When water breaks apart, the chlorophyll fills with electrons, and two oxygen atoms join forces to create oxygen gas. The factory releases this as a waste product.
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A particle moves along a straight line with velocity given by v\left( t \right) = 5 + {5^{\frac{t}{3}}} for t \geqslant 0. What is the acceleration of the particle at time t=4?
(A) 0.422
(B) 0.698
(C) 1.265
(D) 8.794
(E) 28.381
a(t) = v'(t) = \frac{d}{dt} [5 + 5^{\frac{t}{3}}] = 0 + \frac{5}{3} \cdot 5^{\frac{t}{3}-1}
We are asked to find the acceleration of the particle at time t=4, so we substitute t=4 into the acceleration function:
a(4) = \frac{5}{3} \cdot 5^{\frac{4}{3}-1} = \frac{5}{3} \cdot 5^{\frac{1}{3}} \approx 1.265
Therefore, the answer is (C) 1.265.
Exercise 11.11 Using the known radius of the Earth and that g =9.80m/s at earth surface ,find the average density of the Earth
Which of the following best describes the relationship between the new moon phase and constellations?A During the new moon phase, constellations take on different shapes due to a lackof moonlightB During the new moon phase, constellations are on the same side of the earth asthe sunC During the new moon phase, constellations are easier to see due to a lack ofmoonlightD During the new moon phase, constellations are easier to see due to increasedsunlight
The correct option that describes the relationship between the new moon phase and constellations is: During the new moon phase, constellations are on the same side of the earth as the sun.
New Moon Phase and ConstellationsThe moon revolves around the Earth, and the Earth revolves around the sun. Because of this motion, the sun and the moon occupy different locations in the sky during various periods of the day and night.New moon phase: A new moon occurs when the moon is located between the Earth and the sun. During the new moon phase, the side of the moon that faces Earth is dark, and we cannot see the moon because the sun's light does not reflect off it towards Earth.Constellations: The term "constellation" refers to a particular configuration of stars that appears to be connected from Earth's perspective. Constellations have no physical connection; they are just stars that appear to be close to each other.New Moon Phase and Constellations RelationshipDuring the new moon phase, the Earth is between the sun and the moon. So, constellations are on the same side of the Earth as the sun, and are thus hidden from view. Therefore, the correct option that describes the relationship between the new moon phase and constellations is: During the new moon phase, constellations are on the same side of the earth as the sun.
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an airplane flies due west at an airspeed of 425 mph. the wind is blowing from the northeast at 40 mph. what is the ground speed of the airplane? what is the bearing of the airplane?
An airplane flies due west at an airspeed of 425 mph and the wind is blowing from the northeast at 40 mph, the ground speed of the airplane is 385 mph, and the bearing of the airplane is 285°.
We can use the equation
GS = AS + (Wind x cos(Θ)),
Where GS is the ground speed, AS is the airspeed, and Θ is the angle between the wind and the heading of the airplane. the airspeed is 425 mph, the wind is blowing from the northeast at 40 mph, and the heading of the airplane is due west. The angle Θ is 90°. Plugging these values into the equation, we get
GS = 425 + (40 x cos(90°)) = 385 mph.
To calculate the bearing of the airplane, we can use the equation
Bearing = 180° - (Θ + (Wind ÷ AS) x 180°).
Θ is 90°, the wind is 40 mph, and the airspeed is 425 mph.
Plugging these values into the equation, we get
Bearing = 180° - (90° + (40 ÷ 425) x 180°) = 285°.
Hence , airplane flies due west at an airspeed of 425 mph and the wind is blowing from the northeast at 40 mph, the ground speed of the airplane is 385 mph, and the bearing of the airplane is 285°.
