Wow ! This is not simple. At first, it looks like there's not enough information, because we don't know the mass of the cars. But I"m pretty sure it turns out that we don't need to know it.
At the top of the first hill, the car's potential energy is
PE = (mass) x (gravity) x (height) .
At the bottom, the car's kinetic energy is
KE = (1/2) (mass) (speed²) .
You said that the car's speed is 70 m/s at the bottom of the hill,
and you also said that 10% of the energy will be lost on the way
down. So now, here comes the big jump. Put a comment under
my answer if you don't see where I got this equation:
KE = 0.9 PE
(1/2) (mass) (70 m/s)² = (0.9) (mass) (gravity) (height)
Divide each side by (mass):
(0.5) (4900 m²/s²) = (0.9) (9.8 m/s²) (height)
(There goes the mass. As long as the whole thing is 90% efficient,
the solution will be the same for any number of cars, loaded with
any number of passengers.)
Divide each side by (0.9):
(0.5/0.9) (4900 m²/s²) = (9.8 m/s²) (height)
Divide each side by (9.8 m/s²):
Height = (5/9)(4900 m²/s²) / (9.8 m/s²)
= (5 x 4900 m²/s²) / (9 x 9.8 m/s²)
= (24,500 / 88.2) (m²/s²) / (m/s²)
= 277-7/9 meters
(about 911 feet)
By Newton's 2nd law of motion, F = ma, where F is force, m is mass, and a is acceleration.
Rearranging this equation to find acceleration would give us:
a = F/m
The horizontal force to the right is 10N, because the box is pushed to the right with a force of 20N, and the friction force of 10N opposes that, so:
20N - 10N = 10N
The mass is 2kg.
Putting these values into the equation gives us:
a = F/m
= 10/2
= 5ms^-2
The acceleration of the box is 5ms^-2
Answer:
F' = (3/2)F
Explanation:
the formula for the electric field strength is given as follows:
E = F/q
where,
E = Electric Field Strength
F = Force due to the electric field
q = magnitude of charge experiencing the force
Therefore,
F = E q ---------------- equation (1)
Now, if we half the electric field strength and make the magnitude of charge triple its initial value. Then the force will become:
F' = (E/2)(3 q)
F' = (3/2)(E q)
using equation (1)
<u>F' = (3/2)F</u>
