Answer:
Final speed of the car, v = 24.49 m/s
Explanation:
It is given that,
Initial velocity of the car, u = 0
Acceleration, 
Time taken, t = 7.9 s
We need to find the final velocity of the car. Let it is given by v. It can be calculated using first equation of motion as :
v = 24.49 m/s
So, the final speed of the car is 24.49 m/s. Hence, this is the required solution.
Answer:
To write a number in scientific notation. First write a decimal point in the numbers so that there's only one digit to the left of the decimal point.
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)
Linear momentum has to be conserved. It was zero before the thread eas burned ... when nothing was moving ... so the momentum of the masses moving in opposite directions has to add up to zero. ... Momentum = mass times speed. ... In one direction, you have 5 kg times 1/5 m/s= 1 kg-m/s. ... We need 1 kg-m/s in the other direction. ... 7 kg times speed = 1 kg-m/s. ... Can you finish it from here ?
