How far will a free falling object fall in 8.7 secs if it started from rest? Remember acceleration is negative for free fall. Do
1 answer:
Answer:
h~=371.26m
Explanation:
when an object falls we use the equations of accelerated motion. There is only one that gives distance.
Since we have no initial velocity (started from rest) we can get rid of the (ut) term
where a we substitute g (gravitational acceleration, constant for given heights and almost 9.81m/s^2).
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Answer:
The induced emf in the coil is 0.522 volts.
Explanation:
Given that,
Radius of the circular loop, r = 9.65 cm
It is placed with its plane perpendicular to a uniform 1.14 T magnetic field.
The radius of the loop starts to shrink at an instantaneous rate of 75.6 cm/s ,
Due to the shrinking of radius of the loop, an emf induced in it. It is given by :
So, the induced emf in the coil is 0.522 volts.
Answer:
The speed of the ball is approximately 5.94 m/s
The Tension of the string at the bottom is 3.92 N
Explanation:
We need to find the speed of the ball, which is constant due to the fact that we are in a uniform circular motion. Notice as well that the speed of the ball is the magnitude of the tangential velocity "" (vector that changes direction with the position of the ball but doesn't change magnitude in this case).
We analyze first the top position of the circular motion, for which information on the tension of the string is given (see first free body diagram in the attached picture). We are told that the tension at the top of the movement equals twice the force of gravity on the ball's mass: T - 2*m*g = 1.96 N. And we know that there are two forces acting on the ball in that position (illustrated with the green arrows pointing down): one is the ball's weight due to gravity, and the other is the string's tension. So we can write Newton's second law for this situation:
Newton's second law tells us that the net force should equal the mass of the ball times its acceleration (F = m * a), and in this motion, the acceleration is the centripetal acceleration. Therefore weuse this equation to solve for the centripetal acceleration of the ball:
The centripetal acceleration is defined as the square of the tangential velocity divided the radius of the circular motion. Then we use it to derive the magnitude of the tangential velocity (speed of the ball):
So we have found the speed of the ball.
Now we focus our attention to the bottom of the motion, and again use Newton's second law to solve for the string tension (see second free body diagram in the attached picture).
We notice here that the tension and the weight are acting in opposite directions, so we have such into account when finding the net force on the ball, and then solve for the tension knowing the value of the centripetal acceleration (recall that the magnitude of the tangential velocity is the same because of the uniform circular motion).
