mass gram, time sec, temp kelvin, vol liter, dens grams/cm3
The first thing you should know for this case is the definition of distance.
d = v * t
Where,
v = speed
t = time
We have then:
d = v * t
d = 9 * 12 = 108 m
The kinetic energy is:
K = ½mv²
Where,
m: mass
v: speed
K = ½ * 1500 * (18) ² = 2.43 * 10 ^ 5 J
The work due to friction is
w = F * d
Where,
F = Force
d = distance:
w = 400 * 108 = 4.32 * 10 ^ 4
The power will be:
P = (K + work) / t
Where,
t: time
P = 2.86 * 10 ^ 5/12 = 23.9 kW
answer:
the average power developed by the engine is 23.9 kW
The acceleration is the correct answer
Answer:
W = 157.5kJ
Explanation:
Assuming it moves the container at constant speed, the work done by the crane will be equal to the variation of the potential gratitational energy on the container:
where h2= -8m and h1=0m
Wc = 157.5kJ
We can use the formula,
1/R = 1/r1 + 1/r2 + 1/r3 + ....
Hope this helps!