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scZoUnD [109]
3 years ago
6

A rocket achieves a lift-off velocity of 500.0 m/s from rest in 30.0 seconds. Calculate the average acceleration of the rocket.

Physics
1 answer:
Anettt [7]3 years ago
6 0

Answer:

The average acceleration is 16.6 m/s² ⇒ 1st answer

Explanation:

A rocket achieves a lift-off velocity of 500.0 m/s from rest in

30.0 seconds

The given is:

→ The initial velocity = 0

→ The final velocity = 500 meters per seconds

→ The time is 30 seconds

Acceleration is the rate of change of velocity of the rocket

→ a=\frac{v-u}{t}

where a is the acceleration, v is the final velocity, u is the initial velocity

and t is the time

→ u = 0 , v = 500 m/s , t = 30 s

Substitute these values in the rule

→ a=\frac{500-0}{30}=\frac{500}{30}=16.6 m/s²

<em>The average acceleration is 16.6 m/s²</em>

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<h3>What is Kinetic energy?</h3>

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3 years ago
At the same instant that a 0.50-kg ball is dropped from 25 m above Earth, a second ball, with a mass of 0.25 kg, is thrown strai
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Answer:

The center of mass of the two-ball system is 7.05 m above ground.

Explanation:

<u>Motion of 0.50 kg ball:</u>

Initial speed, u = 0 m/s

Time = 2 s

Acceleration = 9.81 m/s²

Initial height = 25 m

Substituting in equation s = ut + 0.5 at²

                 s = 0 x 2 + 0.5 x 9.81 x 2² = 19.62 m

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<u>Motion of 0.25 kg ball:</u>

Initial speed, u = 15 m/s

Time = 2 s

Acceleration = -9.81 m/s²

Substituting in equation s = ut + 0.5 at²

                 s = 15 x 2 - 0.5 x 9.81 x 2² = 10.38 m

Height above ground = 10.38 m

We have equation for center of gravity

          \bar{x}=\frac{m_1x_1+m_2x_2}{m_1+m_2}

          m₁ = 0.50 kg

          x₁ = 5.38 m        

          m₂ = 0.25 kg

          x₂ = 10.38 m    

Substituting

         \bar{x}=\frac{0.50\times 5.38+0.25\times 10.38}{0.50+0.25}=7.05m

The center of mass of the two-ball system is 7.05 m above ground.  

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