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MAXImum [283]
2 years ago
11

G A dragster starts from rest and accelerates at 35 m/s2 m / s 2 . How fast is it going after t t

Physics
1 answer:
bija089 [108]2 years ago
4 0

Complete question:

A dragster starts from rest and accelerates at 35 m/s2 m / s 2 . How fast is it going after 7s.

Answer:

The final velocity of the dragster is 245 m/s.

Explanation:

Given;

acceleration of the dragster, a = 35 m/s²

initial velocity of the dragster, u = 0

time of motion, t = 7 s

the final velocity of the dragster after 7s is given by;

v = u + at

where;

v is the final velocity

u is the initial velocity

v = 0 + (35 x 7)

v = 245 m/s

Therefore, the final velocity of the dragster is 245 m/s.

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A geosynchronous satellite moves in a circular orbit around the Earth and completes one circle in the same time T during which t
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7 0
3 years ago
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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
zhannawk [14.2K]

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

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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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