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)
Answer:

Explanation:
consider the mass of each train car be m
m₁ = m₂ = m₃ = m
speed of the three identical train
u₁ = u₂ = u₃ = 1.8 m/s
m₄ = m u₄ = 4.5 m/s
m₅ = m u₅ = 0 (initial velocity )
final velocity
v₁ = v₂ = v₃ = v₄ = v₅ = v
using conservation of momentum
m₁u₁ + m₂u₂ + m₃u₃ + m₄u₄ + m₅u₅ = m₁v₁ + m₂v₂ + m₃v₃ + m₄v₄ + m₅v₅
m (1.8 + 1.8 + 1.8 +4.5) = 5 m v


Answer:
The height of the building is 8,302.5 m
Explanation:
Given;
velocity of the projectile, u = 36 m/s
time of motion, t = 45 s
Let the upward direction of the bullet be negative,
The height of the building is calculated as;

The energy conservation and trigonometry we can find the results for the questions about the movement of the acrobat are;
a) The maximum speed is v = 4.89 m / s
b) The maximum height is h = 1.22 m
The energy conservation is one of the most fundamental principles of physics, stable that if there are no friction forces the mechanistic energy remains constant. Mechanical energy is the sum of the kinetic energy plus the potential energies.
Em = K + U
Let's write the energy in two points.
Starting point. Highest part of the oscillation
Em₀ = U = m g h
Final point. Lower part of the movement
= K = ½ m v²
Energy is conserved.
Emo =
m g h = ½ m v²
v² = 2 gh
Let's use trigonometry to find the height, see attached.
h = L - L cos θ
h = L (1- cos θ)
They indicate that the initial angle is tea = 48º and the length is L = 3.7 m, let's calculate.
h = 3.7 (1- cos 48)
h = 1.22 m
this is the maximum height of the movement.
Let's calculate the velocity.
v = 4.89 m / s
In conclusion using the conservation of energy and trigonometry we can find the results for the questions about the movement of the acrobat are;
a) The maximum speed is v = 4.89 m / s
b) The maximum height is h = 1.22 m
Learn more here: brainly.com/question/13010190