Answer: 219,344.5 j
Explanation:
from the question we have the following parameters
mass of barge (m) = 5 x 10^4 kg
angle of ropes (θ) = 45 degrees
force by each mule (F1) = 1.10 kN = 1100 N
force by each mule (F2) = 1.10 kN = 1100 N
distance (s) = 141 m
work done by the mules = force x cos θ x distance
since we are calculating the force of both mules we have to both mules before we can use in the equation
force = 1100 + 1100 = 2200 N
work done = 2200 x cos (45) x 141
work done = 219,344.5 j
Answer:
5.886 J
Explanation:
Given:
The mass of the book is, 
kg
Height of lift is, 
m
Acceleration due to gravity is, 
m/s²
Now, gain in gravitational potential energy is a function of change in position and is given as:
Here, 
is the change in gravitational potential energy.
Plug in 0.5 kg for 
, 9.81 for 
and 1.2 for 
. Solve for
Therefore, the gain in gravitational energy of the book is 5.886 J.
Average speed = (distance covered) / (time to cover the distance).
Distance covered = 437 km = 437,000 meters
Time to cover the distance = (7.27 hrs) x (3600 sec/hr) = 26,172 seconds
Average speed = (437,000 meters) / (26,172 seconds)
Average speed = 16.7 m/s
Answer:
Explanation:
Let's call the large box "1" and the smaller box "2". The equation for this works so that momentum is conserved, as the Law states.
![[(m_1v_1+m_2v_2)]_b=[m_1+m_2)v]_a](https://tex.z-dn.net/?f=%5B%28m_1v_1%2Bm_2v_2%29%5D_b%3D%5Bm_1%2Bm_2%29v%5D_a)
That says that the momentum of the larger box plus the momentum of the smaller box before the collision has to equal the momentum of the combined masses since they stick together. Filling in:
[(4(12) + 2(0)] = [(4 + 2)v] and
48 + 0 = 6v so
v = 8 m/s to the right
