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creativ13 [48]
2 years ago
5

If a car is moving at a constant velocity of 10 m/s, what is its acceleration?

Physics
1 answer:
marin [14]2 years ago
4 0

Answer:

If a car is moving at a <em><u>constant velocity of 10 m/s,</u></em> there will be no change in the velocity per time.

Acceleration is the rate of change of velocity

Acceleration=(Final velocity-Initial velocity)/Time

Acceleration=(10m/s-10m/s)/Time= <em><u>0 m/s²</u></em>

<h3>★ <u>0 m/s²</u> is the right answer. </h3>
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Answer:

Explanation:

Blue: 10/30

Red: 5/30

Yellow: 15/30

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2 years ago
Determine the change in velocity of a car that starts at rest and has a final velocity of 20 miles per second North
Katen [24]

Answer: The change in velocity is 20mph

Explanation: The change in velocity is the difference between the final velocity and the initial velocity.

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dV=20-0

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The potential energy stored in the compressed spring of a dart gun, with a spring constant of 32.50 N/m, is 0.640 J. Find by how
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Answer:

A

   x = 0.198456 \ m

B

    h  =  1.3061 \  m

C

 v =  5.06 \  m/s

D

  d = 4.0273 \  m

Explanation:

Considering the first question

From the question we are told that

   The spring constant is  k  =  32.50 N/m

    The potential energy is  PE  =  0.640 \ J

Generally the potential  energy stored in spring  is mathematically represented as   PE  =  \frac{1}{2}  *  k  *  x^2

=>    0.640=  \frac{1}{2}  * 32.50  *  x^2  

=>    x = \sqrt{0.03938}  

=>    x = 0.198456 \ m  

Considering the second question

 From the question we are told that

   The mass of the dart is  m =  0.050 kg

Generally from the law of energy conservation

         PE =  mgh

=>       0.640   =  0.050 *  9.8  *  h

=>      h  =  1.3061 \  m

Considering the third  question

   The height at which the dart was fired horizontally is  H  =   3.90\  m

Generally  from the law of energy conservation

         PE = KE

Here  KE is kinetic energy of the dart which is mathematical represented as

     KE  =  \frac{1}{2}  *  mv^2

=>      0.640 =  \frac{1}{2}  * 0.050 *  v^2

=>       v^2 = 25.6

=>       v =  5.06 \  m/s

Considering the fourth question

Generally the total time of flight of the dart is mathematically represented as

       t  =  \frac{ 2 *  H }{g}

=>     t  =  \frac{ 2 * 3.90 }{9.8 }

=>     t  =  0.7959 \ s

Generally the  horizontal distance from the equilibrium position to the ground is  mathematically represented as

       d =  v  *   t

=>     d = 5.06  *   0.7959

=>     d = 4.0273 \  m

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