Answer:
557.7Nm
Explanation:
We are given that
Radius,r=1.9 m
Angular speed,
Time,t=3.5 s
Initial angular velocity,
We have to find the large torque would have to be exerted .
Where
Substitute the values
Answer:
Explanation:
The distance travelled in the free fall is H - h
Since the apple originally started from rest we can use v^2 = u^2 + 2 x g x s where v is the final velocity, g the accln due to gravity and s the distance travelled and u is the initial velocity = 0
So the velocity just before it enters the grass is sq rt [2 x g x (H - h)]
Once in the grass, it slows down at a constant rate which means that the acceleration (a) during this period is constant.
So once again using the same formula we have v = O and u = sq rt[2 x g x (H-h)]
so since v^2 = u^2 + 2 x a x s then
O^2 = 2 x g x (H-h) + 2 x a x h
{O^2 - 2 x g x (H - h)}/(2 x h) = a
Answer:
I think the answer might be C
Show a common scale and convert each choice to it ( I will choose Celsuis):
A)100C
B) 100F = 37 C
C) 100 - 273 = -173 C
so the answer is A
OK, the wedge is accelerating (a) at Theta = 180 degrees (to the right) and the wedge is inclined theta = 75 degrees. For the m = 2 kg block to remain at rest all we need is a net force f = W cos(theta) - F sin(theta) = 0; where F = ma and W = mg the weight of the block. That is, the weight component along the incline is offset by the acceleration component along the surface; so the block does not slide.
Solving we have W cos(theta) = mg cos(theta) = ma sin(theta) = F sin(theta); such that a = g cos(theta)/sin(theta) = g cot(theta). Assuming g ~ 9.81 m/sec^2, you can now plug and chug to find the answer.
<span>The physics is this...when the net force on a body is f = 0, that body will not accelerate and start to move if it is already still. So when the block's weight component along the surface of the wedge is offset by the equal but opposite force along the surface of the accelerating wedge, the still block will not move.
I hope my answer has come to your help. Thank you for posting your question here in Brainly.
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