2 years ago

What happens to light that shines on an object but is not transmitted?

Physics

1 answer:

bekas [8.4K]2 years ago

7 0

The answer is it is reflected or absorbed, so D.

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Suppose that you'd like to find out if a distant star is moving relative to the earth. The star is much too far away to detect a

rjkz [21]

Answer:

The speed will be "18km/s". A further explanation is given below.

Explanation:

According to the question, the values are:

Wavelength,

$\lambda = 656.46 \ nm$

$\Delta \lambda = 0.04$

$c=3\times 10^8$

As we know,

⇒ $\frac{\Delta \lambda}{\lambda} =\frac{v}{c}$

On substituting the values, we get

⇒ $\frac{656.46}{0.04} =\frac{v}{3\times 10^8}$

⇒ $v=\frac{656.46}{0.04} (3\times 10^8)$

⇒ $=16411.5\times 3\times 10^8$

⇒ $=18280 \ m/s$

or,

⇒ $=18 \ km/s$

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2 years ago

23692U = 14456Ba + 89 36Kr + 3n is a

alekssr [168]

I think The answer is c

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2 years ago

How far is the Earth from the Sun?

ohaa [14]

Answer:93.431 million mi

Explanation:

google lol

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2 years ago

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Jordana pulled a muscle while running at a track meet. Her coach gave her a self-heating pack to treat the injury. The instructi

pochemuha

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2 years ago

If the car's speed decreases at a constant rate from 77 mi/h to 50 mi/h in 3.0 s, what is the magnitude of its acceleration, ass

Lostsunrise [7]

The acceleration of the car,

$a = \frac{v-u}{t}$

Here, v is final velocity, u is initial velocity and t is time taken by the car.

Given $u =77 \ mil/h$ , $v = 50 mi/h$ and $t = 3.0 s = 3.0 \times \frac{1 \ h}{3600} = 8.3 \times 10^{-4} h$

Therefore, from above equation

$a = \frac{50 \ mi/h -77 \ mi/h}{8.33 \times 10^{-4} h} = - \frac{27 \ mi/h }{8.33 \times 10^{-4} h} = - 3.2 \times 10^{4} \ mi/h^2$ .

Here, negative sign shows deceleration of a car.

Thus the the magnitude of car acceleration is $3.2 \times 10^{4} \ mi/h^2$ .

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2 years ago