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Kamila [148]
3 years ago
13

Which speed is measured in speedometer to track speed violation?

Physics
1 answer:
damaskus [11]3 years ago
4 0

Answer:

Infrared radiations in radar gun

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Jupiter is composed almost entirely of gases and liquids, mostly
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<span>Jupiter is composed almost entirely of gases and liquids, mostly C. hydrogen and helium.</span>
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A 1 kg ball and a 10 kg ball are dropped from a height of 10 m at the same time. In the absence of air resistance,
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Answer:

the two balls will hit the ground at the same time.

Explanation:

The time of dropping, in the following equation, is related to both the distance travel s and the gravitational acceleration g, which are the same for both ball (if we neglect air resistance), no matter what their mass are.

s = gt^2/2

t^2 = 2s/g

So the time it takes to drop 2 balls are the same. They will hit the ground at the same time.

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3 years ago
A focal arrangement that has a thin lens that the light passes through before traveling down the tube to the objective mirror is
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The correct answer for the question that is being presented above is this one: "Schmidt-Cassegrain focus." A focal arrangement that has a thin lens that the light passes through before traveling down the tube to the objective mirror is a Schmidt-Cassegrain focus.
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b. Newtonian focus
c. Schmidt-Cassegrain focus
<span>d. Schmidt focus</span>
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3 years ago
Can you give me the definition of sound wave
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A sound wave is a wave of compression and rarefaction, by which sound is propagated in an elastic medium such as air.

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a painting in an art gallery has height h and is hung so that its lower edge is a distance d above the eye of an observer. How f
harkovskaia [24]

Solution:

With reference to Fig. 1

Let 'x' be the distance from the wall

Then for \DeltaDAC:

tan\theta = \frac{d}{x}

⇒ \theta = tan^{-1} \frac{d}{x}

Now for the \DeltaBAC:

tan\theta = \frac{d + h}{x}

⇒ \theta = tan^{-1} \frac{d + h}{x}

Now, differentiating w.r.t x:

\frac{d\theta }{dx} = \frac{d}{dx}[tan^{-1} \frac{d + h}{x} -  tan^{-1} \frac{d}{x}]

For maximum angle, \frac{d\theta }{dx} = 0

Now,

0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} -  tan^{-1} \frac{d}{x}][/tex]

0 = \frac{-(d + h)}{(d + h)^{2} + x^{2}} -\frac{-d}{x^{2} + d^{2}}

\frac{-(d + h)}{(d + h)^{2} + x^{2}} = \frac{{d}{x^{2} + d^{2}}

After solving the above eqn, we get

x = \sqrt{\frac{d}{d + h}}

The observer should stand at a distance equal to x = \sqrt{\frac{d}{d + h}}

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