The energy transformations in a car’s engine are an example of multiple transformations. Please select the best answer from the
1 answer:
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Answer:
<em>60008.4 J</em>
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Explanation:
The mass of each kid = 30 kg
mass of the cart = 20 kg
The speed of the cart down the hill = 30 km/hr = 30 x 1000/3600 = 8.33 m/s
The height of the hill = 80 m
The potential energy of the boys at the top of the hill = mgh
where
m is the total mass of the kids and the cart = (30 x 2) + 20 = 80 kg
g is the acceleration due to gravity = 9.81 m/s^2
h is their height above the ground = 80 m (on the top of the hill)
substituting, we have
potential energy PE = 80 x 9.81 x 80 = 62784 J
At an instance at the bottom of the hill
their kinetic energy =
where
v is their velocity = 8.33 m/s
m is their total mass = 80 kg
substituting, we have
kinetic energy KE = = 2775.6 J
Total work done on the cart is equal to the energy lost by the cart when it reached the bottom of the hill
work done by friction = PE - KE = 62784 - 2775.6 = <em>60008.4 J</em>
Answer:
Explanation:
The two charges are q and Q - q. Let the distance between them is r
Use the formula for coulomb's law for the force between the two charges
So, the force between the charges q and Q - q is given by
For maxima and minima, differentiate the force with respect to q.
For maxima and minima, the value of dF/dq = 0
So, we get
q = Q /2
Now
the double derivate is negative, so the force is maxima when q = Q / 2 .