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kozerog [31]
1 year ago
6

A 1100 kg racing car accelerates from rest at a constant rate and covers a distance of 50 m in 5 s. what is the car's accelerati

on? (in m/s2)
Physics
1 answer:
allsm [11]1 year ago
8 0

The car's acceleration, given the data from the question is 4 m/s²

<h3>Data obtained from the question</h3>

The following data were obtained from the question:

  • Initial velocity (u) = 13 m/s
  • Distance (s) = 50 m
  • Time (t) = 5 s
  • Acceleration (a) = 9.8

<h3>How to determine the acceleration of the car</h3>

The acceleration of the car can be obtained as illustrated below:

s = ut + ½at²

50 = (0 × 5) + (½ × a × 5²)

50 = 0 + (½ × a × 25)

50 = 0 + 12.5a

50 = 12.5a

Divide both sides by 12.5

a = 50 /12.5

a = 4 m/s²

Thus, the acceleration of the car is 4 m/s²

Learn more about acceleration:

brainly.com/question/491732

#SPJ1

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Explanation:

7 0
3 years ago
Please help! This is due tomorrow and I absolutely need help.
zysi [14]

Answer:

Correct answer:  11. Total distance d = 200m ; 12. Vav = 3.63m/s ;

13. Total displacement Dt = 0m ; 14. V₂(10s-15s) = 0 m/s ;

15. V₃(15s-40s) = 4 m/s ; 16. V₁(0s-10s) = 6 m/s > V₄(40s-55s) = 2.67 m/s

Explanation:

The whole movement can be divided into four stages.

In the first stage the subject moves 60m in a positive direction for 10s,

in the other it is stationary for 5s, in the third it moves 100m in the opposite (negative) direction for 25s and in the fourth in the positive 40m for 15s.

11. Total distance = 60 + 0 + 100 + 40 = 200m

12. The formula for calculating the average speed (velocity) is

Vav = (S₁ + S₂ + S₃ + S₄) / (t₁ + t₂ + t₃ + t₄)

Vav = (60 + 0 + 100 + 40)/ (10 + 5 + 25 + 15) = 200/55 = 3.63 m/s

13. The movement started from the origin and ended at the origin

Total displacement is zero meters.

14. The speed between 10s and 15s is zero, because he did not move.

15. V₃ = S₃/t₃ = 100/25 = 4 m/s

16. V₁ = S₁/t₁ = 60/10 = 6 m/s   and V₄ = S₄/t₄ = 40/15 = 2.67 m/s

V₁ > V₄

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3 years ago
All organisms need energy to function. Which cell organelle provides this energy
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3 years ago
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A child on a tricycle is moving at a speed of 1.40 m/s at the start of a 2.25 m high and 12.4 m long incline. The total mass is
goblinko [34]

Answer:

The work done by the child as the tricycle travels down the incline is 416.96 J

Explanation:

Given;

initial velocity of the child, v_i = 1.4 m/s

final velocity of the child, v_f = 6.5 m/s

initial height of the inclined plane, h = 2.25 m

length of the inclined plane, L = 12.4 m

total mass, m = 48 kg

frictional force, f_k = 41 N

The work done by the child is calculated as;

\Delta E_{mech} = W - f_{k} \Delta L\\\\W = \Delta E_{mech}  + f_{k} \Delta L\\\\W = (K.E_f - K.E_i) + (P.E_f - P.E_i) + f_{k} \Delta L\\\\W = \frac{1}{2} m(v_f^2 - v_i^2) + mg(h_f - h_i) + f_{k} \Delta L\\\\W = \frac{1}{2} \times 48(6.5^2 - 1.4^2) + 48\times 9.8(0-2.25) + (41\times 12.4)\\\\W = 966.96  \ - \ 1058.4 \ + \ 508.4\\\\W = 416.96 \ J

Therefore, the work done by the child as the tricycle travels down the incline is 416.96 J

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