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2 years ago

What is the kinetic energy of a 2000 kg car traveling at a speed of 30 m/s (?65 mph)?

Physics

1 answer:

Brrunno [24]2 years ago

7 0

KE = 1/2mv^2

m = 2000

v = 30m/s

1/2*2000*30^2=

1000*900 = 900,000kgm^2/s^2

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The radar system of a navy cruiser transmits at a wavelength of 1.4 cm, from a circular antenna with a diameter of 2.7 m. At a r

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Answer:

Distance will be 49.34 m

Explanation:

We have given wavelength $\lambda =1.4cm =0.014m$

Diameter of the antenna d = 2.7 m

Range L = 7.8 km = 7800 m

We have to find the smallest distance hat two speedboats can be from each other and still be resolved as two separate objects D

We know that distance is given by $D=\frac{L1.22\lambda }{d}=\frac{7800\times 1.22\times 0.014}{2.7}=49.3422m$

So distance D will be 49.34 m

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3 years ago

Why do different substances have diffrent properties

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Different elements require different levels of energy to make or break a bond

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3 years ago

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A 20 ft ladder leans against a wall. The bottom of the ladder is 3 ft from the wall at time t=0 and slides away from the wall at

stellarik [79]

Answer: 0.516 ft/s

Explanation:

Given

Length of ladder L=20 ft

The speed at which the ladder moving away is v=2 ft/s

after 1 sec, the ladder is 5 ft away from the wall

So, the other end of the ladder is at

$\Rightarrow y=\sqrt{20^2-5^2}=19.36\ ft$

Also, at any instant t

$\Rightarrow l^2=x^2+y^2$

differentiate w.r.t.

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3 years ago

A 0.750 kg block is attached to a spring with spring constant 13.0 N/m . While the block is sitting at rest, a student hits it w

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To solve this problem we will apply the concepts related to energy conservation. From this conservation we will find the magnitude of the amplitude. Later for the second part, we will need to find the period, from which it will be possible to obtain the speed of the body.

A) Conservation of Energy,

$KE = PE$

$\frac{1}{2} mv ^2 = \frac{1}{2} k A^2$

Here,

m = Mass

v = Velocity

k = Spring constant

A = Amplitude

Rearranging to find the Amplitude we have,

$A = \sqrt{\frac{mv^2}{k}}$

Replacing,

$A = \sqrt{\frac{(0.750)(31*10^{-2})^2}{13}}$

$A = 0.0744m$

(B) For this part we will begin by applying the concept of Period, this in order to find the speed defined in the mass-spring systems.

The Period is defined as

$T = 2\pi \sqrt{\frac{m}{k}}$

Replacing,

$T = 2\pi \sqrt{\frac{0.750}{13}}$

$T= 1.509s$

Now the velocity is described as,

$v = \frac{2\pi}{T} * \sqrt{A^2-x^2}$

$v = \frac{2\pi}{T} * \sqrt{A^2-0.75A^2}$

We have all the values, then replacing,

$v = \frac{2\pi}{1.509}\sqrt{(0.0744)^2-(0.750(0.0744))^2}$

$v = 0.2049m/s$

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2 years ago

A rifle is fired and recoils when the bullet leaves the gun. This is an example of Newton's third law. The acceleration of the b

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<span>The acceleration of the bullet is called ACTION.

</span>Formally stated, Newton's third law<span> is: For every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object.
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2 years ago

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