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avanturin [10]
3 years ago
13

A car slows down from 27.7 m/s to 10.9 m/s in 2.37 s. What is its acceleration?

Physics
2 answers:
frosja888 [35]3 years ago
7 0

Answer:

\boxed {\boxed {\sf  -7.09 \ m/s^2}}

Explanation:

We are asked to find the acceleration of a car. Acceleration is the change in velocity with respect to time. We will use the following formula:

a= \frac {v_f-v_i}{t}

In this formula, v_f is the final velocity, v_i is the initial velocity, and t is the time. The car slows down from 27.7 meters per second to 10.9 meters per second in 2.37 seconds. Therefore:

  • v_f= 10.9 m/s
  • v_i= 27.7 m/s
  • t= 2.37 s

Substitute the values into the formula.

a= \frac{ 10.9  m/s-27.7 m/s}{2.37  \ s}

Solve the numerator.

  • 10.9 m/s - 27.7 m/s= -16.8 m/s

a= \frac{-16.8 \ m/s}{2.37 \ s}

Divide.

a= -7.088607595 \ m/s/s

a= -7.088607595 \ m/s^2

The original measurements of velocity and time have 3 significant figures, so our answer must have the same. For the number we found, that is the hundredth place. The 8 in the thousandth place tells us to round the 8 in the hundredth place up to a 9.

a \approx - 7.09 \ m/s^2

The acceleration of the car is approximately <u>-7.09 meters per second squared. </u>The acceleration is negative because the car is slowing down.

Sunny_sXe [5.5K]3 years ago
4 0

Answer:

- 7.088 m/s²

Explanation:

As we know that,

★ Acceleration = Change in velocity/Time

→ a = (v - u)/t

Here,

  • Initial velocity (u) = 27.7 m/s
  • Final velocity (v) = 10.9 m/s

→ a = (10.9 m/s - 27.7 m/s)/2.37 s

→ a = -16.8/2.37 m/s²

→ <u>a</u><u> </u><u>=</u><u> </u><u>-</u><u>7</u><u>.</u><u>0</u><u>8</u><u>8</u><u> </u><u>m/s²</u> [Answer]

Negative sign denotes that the velocity is decreasing.

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Two trains on separate tracks move toward each other. Train 1 has a speed of 145 km/h; train 2, a speed of 72.0 km/h. Train 2 bl
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Answer:

Therefore,

The frequency heard by the engineer on train 1

f_{o}=603\ Hz

Explanation:

Given:

Two trains on separate tracks move toward each other

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v_{o}=145\ km/h=145\times \dfrac{1000}{3600}=40.28\ m/s

For Train 2 Velocity of the Source,

v_{s}=90\ km/h=90\times \dfrac{1000}{3600}=25\ m/s

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To Find:

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f_{o}=?  (frequency heard by the engineer on train 1)

Solution:

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The Equation is

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Where,

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Substituting the values we get

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Therefore,

The frequency heard by the engineer on train 1

f_{o}=603\ Hz

7 0
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