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Alex17521 [72]
2 years ago
15

If the force acting on a body of mass 40 kg is doubled. By how much will the acceleration change.

SAT
1 answer:
mafiozo [28]2 years ago
3 0
F = ma

Since F doubles and m is constant at 40kg, a doubles
Thus acceleration doubles
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A child riding a bicycle at 15 m/s accelerates at 3. 0 m/s/s for 4. 0 s. What is the child's speed at the end of this 4. 0 s int
mamaluj [8]

Answer:

The final speed of bicycle would be <u>27 m/s</u>².

Step-by-step explanation:

<u>Given</u> :

  • \small\red\bull Initial velocity (u) = 15 m/s
  • \small\red\bull Acceleration (a) = 3.0 m/s²
  • \small\red\bull Time = 4 seconds

<u>To</u><u> </u><u>Find</u> :

  • \small\red\bull Final velocity (v)

<u>Using</u><u> </u><u>Formula</u> :

\longrightarrow{\pmb{\sf{v = u + at}}}

  • \small\blue\star v = final velocity
  • \small\blue\star u = initial velocity
  • \small\blue\star a = acceleration
  • \small\blue\star t = time taken

<u>Solution</u> :

Substituting the values in the formula to find the final velocity :

\longrightarrow{\sf{v = u + at}}

\longrightarrow{\sf{v = 15 + (3.0 \times 4)}}

\longrightarrow{\sf{v = 15 + (3 \times 4)}}

\longrightarrow{\sf{v = 15 +(12)}}

\longrightarrow{\sf{v = 15 +12}}

\longrightarrow{\sf{v = 27 \:  {m/s}^{2}}}

\star{\underline{\boxed{\red{\sf{v = 27 \:  {m/s}^{2}}}}}}

Hence, the final speed of bicycle would be 27 m/s².

\rule{300}{1.5}

3 0
2 years ago
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