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3 years ago

The photoelectric effect tells us that __________.

Physics

1 answer:

Ivanshal [37]3 years ago

6 0

It describes the process by which surface electron's are emitted from a metal when light is shined on it.
And it tells us that light at a higher intensity must contain more quanta of energy, known as protons.

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Earthquakes often occur along _____ as a result of the build up of stress

ExtremeBDS [4]

Hello Lauren, The answer to this question is D:Plate boundaries.

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3 years ago

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Using the following equation, find in seconds 145m = 9 m/s x t-35m

Zina [86]

Answer:

t = 20 s

Explanation:

145 m = 9 m/s × t – 35m

We can obtain the value of t in seconds by doing the following:

145 m = 9 m/s × t – 35m

Add 35 m to both side as shown below:

145 m + 35 m = 9 m/s × t – 35m + 35 m

180 m = 9 m/s × t

Divide both side by 9 m/s

t = 180 m / 9 m/s

t = 20 s

Therefore, the value of t is 20 seconds.

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3 years ago

Creating laws outlawing discrimination has done little to change prejudiced views or behaviors.

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True.........All kinds of bad stuff hidden from cops

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3 years ago

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A clarinet behaves like a tube closed at one end. If its length is 1.0 m, and the velocity of sound is 344 m/s, what is its fund

miv72 [106K]

Answer:

Fundamental frequency = 1376Hz

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4 years ago

mass of the planet is 12 times that of earth and its radius is thrice that of earth , then find the escape velocity on that plan

Over [174]

Answer:

The escape velocity on the planet is approximately 178.976 km/s

Explanation:

The escape velocity for Earth is therefore given as follows

The formula for escape velocity, $v_e$ , for the planet is $v_e = \sqrt{\dfrac{2 \cdot G \cdot m}{r} }$

Where;

$v_e$ = The escape velocity on the planet

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

m = The mass of the planet = 12 × The mass of Earth, $M_E$

r = The radius of the planet = 3 × The radius of Earth, $R_E$

The escape velocity for Earth, $v_e_E$ , is therefore given as follows;

$v_e_E = \sqrt{\dfrac{2 \cdot G \cdot M_E}{R_E} }$

$\therefore v_e = \sqrt{\dfrac{2 \times G \times 12 \times M}{3 \times R} } = \sqrt{\dfrac{2 \times G \times 4 \times M}{R} } = 16 \times \sqrt{\dfrac{2 \times G \times M}{R} } = 16 \times v_e_E$

$v_e$ = 16 × $v_e_E$

Given that the escape velocity for Earth, $v_e_E$ ≈ 11,186 m/s, we have;

The escape velocity on the planet = $v_e$ ≈ 16 × 11,186 ≈ 178976 m/s ≈ 178.976 km/s.

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3 years ago