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2 years ago

A 1400-kg car is moving at a speed of 25m/s. how much kinetic energy does the car have?

Physics

1 answer:

Luba_88 [7]2 years ago

3 0

Answer:

437.5Kjoules

Explanation:

K.E=half multiply by mass multiply by square of velocity

=437.5Kjoules

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A top fuel dragster with a mass of 500.0 kg starts from rest and completes a quarter mile (402 m) race in a time of 5.0 s. The d

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The average power needed to produce this final speed is 1069.1 hp.

Mass of the dragster, m = 500.0 kg,

Displacement travelled by the dragster, s = 402 m,

Time taken in this travel, t = 5.0 s,

Final velocity of the dragster, v = 130 m/s.

Let the initial velocity of the dragster be u and acceleration be a.

Using kinematical equation, s = ut + (1/2)at^2.

402 = u*5 + (1/2)*a*5^2

10*u + 25*a = 804. ...........(1)

Using kinematical equation, v = u +at.

130 = u + 5*a

5*u + 25*a = 650. .............(2)

Solving (1)and (2), we get,

u = 30.8 m/s.

According to work-energy theorem,

Work done = change in kinetic energy

W = (1/2)*m*(v^2 - u^2)

W = (1/2)*500*(130^2 - 30.8^2)

W = 3987840. J

Therefore power rating of the dragster is given by,

P ⇒ W/t. = 3987840/5 = 797568 watt.

P ⇒ 797568/746 = 1069.1 hp.

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The turntable in a microwave oven has a moment of inertia of 0.039 kg⋅m2 and is rotating once every 4.4 s . Part A What is its k

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To solve this problem it is necessary to apply the concepts related to Kinetic Energy, specifically, since it is a body with angular movement, the kinetic rotational energy. Recall that kinetic energy is defined as the work necessary to accelerate a body of a given mass from rest to the indicated speed.

Mathematically it can be expressed as,

$KE = \frac{1}{2} I\omega^2$

Where

I = Moment of Inertia

$\omega =$ Angular velocity

Our values are given as

$I = 0.039kg\cdot m^2$

A revolution is made every 4.4 seconds.

$\theta = 1 rev \rightarrow 4.4s$

$\Rightarrow \omega = \frac{1rev}{4.4s}$

If the angular velocity is equivalent to the displacement over the time it takes to perform it then

$\omega = 0.2272rev/s(\frac{2\pi rad}{1rev})$

$\omega = 1.42rad/s$

Replacing at our previous equation we have,

$KE = \frac{1}{2} I\omega^2$

$KE = \frac{1}{2} (0.039)(1.42)^2$

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