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:
A. At warm tempetures, molecules move around more.
Explanation:
I'm at k12 and I just took the test got it right. Physical Science: Unit 2 Test
Answer:
0.231 m/s
Explanation:
m = mass attached to the spring = 0.405 kg
k = spring constant of spring = 26.3 N/m
x₀ = initial position = 3.31 cm = 0.0331 m
x = final position = (0.5) x₀ = (0.5) (0.0331) = 0.01655 m
v₀ = initial speed = 0 m/s
v = final speed = ?
Using conservation of energy
Initial kinetic energy + initial spring energy = Final kinetic energy + final spring energy
(0.5) m v₀² + (0.5) k x₀² = (0.5) m v² + (0.5) k x²
m v₀² + k x₀² = m v² + k x²
(0.405) (0)² + (26.3) (0.0331)² = (0.405) v² + (26.3) (0.01655)²
v = 0.231 m/s
Answer:
a) 42 m/s, positive direction (to the east), b) 42 m/s, negative direction (to the west).
Explanation:
a) Let consider that Car A is moving at positive direction. Then, the relative velocity of Car A as seen by the driver of Car B is:
42 m/s, positive direction (to the east).
b) The relative velocity of Car B as seen by the drive of Car A is:
42 m/s, negative direction (to the west).