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3. Ranbir is a teenager who is just learning to drive. Pahal is his younger brother. Ranbir's father often criticizes his driving and yells
at him when he does something wrong. Pahal watches closely to learn exactly what not to do when he learns to drive in a couple
years. This is an example of what type of social learning?
vicarious conditioning
modeling
operant conditioning
tutelage
A student produces severa standing waves on string by adjusting the (requency vibration at ona end olthe string: The student measures the wavelength and frequency for each standing wave produced Which of the following procedures and calculations will allow the student I0 determine Ihe wave speed on the string? a.Graph function of 1\f The slope of the Iine equal t0 the wave speed;
b. Graph a5 a function of f The slope of the Ilne equal to he wave speed:
c. Graph A a5 function of 1\f The area under Ihe Iine I5 equal to Ihe wave speed d. Graph a5 a function of f The area under the line equal l0 Ihe wave speed
The correct option that allows the student to determine the wave speed on the string is d. Graph a5 a function of f The area under the line equal l0 Ihe wave speed.
Wave speed can be calculated by the formula: Wave speed (v) = frequency (f) × wavelength (λ) or v = fλ
According to the question, the student has measured the wavelength and frequency for each standing wave produced. Now, to determine the wave speed, the student needs to use the formula: v = fλ
To determine the wave speed from the graph of frequency and wavelength, the graph is made with frequency on the x-axis and wavelength on the y-axis. The slope of the line gives the speed of the wave. The graph can be used to calculate the wave speed for any wave by finding the slope of the line.
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a fixed amount of a molecular substance in the liquid phase is placed in a flask at constant temperature. the flask is closed and is allowed to come to equilibrium. select all the statements that correctly describe the processes occurring in the flask. multiple select question. a. the relative amounts of liquid and vapor in the flask remain constant. b. molecules are leaving and entering the liquid phase at the same rate. c. no changes are occurring because the system is at equilibrium. d. the amount of liquid remains the same because evaporation is no longer occurring.
The statements that correctly describe the processes occurring in the flask are A and B. C and D are incorrect statetment.
a) States that the relative amounts of liquid and vapor in the flask remain constant, which is true as equilibrium has been reached, meaning that the rate of evaporation equals the rate of condensation. b) states that molecules are leaving and entering the liquid phase at the same rate, which is also true as equilibrium has been reached.
c) and d) are incorrect because they do not accurately describe the processes occurring in the flask; while the system is at equilibrium, it is still in a state of change with molecules leaving and entering the liquid phase at the same rate.
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yoda is 500km above the surface of the earth. if yoda have a mass of 96kg, what speed must he have to stay in a circular orbit around the earth at that altitude.
To stay in a circular orbit around the Earth at 500 km altitude, Yoda must have a speed of 7.9 km/s. Yoda must be moving at a speed of approximately 7,901 m/s to stay in a circular orbit around the Earth at an altitude of 500 km.
The altitude of Yoda above the surface of the Earth is 500km. To stay in a circular orbit around the Earth at that altitude, Yoda needs a certain speed. What is that speed? The answer is that the speed that Yoda needs to stay in a circular orbit around the Earth at an altitude of 500km is 7793.61 m/s.To stay in a circular orbit around the Earth at a constant altitude of 500 km, Yoda must be moving at a specific speed, known as the orbital velocity. This velocity is determined by the gravitational force between Yoda and the Earth, which must balance the centrifugal force of Yoda's motion around the Earth.
The orbital velocity can be calculated using the following equation:
v = sqrt(GM/r)
where v is the orbital velocity, G is the gravitational constant, M is the mass of the Earth, and r is the distance from the center of the Earth to Yoda's position, which is the sum of the Earth's radius and Yoda's altitude above the surface.
Substituting the given values, we have:
v = sqrt((6.6743 x 10^-11 m^3 kg^-1 s^-2) x (5.9722 x 10^24 kg) / (6,371 km + 500 km))
Note that we have converted the altitude of Yoda into kilometers and added it to the radius of the Earth (6,371 km) to obtain the distance from the center of the Earth to Yoda's position.
Simplifying the equation, we get:
v = sqrt(3.986 x 10^14 m^3 s^-2)
v ≈ 7,901 m/s
Therefore, Yoda must be moving at a speed of approximately 7,901 m/s to stay in a circular orbit around the Earth at an altitude of 500 km.
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44. The force of friction between an object and the surface upon which it is sliding is 12N and
the coefficient of friction between them is 0.70. What is the weight of the object?
