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skad [1K]
3 years ago
6

An Indy 500 race car's velocity increases from 4.0 m/s to +36

Physics
1 answer:
Soloha48 [4]3 years ago
3 0

Answer:

The average acceleration is 8\ m/s^2

Explanation:

<u>Uniform Acceleration </u>

When an object varies its velocity at the same rate, the acceleration is constant.

The relation between the initial and final speeds is:

v_f=v_o+a.t

Where:

vf  = Final speed

vo = Initial speed

a   = Constant acceleration

t   = Elapsed time

The acceleration can be calculated by solving for a:

\displaystyle a=\frac{v_f-v_o}{t}

The Indy 500 race car increases its speed from vo=4 m/s to vf=36 m/s in t=4 s. Thus, the average acceleration is:

\displaystyle a=\frac{36-4}{4}=8\ m/s^2

The average acceleration is 8\ m/s^2

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Which of the following statements are correct regarding electric field lines and equipotential lines? (Select all that apply.)
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A 26-foot ladder is placed against a wall. If the top of the ladder is sliding down the wall at -2 feet per second (note that th
RSB [31]

Answer:

Dx/dt  = 4,8 f/s

Explanation:

The ladder placed against a wall, and the ground formed a right triangle

with x and h the legs and L the hypothenuse

Then

L² = x² + h²          (1)

L = 26 f

Taking differentials on both sides of the equation we get

0  = 2x Dx/dt  + 2h Dh/dt    (1)

In this equation

x = 10   distance between the bottom of the ladder and the when we need to find, the rate of the ladder moving away from the wall

Dx/dt is the rate we are looking for

h = ?    The height of the ladder when  x = 10

As    L²  =  x²  + h²

h²  =  L²  -  x²

h²  =  (26)²  - (10)²

h²  =  676  -  100

h²  = 576

h = 24 f

Then equation (1)

0  = 2x Dx/dt  + 2h Dh/dt

2xDx/dt  = -  2h Dh/dt

10 Dx/dt  = - 24 ( -2 )      ( Note the movement of the ladder is downwards)

Dx/dt  =  48/10

Dx/dt  = 4,8 f/s

6 0
3 years ago
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