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nikitadnepr [17]
3 years ago
15

Which of the following is an example of diffraction

Physics
1 answer:
Leno4ka [110]3 years ago
7 0

Answer:

the answer would be D. Yelling out in a cave and hearing your voice call back to you

Explanation:

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A. The English system uses one unit for each category of measurement.

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At which of the following angles will the sunlight received at a location on Earth spread out over the largest area?
AleksandrR [38]

The angle at which the sunlight received at a location on Earth spread out over the largest area is 10°. Last option is correct.

<h3>What is sunlight?</h3>

The light coming from the Sun reaching the Earth's surface is called as Sunlight.

When the sun is overhead, the intensity is high because sun's rays are perpendicular to the earth's surface, so the energy spreads over a small area and the heat is too high in that region.

When, the angle is smaller, the sunlight will spread out over a larger area.

Thus, at 10° the sunlight received at a location on Earth spread out over the largest area. Last option is correct.

Learn more about Sunlight.

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2 years ago
Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its te
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Answer:

Please see attachment

Explanation:

Please see attachment

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3 years ago
When a piece of copper is taken to the moon , a change will be observed in it's ?​
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Weight. Because there is less gravity on the moon.
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3 years ago
Two parallel plate capacitors 1 and 2 are identical except that capacitor 1 has charge +q on one plate and charge −q on the othe
Grace [21]

Answer:

a) the capacitance is the same for both capacitors.

b) The potential difference between the plates for the capacitor with charge +2q, is double of the one for the capacitor  with charge +q.

c) The electric field magnitude between the plates for the capacitor with charge +2q, is double of the one for the capacitor with charge +q

d) The energy stored between the plates for the capacitor with charge +2q, is 4 times the value for the one with charge +q  

Explanation:

a) The capacitance of a capacitor, by definition, is as follows:

C = \frac{q}{V}

Appying Gauss' Law to one of plates, it can be showed, that the capacitance (for a parallel plates capacitor) can be  expressed as follows:

C = ε*A / d

As it can be seen, it does not depend on the charge. so we conclude that the capacitance must be the same for both capacitors, due to they are identical except for the value of the charge on the plates.

b) By definition, as we said above, the capacitance is equal to the proportion between the charge of one of the plates, and the potential difference between them.

If this proportion must remain the same, and one of the capacitors has the double of  the charge than the other, the potential difference must be the double also.

c) Applying Gauss' law, to the surface of one of  the plates, and assuming a constant surface charge density σ, it can be  showed that the  electric field can be calculated as follows:

E*A = Q/ε₀ as σ=Q/A

⇒ E = σ/ε₀

As σ is directly proportional to the charge (being the area A the same), we conclude that the electric field for the capacitor with charge +2q must be the double than the one for the capacitior with charge +q.

c) The electric potential energy, stored between plates of a capacitor, can be written as follows:

Ue = \frac{1}{2} *\frac{q^{2}}{C}

If the capacitance remains the same, we can conclude that the electric potential energy for the capacitor with charge +2q, as the charge is raised to the 2nd power, must be 4 times the one for the capacitor with charge +q.

4 0
3 years ago
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