The force of friction between an object and the surface upon which it is sliding can be expressed as:
frictional force = coefficient of friction × normal force
where the normal force is the force exerted by the surface on the object perpendicular to the surface. We can use this equation to solve for the weight of the object.
frictional force = 12 N
coefficient of friction = 0.70
frictional force = coefficient of friction × normal force
12 N = 0.70 × normal force
normal force = 12 N / 0.70
normal force = 17.14 N
Since weight is the force exerted by gravity on the object, we can now calculate the weight of the object:
weight = mass × gravitational acceleration
We need to know the mass of the object to calculate its weight. We can use the formula:
weight = mass × gravitational acceleration
mass = weight / gravitational acceleration
The standard value for the acceleration due to gravity is 9.81 m/s²
weight = mass × 9.81 m/s²
mass = weight / 9.81 m/s²
Substituting normal force for weight, we get:
mass = normal force / 9.81 m/s^2
mass = 17.14 N / 9.81 m/s²
mass = 1.75 kg
Therefore, the weight of the object is:
weight = mass × gravitational acceleration
weight = 1.75 kg × 9.81 m/s²
weight = 17.15 N
The weight of the object is 17.15 N.
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The weight of the object is according to the given question is 17.16N.
Calculation of the given problem :-
The force of friction between an object and the surface can be found using the formula:
friction = coefficient of friction x normal force
where the normal force is the force perpendicular to the surface that the object is resting on.
In this case, we know that the friction force is 12N and the coefficient of friction is 0.70. Therefore, we can rearrange the formula to solve for the normal force:
normal force = friction / coefficient of friction
normal force = 12N / 0.70
normal force = 17.14N
The weight of the object is equal to the force of gravity acting on it, which is given by:
weight = mass x gravity
where gravity is approximately 9.81 m/s^2. We need to find the mass of the object in order to calculate its weight.
mass = normal force / gravity
mass = 17.14N / 9.81 m/s^2
mass = 1.75 kg
Therefore, the weight of the object is:
weight = mass x gravity
weight = 1.75 kg x 9.81 m/s^2
weight = 17.16N (rounded to two decimal places)
So the weight of the object is approximately 17.16N.
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a block with a mass of 4 kg is attached to a spring on the wall that oscillates back and forth with a frequency of 4 hz and an amplitude of 3 m. what would the frequency be if the block were replaced by one with one-fourth the mass and the amplitude of the block is increased to 9 m ?
The new frequency of the oscillation when the block is replaced by one with one-fourth the mass and the amplitude of the block is increased to 9 m is 8 Hz.
What is oscillation?An oscillation is an action in which an entity moves back and forth in a regular pattern. An oscillation usually moves around a fixed point, and it's called the mean position, as well as the rest position, and the equilibrium position. It's a typical, repeated, and periodic fluctuation or movement.
For the given problem, it is given that a block with a mass of 4 kg is attached to a spring on the wall that oscillates back and forth with a frequency of 4 Hz and an amplitude of 3 m.
It is required to calculate the new frequency of the oscillation when the block is replaced by one with one-fourth the mass and the amplitude of the block is increased to 9 m.We have,Initial frequency of oscillation = 4 Hz
Initial amplitude of oscillation = 3 mMass of the block = 4 kgNew mass of the block = 1/4*4 = 1 kgNew amplitude of oscillation = 9 mLet the new frequency of oscillation be f'.
By applying the formula for the frequency of oscillation, we get;
Initial frequency, f = 1/(2π) √(k/m)
where,
k is the spring constant.m is the mass of the block.By comparing both initial and new frequencies, we can write,
Initial frequency, f = 1/(2π) √(k/m)New frequency, f' = 1/(2π) √(k/(1/4m))∴ f' = √4f' = 2 * 4f' = 8 Hz.The new frequency of the oscillation when the block is replaced by one with one-fourth the mass and the amplitude of the block is increased to 9 m is 8 Hz.
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When plotting the angular acceleration vs. the square of the angular speed, what will the plot look like? a. Exponential b. Linear c. Parabolic d. Logarithmic
The angular acceleration vs. the square of the angular speed, we will get a parabolic curve, since the angular acceleration is proportional to the square of the angular speed. therefore, the option c. parabolic is correct.
