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
To find the impulse you multiply the mass by the change in velocity (impulse=mass×Δvelocity). So in this case, 3 kg × 12 m/s ("12" because the object went from zero m/s to 12 m/s).
The answer is 36 kg m/s
Answer: 2.068*
m
Explanation: According to work energy-theorem , the workdone in accelerating the electron equals the energy it would give off in terms of light.
workdone= qV
energy = hc/λ
q=magnitude of an electronic charge= 1.602*
h= planck constant = 6.626*
c= speed of light =2.998* 
v= potential difference= 6*
λ= wavelength=unknown
by making λ subject of formulae we have that
λ= 
λ = 6.626* * 2.998* / 1.602* * 6*
λ = 
by doing the necessary calculations, we have that
λ = 2.068*m
F=m*a
F=65 kg *9.8 m/s^2
F=637 N (Newtons) — this is the weight
