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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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Explanation:

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Insulators have very high resistance and are used to protect us from the flow of electricity.

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Like-charged bodies, when brought closer together, will: repel more strongly be neutralized repel less strongly attract more str
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Answer:

That's almost the true

Explanation:

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3 years ago
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A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The
Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}

Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

\sum {F = ma}

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg

Here, {\mu _{\rm{s}}} is the coefficient of static friction, mm is mass of the object, and g is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass {m_1}

Substitute  for in the expression of maximum static friction {F_{\rm{f}}} = {\mu _{\rm{s}}}mg

{F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute  for , [/tex]{m_1}[/tex] for in the equation .

{F_{\rm{f}}} = {m_1}a

Substitute {\mu _{\rm{s}}}{m_1}g for {F_{\rm{f}}} in the equation {F_{\rm{f}}} = {m_1}a

{\mu _{\rm{s}}}{m_1}g = {m_1}a

Rearrange for a.

a = {\mu _{\rm{s}}}g

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

{F_{\min }} = \left( {{m_1} + {m_2}} \right)a

Substitute for in the above equation .

{F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

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Two balls are thrown against a wall. Ball 1 has a much higher speed than ball 2.
Sunny_sXe [5.5K]

Let both the balls have the same mass equals to m.

Let v_1 and v_2 be the speed of the ball1 and the ball2 respectively, such that

v_1>v_2\;\cdots(i)

Assuming that both the balls are at the same level with respect to the ground, so let h be the height from the ground.

The total energy of ball1= Kinetic energy of ball1 + Potential energy of ball1. The Kinetic energy of any object moving with speed, v, is \frac 12 m v^2

and the potential energy is due to the change in height is mgh [where g is the acceleration due to gravity]

So, the total energy of ball1,

=\frac 12 m v_1^2 + mgh\;\cdots(ii)

and the total energy of ball1,

=\frac 12 m v_2^2 + mgh\;\cdots(iii).

Here, the potential energy for both the balls are the same, but the kinetic energy of the ball1 is higher the ball2 as the ball1 have the higher speed, refer equation (i)

So, \frac 12 m v_1^2 >\frac 12 m v_2^2

Now, from equations (ii) and (iii)

The total energy of ball1 hi higher than the total energy of ball2.

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Answer:

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Explanation:

Let the distance moved by Joe is "d"

so the time taken by him to drove it by speed 45 mph is given as

t_1 = \frac{d}{v_1}

t_1 = \frac{d}{45}

now the same distance is traveled by him with speed 55 mph

so the time taken by him

t_2 = \frac{d}{55}

so total time taken by him for complete distance 2d

t = t_1 + t_2

t = \frac{d}{45} + \frac{d}{55}

t = 0.0404 d

now the average speed is given as

v_{avg} = \frac{2d}{t}

v_{avg} = \frac{2d}{0.0404d}

v_{avg} = 49.5 mph

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3 years ago
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