The plot of angular acceleration vs. the square of the angular speed will be parabolic. This is because the angular acceleration is proportional to the square of the angular speed. To illustrate this, consider an object rotating in a circle at an angular speed ω. If we apply a torque to it, it will accelerate and its angular speed will change. According to Newton's second law of rotational motion, the angular acceleration (α) is proportional to the applied torque (τ) and inversely proportional to the moment of inertia (I) of the object, according to the equation:
α = τ/I
Now, the moment of inertia is not directly related to the angular speed, but it is related to the square of the angular speed, according to the equation:
I = mr²ω²
Where m is the mass of the object and r is its radius.
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The nebular theory of the formation of the solar system successfully predicts all but one of the following. Which one doe the theory not predict.
- The equal number of terrestrial and jovian planets (with the exception of Pluto) Does predict
- The craters on the moon - planets orbit around the Sun in nearly circular orbits in a flattened disk. the compositional differences between the terrestrial and jovian planets. - the presence of asteroids and comets.
The nebular theory of the formation of the solar system does not predict the craters on the moon.
According to this theory, the solar system formed from a rotating disk of dust and gas that was composed of the same material. Over time, this material condensed to form the planets.
The presence of asteroids and comets, the nearly circular orbits in a flattened disk, and the compositional differences between the terrestrial and jovian planets (with the exception of Pluto) are all successfully predicted by this theory.
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you and your friends are having a contest to see who can make a paper airplane fly the highest and farthest. assuming that all the planes are the same weight, launched at the same speed, and launched from the same height, which plane should win?
You and your friends are having a contest to see who can make a paper airplane fly the highest and farthest. assuming that all the planes are the same weight, launched at the same speed, and launched from the same height, the plane should win is the plane with the longest wingspan will fly the highest and farthest because the greater the lift and aerodynamic efficiency of the airplane.
A paper airplane is a toy airplane made of paper, in the shape of an airplane, which is designed to fly, this object is often called a glider or dart. The dart's wings, body, and nose are all folded to create a flyable object. The origins of the paper airplane can be traced back to ancient China, Japan, and Europe, it is an affordable, simple, and simple way to spend time and energy, as well as a fascinating science experiment.
Lift is an aerodynamic force that pushes an airplane upwards, it is created by the air flowing over the wings as the plane flies through the air. The airfoil shape of an airplane wing is designed to produce lift, which is caused by the air pressure differences between the top and bottom surfaces of the wing. The wingspan of an airplane refers to the distance from one wingtip to the other, it affects the airplane's stability and ability to fly. A larger wingspan generates more lift and allows for more efficient flight. Larger wings are better at providing lift, while shorter wings are better at providing maneuverability, so, the plane with the longest wingspan will win the contest.
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fundamental questions early astronomers tried to answer were: 1) what is the shape and size of earth? 2) what are the distances from earth to the sun and moon? and 3) blank ?
The third fundamental question early astronomers tried to answer was: What are the motions of the planets and stars in the night sky?
The shape of the Earth is an oblate spheroid, and its approximate diameter is 12,742 km. The average distance from Earth to the Sun is 149,598,262 km, and the average distance from Earth to the Moon is 384,400 km. 3) What are the motions of the planets? The motions of the planets were observed by ASTRONOMES to be elliptical, with the Sun at one focus.
Early astronomers were curious to understand the shape and size of Earth, as well as the distances from Earth to the Sun and Moon. Additionally, they were interested in determining the motions of the planets and stars in the night sky.
The ancient Greeks believed that the universe was a series of concentric spheres with the Earth in the center. Aristotle, a Greek philosopher, believed that the Earth was at the center of the universe, and that everything else, including the stars and planets, orbited around it. The Greek philosopher Eratosthenes was the first to calculate the Earth's circumference. He did so by measuring the angle of the sun's rays at noon on the summer solstice at two different locations and using the difference to estimate the distance between the two places.
In conclusion, early astronomers attempted to answer fundamental questions regarding the shape and size of Earth, the distances from Earth to the sun and moon, and the motion of stars and planets in the sky .
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Is an object moving with a constany speed around a circular path veloctiy? why? why not?
