Answer:
The velocity of the hay bale is - 0.5 ft/s and the acceleration is 
Solution:
As per the question:
Constant velocity of the horse in the horizontal, 
Distance of the horse on the horizontal axis, x = 10 ft
Vertical distance, y = 20 ft
Now,
Apply Pythagoras theorem to find the length:


Now,
(1)
Differentiating equation (1) w.r.t 't':


where
= Rate of change of displacement along the horizontal
= Rate of change of displacement along the vertical
= velocity along the x-axis.
= velocity along the y-axis



Acceleration of the hay bale is given by the kinematic equation:





Answer:
41.8m/s^2
Explanation:
Since the dragster starts from rest, initial velocity (u) = 0m/s, final velocity (v) = 25.9m/s, time (t) = 0.62s
From the equations of motion, v = u + at
a = (v - u)/t = (25.9 - 0)/0.62 = 25.9/0.62 = 41.8m/s^2
(a) 0.249 (24.9 %)
The maximum efficiency of a heat engine is given by

where
Tc is the low-temperature reservoir
Th is the high-temperature reservoir
For the engine in this problem,


Therefore the maximum efficiency is

(b-c) 0.221 (22.1 %)
The second steam engine operates using the exhaust of the first. So we have:
is the high-temperature reservoir
is the low-temperature reservoir
If we apply again the formula of the efficiency

The maximum efficiency of the second engine is

Answer:
(A) Velocity will be 1.88 m/sec
(b) Force will be 187.45 N
Explanation:
We have given work done = 4780 j
Distance d = 25.5 m
(A) Mass of the truck m = 
We know that kinetic energy is given by

So 
(B) We know that work done is given by
W = Fd
So 