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konstantin123 [22]
2 years ago
13

Explain why the rocket can move at constant speed in outer space.

Physics
2 answers:
iren [92.7K]2 years ago
4 0

Answer:

When the rocket's engines are fired up, the force of gravity is disturbed, and the rocket takes off in the air. Later, as the rocket's fuel runs out, it slows down, comes to a halt, and eventually plummets to Earth's surface. Forces can affect objects in space as well. Spaceships are constantly in motion while traversing the solar system.

Explanation:

Please read!

disa [49]2 years ago
3 0

Answer:

Explanation:

The rocket works because of the law of conservation of linear momentum. The law of conservation of linear momentum is very important in physics. Momentum is defined as the mass of an object times its velocity.

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The total resistance of a parallel circuit is 25 ohms. If the total current is 100mA, how much current is through a 220 ohm resi
gulaghasi [49]

Answer:

The current across the resistance is 0.011 A.

Explanation:

Total resistance, R = 25 ohms

Total current, I = 100 mA = 0.1 A

Let the voltage is V.

By the Ohm's law

V = I R

V = 0.1 x 25 = 2.5 V

Now the resistance is R' = 220 ohm

As they are in parallel so the voltage is same. Let the current is I'.

V = I' x R'

2.5 = I' x 220

I' = 0.011 A

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3 years ago
How would you compare and contrast the 3 types of friction?
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sliding friction.
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4 years ago
Read 2 more answers
An aircraft with a mass of 10,000 kg starts from rest at sea level and takes off, then flies to a cruising speed of 620 km/h and
Natasha_Volkova [10]

Answer:

The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

Explanation:

Given that,

Mass of aircraft = 10000 kg

Speed = 620 km/h = 172.22 m/s

Altitude = 10 km = 1000 m

We calculate the change in potential energy

\Delta P.E=mg(h_{2}-h_{1})

\Delta P.E=10000\times9.8\times(10000-0)

\Delta P.E=10000\times9.8\times10000

\Delta P.E=980000000\ J

\Delta P.E=980\ MJ

For g = 10 m/s²,

The change in potential energy will be 1000 MJ.

We calculate the change in kinetic energy

\Delta K.E=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)

\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2-0^2)

\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2)

\Delta K.E=148298642\ J

\Delta K.E=148.3\ MJ

For g = 10 m/s²,

The change in kinetic energy will be 150 MJ.

Hence, The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

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3 years ago
Momentum and different surfaces
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