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inessss [21]
3 years ago
14

How can acceleration be involved if velocity is constant?

Physics
1 answer:
Olin [163]3 years ago
5 0
If a body is in Linear Motion, and if velocity is constant, acceleration is zero,because acceleration is rate of change of velocity. We would need a change in velocity to calculate acceleration. Since it is CONSTANT, no change in velocity, no acceleration. Therefore ACCELERATION is said to be ZERO.

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विद्यार्थी जीवनमा अनुशासनको महत्व essay​
loris [4]

विद्यार्थी जीवनमा अनुशासनको विशेष महत्व हुन्छ। विद्यार्थी जीवन मा, विद्यार्थीहरु अनुशासित भएर आफ्नो जीवन मा सफलता प्राप्त गर्न सक्छन्। राम्रो शिक्षा हरेक विद्यार्थी नियमहरु को पालन गर्नु पर्छ

विद्यार्थी जीवन मा, प्रत्येक विद्यार्थी स्कूल को नियम अनुसार हिड्नु पर्छ। उसले आफ्ना शिक्षकहरुको आदेश पालन गर्नु पर्छ। उहाँको आदेश पालना गरेर, एक चरित्र, एक आदर्श र एक योग्य नागरिक बन्नुहोस्।

तसर्थ, प्रत्येक विद्यार्थी अनुशासन को पालन गर्नु पर्छ आफ्नो जीवन खुशी बनाउन को लागी। यसको साथसाथै, प्रत्येक विद्यार्थीले समय बर्बाद नगरी आफ्नो लक्ष्य हासिल गर्न कडा मेहनत गर्नुपर्छ।

5 0
3 years ago
Read 2 more answers
A football player with a mass of 88 kg and a speed of 2.0 m/s collides head-on with a player from the opposing team whose mass i
Ket [755]

Answer:

Speed of another player, v₂ = 1.47 m/s

Explanation:

It is given that,

Mass of football player, m₁ = 88 kg

Speed of player, v₁ = 2 m/s

Mass of player of opposing team, m₂ = 120 kg

The players stick together and are at rest after the collision. It shows an example of inelastic collision. Using the conservation of linear momentum as :

m_1v_1+m_2v_2=(m_1+m_2)V

V is the final velocity after collision. Here, V = 0 as both players comes to rest after collision.

v_2=-\dfrac{m_1v_1}{m_2}

v_2=-\dfrac{88\ kg\times 2\ m/s}{120\ kg}

v_2=-1.47\ m/s

So, the speed of another player is 1.47 m/s. Hence, this is the required solution.

7 0
4 years ago
Select the correct answer.
Sergeeva-Olga [200]

Answer:

All of the above.

Explanation:

All of these steps are useful towards getting a job.

8 0
3 years ago
How much work does the charge escalator do to move 2.30 μC of charge from the negative terminal to the positive terminal of a 3.
loris [4]

Answer:

6.9\times 10^{-6} J

Explanation:

We are given that

Charge=q=2.3\mu C=2.3\times 10^{-6} C

1\mu C=10^{-6} C

Potential difference=V=3 V

We know that

Work done=V\times q

Using the formula

Work done by charge to move from the negative terminal to the positive terminal of battery=2.3\times 10^{-6}\times 3 J

Work done by charge to move from the negative terminal to the positive terminal of battery=6.9\times 10^{-6} J

6 0
3 years ago
List any similarities between discovered exoplanets and planets in our solar system.
gregori [183]

Answer:

1. Revolve around a point

2. Formed from dust and gas particles

3. Exoplanets and associated star orbit a common center of mass  

4. Composed of gases found in Jupiter and Saturn

3 0
3 years ago
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