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andreev551 [17]
3 years ago
14

A concave mirror brings the sun's rays to a focus in front of the mirror. Suppose the mirror is submerged in a swimming pool but

still pointed up at the sun. Part A Will the sun's rays be focused nearer to, farther from, or at the same distance from the mirror? Explain.
Physics
1 answer:
Karo-lina-s [1.5K]3 years ago
8 0

Answer:

The sun rays will be focused at the same distance from the mirror.

Explanation:

Laws of reflection do not depend on the refractive index n any form. Hence, the sun’s rays will be focused the same distance from the mirror.

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In which state of matter is there most molecular motion
allsm [11]

Answer:

Correct answer is gaseous state

Explanation:

in gaseous state , the bonding between the molecules of matter is less compared to that in other state.

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3 years ago
Explain what happens to the pitch of a cell phone ring when the wavelength of a sound wave increases
max2010maxim [7]

You get a more low sound.  

Conversely, when the wavelength becomes shorter you get a more treble sound.

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4 0
3 years ago
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A light year is approximately 9.5 million km long. 'Barnard's Star' is 6 light years away from Earth. Calculate how many million
hammer [34]

Answer:

6 light years = 57 million km

Explanation:

Given;

A light year = 9.5 million km

To calculate how far is 6 light years;

6 light years = 6 × 1 light year = 6 × 9.5 million km

6 light years = 57 million km

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3 years ago
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In a nuclear power plant, nuclear energy is first changed to ____ energy. (1 point)
Finger [1]
I believe the correct answer from the choices listed above is the second option. In a nuclear power plant, nuclear energy is first changed to thermal energy. Heat energy is first given off in this process. Hope this answers the question. Have a nice day.
4 0
3 years ago
A positive charge +q1 is located to the left of a negative charge -q2. On a line passing through the two charges, there are two
AfilCa [17]

Answer:

please the answer below

Explanation:

(a) If we assume that our origin of coordinates is at the position of charge q1, we have that the potential in both points is

V_1=k\frac{q_1}{r-1.0}-k\frac{q_2}{1.0}=0\\\\V_2=k\frac{q_1}{r+5.2}-k\frac{q_2}{5.2}=0\\\\

k=8.89*10^9

For both cases we have

k\frac{q_1}{r-1.0}=k\frac{q_2}{1.0}\\\\q_1(1.0)=q_2(r-1.0)\\\\r=\frac{q_1+q_2}{q_2}\\\\k\frac{q_1}{r+5.2}=k\frac{q_2}{5.2}\\\\q_1(5.2)=q_2(r+5.2)\\\\r=\frac{5.2q_1-5.2q_2}{q_2}

(b) by replacing this values of r in the expression for V we obtain

k\frac{q_1}{\frac{5.2(q_1-q_2)}{q_2}+5.2}=k\frac{q_2}{5.2}\\\\\frac{q_1}{q_2}=\frac{(q_1-q_2)}{q_2}-1.0=\frac{q_1-q_2-q_2}{q_2}=\frac{q_1-2q_2}{q_2}

hope this helps!!

3 0
3 years ago
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