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podryga [215]
3 years ago
15

It takes a car one minute to go from rest to 30m/s east. Whqt is the acceleration of this car?

Physics
1 answer:
zavuch27 [327]3 years ago
5 0

Answer:

0.5 m/s^{2}

Explanation:

a = \frac{v-u}{t} \\a = \frac{30 m/s}{60 s} \\a = 0.5 m/s^2

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All stars go through a lifecycle.
Komok [63]

Mira is much bigger than the Sun.

Only very massive stars will go through a supernova stage, causing the outer layer to explode away and the core to collapse in on itself, becoming very dense.

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2 years ago
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The data table shows how the amplitude of a mechanical wave varies with the energy it carries. Analyze the data to identify the
Maslowich

The energy of a wave is directly proportional to the square of the amplitude of the wave.

<h3>What is the relationship between energy and amplitude?</h3>

There is direct relationship between energy of the wave and the amplitude of the wave. The energy transported by a wave is directly proportional to the square of the amplitude of the wave. This means if energy is increase the amplitude of wave becomes double and vice versa.

Energy = (amplitude)2

So we can conclude that the energy of a wave is directly proportional to the square of the amplitude of the wave.

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Andru [333]
It would be helpful if you gave me a bit more information on what the cars speed is
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3 years ago
O'Malley is riding on a bus which is moving at 10 m/s, and he throws a ball which he observes to be moving at 10 m/s relative to
Vikki [24]

Answer:

<em>20 m/s in the same direction of the bus.</em>

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<u>Relative Motion </u>

Objects movement is always related to some reference. If you are moving at a constant speed, all the objects moving with you seem to be at rest from your reference, but they are moving at the same speed as you by an external observer.

If we are riding on a bus at 10 m/s and throw a ball which we see moving at 10 m/s in our same direction, then an external observer (called Ophelia) will see the ball moving at our speed plus the relative speed with respect to us, that is, at 20 m/s in the same direction of the bus.

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Kon'nichiwa~<br>please help me with this question!!​
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Let's check the relationship

\\ \rm\hookrightarrow g=\dfrac{GM}{r^2}

\\ \rm\hookrightarrow g\propto G

So

  • Raindrops will fall faster . .
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Option C is wrong by now .Let's check D once

\\ \rm\hookrightarrow T\propto \dfrac{1}{\sqrt{g}}

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