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3 years ago

if the moon were somehow knocked out of orbit and flew off into outer space, what are the possible effects on the earth?

Physics

1 answer:

Anarel [89]3 years ago

5 0

<h2>Answers:</h2>

<u>The possible effects on the earth will be</u><u> 1) No light during night 2) No tides</u>

<h2>Explanation:</h2>

As we know that tides are caused by the differential gravitational force on the equatorial regions by the moon. So if the Moon escapes from Earth, the main effect would be the lack of tides on Earth The region closer to the Moon experiences more gravitational force than the region on Earth away from the Moon. Other effects will be no more Moon to light up the night sky and no more solar eclipses.

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3 years ago

Suppose you drop a 10-pound weight and a 5-pound weight on the Moon, both from the same height at the same time. What will happe

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2 years ago

In a science fiction novel two enemies, Bonzo and Ender, are fighting in outer spce. From stationary positions, they push agains

lara [203]

Answer:

ratio of the masses of the two enemies = 1.7

Explanation:

Applying the law of conservation of momentum,

Momentum fo Bonzo = Momentum of Ender

mv = m'v'................. Equation 1

Where m = mass of Bonzo, v = velocity of Bonzo, m' = mass of Ender, v' = velocity of Ender

m/m' = v'/v............... Equation 2

Where m/m' = ratio of the masses of the two enemies

Given: v = 2.0 m/s, v' = 3.4 m/s

Substitute into equation 2

m/m' = 3.4/2

m/m' = 1.7

Hnece, ratio of the masses of the two enemies = 1.7

7 0

3 years ago

Calculate the ratio of the drag force on a passenger jet flying with a speed of 1200 km/h at an altitude of 10 km to the drag fo

Sonbull [250]

Answer:

$2.267$

Explanation:

Drag force is given by

$F=\dfrac{1}{2}\rho Av^2C$

C = Drag coefficient is constant

A = Area is constant

$v_1$ = Velocity of the passenger jet = 1200 km/h = $\dfrac{1200}{3.6}\ \text{m/s}$

$v_2$ = Velocity of the prop plane = $\dfrac{1}{4}v_1$

$\rho_1$ = Density of the air where the jet was flying = $0.38\ \text{kg/m}^3$

$\rho_2$ = Density of the air where the prop plane was flying = $0.67\ \text{kg/m}^3$

$F\propto \rho v^2$

$\dfrac{F_1}{F_2}=\dfrac{\rho_1 v_1^2}{\rho_2 v_2^2}\\\Rightarrow \dfrac{F_1}{F_2}=\dfrac{0.38 v_1^2}{0.67 (\dfrac{1}{4}v_1^2)}\\\Rightarrow \dfrac{F_1}{F_2}=2.267$

The ratio of the drag forces is $2.267$ .

5 0

3 years ago