Answer:
(a) 81.54 N
(b) 570.75 J
(c) - 570.75 J
(d) 0 J, 0 J
(e) 0 J
Explanation:
mass of crate, m = 32 kg
distance, s = 7 m
coefficient of friction = 0.26
(a) As it is moving with constant velocity so the force applied is equal to the friction force.
F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N
(b) The work done on the crate
W = F x s = 81.54 x 7 = 570.75 J
(c) Work done by the friction
W' = - W = - 570.75 J
(d) Work done by the normal force
W'' = m g cos 90 = 0 J
Work done by the gravity
Wg = m g cos 90 = 0 J
(e) The total work done is
Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J
First change km/ s into m/s, then use the formula
Lambda = velocity/ frequency
Neutron is commonly used to initiate a fission chain reaction.
Because acceleration is constant, the acceleration of the car at any time is the same as its average acceleration over the duration. So

Now, we have that

so we end up with a distance traveled of


Answer: Part(a)=0.041 secs, Part(b)=0.041 secs
Explanation: Firstly we assume that only the gravitational acceleration is acting on the basket ball player i.e. there is no air friction
now we know that
a=-9.81 m/s^2 ( negative because it is pulling the player downwards)
we also know that
s=76 cm= 0.76 m ( maximum s)
using kinetic equation

where v is final velocity which is zero at max height and u is it initial
hence


now we can find time in the 15 cm ascent


using quadratic formula

t=0.0409 sec
the answer for the part b will be the same
To find the answer for the part b we can find the velocity at 15 cm height similarly using

where s=0.76-0.15
as the player has traveled the above distance to reach 15cm to the bottom


when the player reaches the bottom it has the same velocity with which it started which is 3.861
hence the time required to reach the bottom 15cm is

t=0.0409