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MaRussiya [10]
2 years ago
15

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

Physics
1 answer:
AleksandrR [38]2 years ago
3 0

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.

brainly.com/question/23504828

#SPJ1

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The volume electric charge density of a solid sphere is given by the following equation: The variable r denotes the distance fro
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Answer:

62.8 μC

Explanation:

Here is the complete question

The volume electric charge density of a solid sphere is given by the following equation: ρ = (0.2 mC/m⁵)r²The variable r denotes the distance from the center of the sphere, in spherical coordinates. What is the net electric charge (in μC) of the sphere if the radius of the sphere is 0.5 m?

Solution

The total charge on the sphere Q = ∫∫∫ρdV where ρ = volume charge density = 0.2r² and dV = volume element in spherical coordinates = r²sinθdθdrdΦ

So,  Q =  ∫∫∫ρdV

Q =  ∫∫∫ρr²sinθdθdrdΦ

Q =  ∫∫∫(0.2r²)r²sinθdθdrdΦ

Q =  ∫∫∫0.2r⁴sinθdθdrdΦ

We integrate from r = 0 to r = 0.5 m, θ = 0 to π and Φ = 0 to 2π

So, Q =  ∫∫∫0.2r⁴sinθdθdrdΦ

Q =  ∫∫∫0.2r⁴[∫sinθdθ]drdΦ

Q =  ∫∫0.2r⁴[-cosθ]drdΦ

Q =  ∫∫0.2r⁴-[cosπ - cos0]drdΦ

Q =  ∫∫∫0.2r⁴-[-1 - 1]drdΦ

Q =  ∫∫0.2r⁴-[- 2]drdΦ

Q =  ∫∫0.2r⁴(2)drdΦ

Q =  ∫∫0.4r⁴drdΦ

Q =  ∫0.4r⁴dr∫dΦ

Q =  ∫0.4r⁴dr[Φ]

Q =  ∫0.4r⁴dr[2π - 0]

Q =  ∫0.4r⁴dr[2π]

Q =  ∫0.8πr⁴dr

Q =  0.8π∫r⁴dr

Q =  0.8π[r⁵/5]

Q = 0.8π[(0.5 m)⁵/5 - (0 m)⁵/5]

Q = 0.8π[0.125 m⁵/5 - 0 m⁵/5]

Q = 0.8π[0.025 m⁵ - 0 m⁵]

Q = 0.8π[0.025 m⁵]

Q = (0.02π mC/m⁵) m⁵

Q = 0.0628 mC

Q = 0.0628 × 10⁻³ C

Q = 62.8 × 10⁻³ × 10⁻³ C

Q = 62.8 × 10⁻⁶ C

Q = 62.8 μC

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A ball is hit so that it accelerates at 10 m/s2. The ball hits a glove with a force of 5 N. What is the mass of the ball?
jenyasd209 [6]

Answer:

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Explanation:

» <u>Concepts</u>

Newton's second law, the Law of Acceleration, states that F = ma, where F = Force in Newtons, m = mass in kg, and a = acceleration in m/s^2.

» <u>Application</u>

We are asked to find the mass of the ball using the equation F = ma. We're also given the force and acceleration, so the equation looks like 5 = 10(m).

» <u>Solution</u>

Step 1: Divide both sides by 10.

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I hope this helps you, have a great day!
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