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mihalych1998 [28]
3 years ago
7

1. A ball rolling down a ramp accelerales from 0 cm/s to

Physics
1 answer:
Tom [10]3 years ago
3 0

Answer:

4 cm/s².

Explanation:

From the question given above, the following data were obtained from it:

Initial velocity (u) = 0 cm/s

Final velocity (v) = 40 cm/s

Time (t) = 10 s

Acceleration (a) =?

Acceleration is simply defined as the rate of change of velocity with time. Mathematically, it is expressed as follow:

Acceleration (a) = Change in velocity /time

Acceleration (a) = [final velocity (v) – initial velocity (u)] /time (t)

a = (v – u) /t

With the above formula, we can obtain the acceleration of the ball as follow:

Initial velocity (u) = 0 cm/s

Final velocity (v) = 40 cm/s

Time (t) = 10 s

Acceleration (a) =?

a = (v – u) /t

a = (40 – 0) / 10

a = 40/10

a = 4 cm/s²

Therefore, the acceleration of the ball is 4 cm/s²

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A car of mass 2400kg accelerates from 10m/s to 30m/s if the force is 12000N what is the distance​
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Explanation:

  • Initial velocity (u) = 10 m/s
  • Final velocity (v) = 30 m/s
  • Mass (m) = 2400 kg
  • Force (F) = 12000 N

Let us find the time taken first.

→ F = ma

  • Acceleration (a) = (v – u)/t

→ 12000 = 2400 × (30 – 10)/t

→ 12000 ÷ 2400 = (20)/t

→ 5 = 20/t

→ 5t = 20

→ t = 20 ÷ 5

→ <u>t</u><u> </u><u>=</u><u> </u><u>4</u><u> </u><u>seconds</u>

Now, find the acceleration.

→ a = (v – u)/t

→ a = (30 – 10)/4

→ a = 20/4

→ <u>a</u><u> </u><u>=</u><u> </u><u>5</u><u> </u><u>m</u><u>/</u><u>s²</u>

Now, by using the third equation of motion,

→ v² – u² = 2as

  • s = distance

→ (30)² – (10)² = 2 × 5 × s

→ 900 – 100 = 10s

→ 800 = 10s

→ 800 ÷ 10 = s

→ <u>8</u><u>0</u><u> </u><u>m</u><u> </u><u>=</u><u> </u><u>s</u>

Therefore, distance travelled is 80 m.

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3 years ago
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