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Alona [7]
3 years ago
13

If the brakes are applied and the speed of the car is reduced to 12 m/s in 15 s , determine the constant deceleration of the car

.
Physics
1 answer:
zavuch27 [327]3 years ago
5 0

Given:

\triangle v = 12 m/s

Time = 15 s

To find:

Deceleration = ?

Formula used:

Deceleration is given by,

a = \frac{\triangle v}{t}

Solution:

Acceleration of the body is rate of increase of velocity. Deceleration of the body is negative acceleration i. e rate of decrease of velocity.

Thus, deceleration of the body is given by,

a = \frac{\triangle v}{t}

where a is the deceleration of the body

v is the velocity

\frac{\triangle v}{t} is rate of change of velocity

t is the time

a = \frac{12}{15}

a = 0.8 m/ s^{2}

Thus, deceleration is 0.8 m/ s^{2}.

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If the frequency of a wave is 400hz and it’s wave length is2.5 what is the velocity ?
gizmo_the_mogwai [7]
To begin with, we can use the formula that links frequency, wavelength and velocity.

Because you already have the wavelength and the frequency, you just need to solve for velocity. You can do this by multiplying each side of the equation by frequency.

Therefore, 400 x 2.5 = 1000m/s.

Hope this helps :)

8 0
2 years ago
State the term using to describe the turning force force exerted by the man
defon

Answer:

The distance between the line of action of force and the axis of rotation (or pivoted point)

Explanation:

The distance between the line of action of force and the axis of rotation (or pivoted point) .

4 0
2 years ago
PLEASE HELP THIS IS FOR SCIENCE!!!!
Oksana_A [137]

Answer:

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Explanation:

fgjcfrcfjcv mghkckgktygftgvkytfyfvky

8 0
2 years ago
A(n) 1700 kg car is moving along a level road at 21 m/s. The driver accelerates, and in the next 10 s the engine provides 22000
allochka39001 [22]

The final speed of the car at the given conditions is 30.1 m/s.

The given parameters:

  • <em>Mass of the car, m = 1700 kg</em>
  • <em>Velocity of the car, v = 21 m/s</em>
  • <em>Time of motion, t = 10 s</em>
  • <em>Additional energy provided by the engine, E₁ = 22,000 J</em>
  • <em>Energy used in overcoming friction, E₂ = 3,666.67 J</em>

The change in the energy applied to the car is calculated as;

\Delta E = E_1 - E_2\\\\\Delta E = 22,000 \ J \ - \ 3,666.67 \ J\\\\\Delta  E = 18,333.33 \ J

The final speed of the car is calculated as follows;

\Delta E = \frac{1}{2} m(v_f^2 - v_0^2)\\\\v_f^2 - v_0^2 = \frac{2\Delta E}{m} \\\\v_f^2  = \frac{2\Delta E}{m} + v_0^2\\\\v_f = \sqrt{ \frac{2\Delta E}{m} + v_0^2} \\\\v_f = \sqrt{ \frac{2\times 18,333.4}{1700} + (21)^2} \\\\v_f = 30.1 \ m/s

Thus, the final speed of the car at the given conditions is 30.1 m/s.

Learn more about change in kinetic energy here: brainly.com/question/6480366

4 0
2 years ago
Determine the energy lost, due to friction, as an 8000 N car that skids to a stop, if its initial velocity was 12 m/s.
chubhunter [2.5K]

Answer:

<em>58,800Joules</em>

Explanation:

The energy lost is equal to the workdone by the car as it skids.

Workdone = Force * Distance

Given

Force = 8000N

Get the distance using the equation of motion

v² = u² - 2gS

0² = 12² - 2(9.8)S

-12² =  - 2(9.8)S

-144 = -19.6S

S = 144/19.6

S = 7.35m

Calculate the required energy

Workdone = 8000 * 7.35

Workdone = 58,800Joules

<em>Hence the energy lost due to friction is 58,800Joules</em>

3 0
3 years ago
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