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A crate with mass 32.5 kg initially at rest on a warehouse floor is acted on by a net horizontal force of 140 N.
a)What acceleration is produced?
b)How far does the crate travel in 10.0 s?
C)What is its speed at the end of
10.0 s?

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meerkat18
Newton's second law of motion may be mathematically expressed as Fnet = m x a.

a. a = F / m. Substituting the known values,
                              a = 140 N / 32.5 kg = 4.3 m/s²
b. d is calculated through the equation, d = Vit + 0.5at². Given that Vi is zero, the distance is,
                                    d = 0.5(4.3 m/s²)(10 s)² = 215 m
c. Velocity is acceleration times time,
                                  v = a x t = (4.3 m/s²)(10 s) = 43 m/s
Eduard22sly

A. The acceleration produced is 4.31 m/s²

B. The distance travelled in 10 s is 215.5 m

C. The speed at the end of 10 s is 43.1 m/s

A. How to determine the acceleration

  • Force (F) = 140 N
  • Mass (m) = 32.5 kg
  • Acceleration (a) =?

F = ma

140 = 32.5 × a

Divide both side by 32.5

a = 140 / 32.5

a = 4.31 m/s²

B. How to determine the distance

  • Initial velocity (u) = 0 m/s
  • Time (t) = 10 s
  • Acceleration (a) = 4.31 m/s²
  • Distance (s) =?

s = ut + ½at²

s = (0 × 10) + ½ × 4.31 × 10²

s = 215.5 m

C. How to determine final velocity

  • Initial velocity (u) = 0 m/s
  • Time (t) = 10 s
  • Acceleration (a) = 4.31 m/s²
  • Final velocity (v) =?

v = u + at

v = 0 + (4.31 × 10)

v = 43.1 m/s

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