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Physics: Post your doubts here!
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The Young modulus of the material of a wire is to be found. The Young modulus E is given by the equation below.
E=4F*l/πd^2*x
The wire is extended by a known force and the following measurements are made. Which measurement has the largest effect on the uncertainty in the value of the calculated Young modulus?
E=4F*l/πd^2*x
The wire is extended by a known force and the following measurements are made. Which measurement has the largest effect on the uncertainty in the value of the calculated Young modulus?
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Oh! am IWelcome.
No,it wasn't.You're just late. XD
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A weight W hangs from a trolley that runs along a rail. The trolley moves horizontally through a distance p and simultaneously raises the weight through a height q.
As a result, the weight moves through a distance r from X to Y. It starts and finishes at rest. How much work is done on the weight during this process?
A Wp
B W(p + q)
C Wq
D Wr
As a result, the weight moves through a distance r from X to Y. It starts and finishes at rest. How much work is done on the weight during this process?
A Wp
B W(p + q)
C Wq
D Wr
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It's C, because work done = Force applied * distance moved in the direction of force
In this case the force acting on the mass is the tension in the string holding the mass which is equal to the weight in this case. And the distance moved by the weight in direction of this force is q.
So,
work done = W *q
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it's the diameter which has the largest effect. Look in the equation, d is a squared quantity. So the %age uncertainty in d would need to be doubled. So it will have the largest effect.
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- Messages
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The diagram shows a trace of a wave on a cathode-ray oscilloscope. The vertical and horizontal gridlines have a spacing of 1.0cm. The voltage scaling is 4Vcm–1 and the time scaling is 5ms cm–1.
What are the amplitude and period of the wave?
What are the amplitude and period of the wave?
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A steel sphere is dropped vertically onto a horizontal metal plate. The sphere hits the plate with a speed u, leaves it at a speed v, and rebounds vertically to half of its original height. Which expression gives the value of v/u?
A 1/2^2
B 1/2
C 1/√2
D 1-1/√2
A 1/2^2
B 1/2
C 1/√2
D 1-1/√2
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It's C.
Take two situations here. First take the situation before the ball hits the metal plate. Let the distance it falls be 'h'.
Now you're given that the ball hits the metal plate with velocity 'u'. So apply an equation of motion here:
Vf^2 = Vi^2 + 2as
final velocity (Vf) is 'u', initial velocity (Vi) is 0 and distance fallen (s) is h. SO substitute these values into the equation, you'll get:
u^2 = 2ah <---- make 'h' the subject of this equation
h = u^2/2a -----equ (1)
Now consider the second situation when the ball rebounds after hitting. When ball rebounds, it leaves the plate with velocity 'v'. So this going to be our initial velocity in this situation. After rising a height of h/2 (as it said that it rises to half it's height of fall), it will come to rest momentarily. so it's final velocity at max height in this situation will be zero. Now again, substitute these values in to equation of motion mentioned above.
0^2 = v^2 -2a(h/2) <--- i have placed a minus sign before 2as, because the ball is decelarating here.
h = v^2/a ---- equ (2)
Now put the two equations: equ (1) and equ (2) equal, and solve.
You'll get v/u = 1/√2
Last edited:
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A ball is released from rest on a smooth slope XY.
It moves down the slope, along a smooth horizontal surface YZ and rebounds inelastically at Z. Then it moves back to Y and comes to rest momentarily somewhere on XY.
Which velocity-time graph represents the motion of the ball?
It moves down the slope, along a smooth horizontal surface YZ and rebounds inelastically at Z. Then it moves back to Y and comes to rest momentarily somewhere on XY.
Which velocity-time graph represents the motion of the ball?
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It's B I think
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Nope the ans is A
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At first I thought it was B too but the ans in ms is A, so I wanna know y the ans is A and y it is not B.
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Yep it's A .
When the ball is released, the weight of the ball causes it accelerate uniformly. So it's speed increases uniformly (constant gradient). When it comes on to horizontal surface, it's speed becomes constant since now it's not accelerating anymore.
Then, it makes an inelastic collision with the wall, and loses it's K.E. Thus it's speed decreases as K.E=1/2mv ^2. So the velocity is smaller as compared to it's initial velocity. As the velocity is smaller now, so it'll take more time. This elimninates the option B.
Now when it again starts rising the slope, the weight plays it's role, decelrating the ball. As the acceleration of free fall remains same, so this gradient should be same as the gradient for the acceleration when the ball was falling down the slope.
Hope you'll understand.
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why wont the answer be C??
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Ofc I understood after all U explained itYep it's A .
When the ball is released, the weight of the ball causes it accelerate uniformly. So it's speed increases uniformly (constant gradient). When it comes on to horizontal surface, it's speed becomes constant since now it's not accelerating anymore.
Then, it makes an inelastic collision with the wall, and loses it's K.E. Thus it's speed decreases as K.E=1/2mv ^2. So the velocity is smaller as compared to it's initial velocity. As the velocity is smaller now, so it'll take more time. This elimninates the option B.
Now when it again starts rising the slope, the weight plays it's role, decelrating the ball. As the acceleration of free fall remains same, so this gradient should be same as the gradient for the acceleration when the ball was falling down the slope.
Hope you'll understand.
and thanks tho
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CoolYep it's A .
When the ball is released, the weight of the ball causes it accelerate uniformly. So it's speed increases uniformly (constant gradient). When it comes on to horizontal surface, it's speed becomes constant since now it's not accelerating anymore.
Then, it makes an inelastic collision with the wall, and loses it's K.E. Thus it's speed decreases as K.E=1/2mv ^2. So the velocity is smaller as compared to it's initial velocity. As the velocity is smaller now, so it'll take more time. This elimninates the option B.
Now when it again starts rising the slope, the weight plays it's role, decelrating the ball. As the acceleration of free fall remains same, so this gradient should be same as the gradient for the acceleration when the ball was falling down the slope.
Hope you'll understand.
Thnx !!
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It's D.
The uncertainty in R can find by first finding the fractional uncertainties of I and V, and then adding them and multiplying by the the value of resistance calculated.
( 0.2/2 + 0.5/15 ) * 7.5
= -+1 ohm