A box of mass 5.0 kg is accelerated from rest by a force a cross a floor at a ra
ID: 1555014 • Letter: A
Question
A box of mass 5.0 kg is accelerated from rest by a force a cross a floor at a rate of 3.0 m/s^2 for 8.0 s. Find the net work done on the box. An atomic nucleus initially moving at 400 m/s emits an alpha particle in the direction of its initial velocity, and the remaining nucleus slows to 280 m/s. If the alpha particle has mass of 4 u and the original nucleus has mass of 200 u, what speed does the alpha particle have when it is emitted? A 0.145 kg baseball pitched horizontally at 27.0 m/s strikes a bat and pops straight up to a height of 31.5 m. If the contact time between bat and ball is 2.5 ms. Calculate the average force between the ball and bat during the contact. Express the following angle in radians (give a numerical values and as a fractions of pi): (a) 45.0 degree, (b) 60.0 degree, (c) 90.0 degree, What is the angular momentum of a 0.30 kg ball revolving on the end of a thin string in a circle of radius 1.35 m at an angular speed of 10.0 rad/s? (a) a grinding wheel 0.35 m in diameter rotates at 2200rpm. Calculate its angular velocity in rad/s. (b) what are the linear speed and acceleration of a point on the edge of the grinding wheel?Explanation / Answer
Total distace travelled is S =1/2 a t2
S = 0.5 X 3 X 64 = 96 m
F = ma = 5 X 3 = 15 N
Work W = F.S = 15 X 96 = 1440J
2)
The momentum of the nucleas remains same before and after the emission of alpha particle. So the momentum of the system is given by
Before, p=mv=400 X 200 =80000 u.m/s
After the emission we have two particles.
The sum of their two momentum must still be 80000 u.m/s
The remaining nucleus has a mass of 200 u and a momentum of:
p=mv=200 X 280=56000 u.m/s
The alpha particle has a momentum of
80000-56000=24000 u.m/s
So you can calculate its speed with
v=p/m=24000/4=6000 m/s.
3)
F = dP/dt = m*dv/dt = 0.145*(27+31.5)/.0025 = 3393 N
6)
(a) The angular velocity, in radians per second, is
2200 rev / min X ( 1 min / 60 s ) ( 2 / 1 rev ) = 230.266 radians / sec
(b) The linear speed on the edge is
v = r = ( 0.35/2 m ) X ( 230.26 radians / s ) =40.29 m/s
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