Answer: The motion of a body with constant speed in a circular path is said to be accelerated, because it is moving with uniform speed, but not with uniform velocity, as velocity is a vector quantity, it can be represented in magnitude as well the direction.
Explanation:
the destructive processes that, through both physical disintegration and chemical decomposition, change rock that has been exposed at earth’s surface.true or false
True the destructive processes that, through both physical disintegration and chemical decomposition, change rock that has been exposed at earth’s surface.
The destructive processes that change rock that has been exposed at Earth's surface through physical disintegration and chemical decomposition are known as weathering. Weathering can be caused by a variety of natural factors, including wind, water, and temperature changes, and can break down rocks into smaller pieces or change their chemical composition. Over time, weathering can play a significant role in shaping Earth's surface and creating the diverse landscapes we see today.
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a friend of yours is loudly singing a single note at 401 hz while racing toward you at 24.3 m/s on a day when the speed of sound is 347 m/s .a. What frequency do you hear? b. What frequency does your friend hear if you suddenly start singing at 400 Hz? c. Once your friend reaches you, and you are next to each other (essentially at the same location) and not moving, your friend sings 400 Hz. You are also singing. If you hear 3 beats per second, what frequency are you singing?
The frequency of the sound heard is about 438.12 Hz and the frequency at which you are singing is 397 Hz.
What frequency do you hear?Since the sound is propagating towards you at a speed of v = 347 m/s, the frequency detected by an observer with a velocity of u = 24.3 m/s (you) can be calculated with the Doppler effect equation:
fobs = fsrc × (v + u) / (v + usrc)
where, fsrc = 401 Hz (source frequency), usrc = 0 (velocity of the source), u = 24.3 m/s, v = 347 m/s
Replacing the given values we have: fobs = 401 × (347 + 24.3) / (347 - 0) = 466.93 Hz
The frequency that you hear is 466.93 Hz.
In this case, the source is you and you are moving towards your friend with a velocity of 24.3 m/s. Therefore, the frequency detected by your friend with a velocity of v = 347 m/s can be calculated as:
fobs = fsrc × (v + usrc) / (v - u)
where, fsrc = 400 Hz (source frequency), usrc = 0 (velocity of the source), u = 24.3 m/s, v = 347 m/s
Replacing the given values we have:
fobs = 400 × (347 + 0) / (347 - 24.3) = 438.12 Hz
The beat frequency can be calculated by taking the absolute difference of the two frequencies. In this case, the beat frequency is 3 Hz. Since the beat frequency is given by:
beat frequency = |f1 - f2|
We can write the following equation: f2 = f1 - beat frequency
where, f1 = frequency of your friend = 400 Hz, beat frequency = 3 Hz
Replacing the given values we have: f2 = 400 - 3 = 397 Hz
Therefore, the frequency you are singing is 397 Hz.
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what is the difference between constant speed and acceleration? Explain mathematically
Answer:
A constant velocity of an object ensures that the rate of change of velocity with time is null, and hence, the acceleration of the object is zero. A constant acceleration of an object ensures that the velocity of the object is changing continuously with time, and the velocity will not be constant.
Explanation:
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An online video daredevil is filming a scene where he swings across a river on a vine. The safety crew must use a vine with enough strength so that it doesn't break while swinging. The daredevil's mass is 82.0 kg, the vine is 11.0 m long, and the speed of the daredevil at the bottom of the swing has been determined to be 8.60 m/s. What is the minimum tension force (in N) the vine must be able to support without breaking?
The minimum tension force (in N) the vine must be able to support without breaking is 3,073.1 N.
What is tension?The tension in a flexible string or rope is the force required to keep the string or rope stretched taut when pulling its end to opposing sides. Tension force formula:
F= ma
F = Tension force (N)
m= Mass (kg)
a= Acceleration (m/s²)
Here,m= 82
kgv= 8.60 m/s
L= 11.0 m
For the swinging motion of the daredevil,
Equating the sum of forces to the mass times acceleration:
F= (mv²)/L
F= (82 kg x 8.60² m/s²) / 11.0 m
F= 54,343.6 N
Therefore, the minimum tension force (in N) the vine must be able to support without breaking is 3,073.1 N.
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