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:
Solution:
- Finding large moons comparable in size to their planets result from impacts of two astro-bodies. The probability of such an event occurring is very rare.
- Even at the best luck, one moon can be made from the result of giant impact. While the probability of 6 planets having moons of comparable sizes is close to impossible.
Answer:
in the air the freeze when it’s freezing outside
Explanation:
I hope this helps
Answer:
λ₁ = 2.50 10⁻² m, λ₂ = 1.66 10⁻² m
Explanation:
Microwave communication is very efficient because it does not have atmospheric interference, for which it is widely used and has been regulated to avoid interference, the ku band is in the range between 12 and 18 GHz.
Let's calculate the wavelength for the two extreme frequencies of this band
wavelength and frequency are related
c = λ f
λ = c / f
f₁ = 12 GHz = 12 10⁹ Hz
λ₁ = 3 10⁸ /12 10⁹
λ₁ = 2.50 10⁻² m
f₂ = 18 GHz = 18 10⁹ Hz
λ₂ = 3 10⁸ /18 10⁹
λ₂ = 1.66 10⁻² m
Unfortunately in your exercise the specific frequency is not fired, for significant figures they must be the same number as the figures of the frequency, in general the frequency has 3 or 4 significant figures
