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Ratling [72]
2 years ago
5

A 40kg dog is sitting on top of a hillside and has a potential energy of 1,568 J. What is the height of the hill side?

Physics
2 answers:
bagirrra123 [75]2 years ago
5 0

Answer:

<h2>39.2 m</h2>

Explanation:

The height of the hill side can be found by using the formula

h =  \frac{p}{m}  \\

p is the potential energy

m is the mass

From the question we have

h =  \frac{1568}{40}  =  \frac{196}{5}  \\

We have the final answer as

<h3>39.2 m</h3>

Hope this helps you

Goshia [24]2 years ago
4 0

Answer: 4.0m

Explanation:

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Answer:

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Explanation:

The x component of the resultant vector is:

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x = 4.07

The y component of the resultant vector is:

y = 3.14 sin(30.0°) + 2.71 sin(-60.0°)

y = -0.777

Therefore, the angle between the resultant vector and the +x axis is:

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3 years ago
An orange light (f = 5.2 * 10'4Hz) is
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Answer:

2.145×10^-10 V or 0.2145nV

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f= frequency of the electromagnetic wave = 5.2×10^4 Hz

e= electronic charge= 1.6×10^-19 C

V= voltage

V= hf/e

V= 6.6×10^-34JS × 5.2×10^4 Hz/ 1.6×10^-19 C

V= 2.145×10^-10 V or 0.2145nV

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Answer:

E=\frac{KQ}{2\sqrt 2a^2}

Explanation:

We are given that

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We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

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Substitute x=a and R=a

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E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}

E=\frac{KQa}{2\sqrt 2a^3}

E=\frac{KQ}{2\sqrt 2a^2}

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Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=\frac{KQ}{2\sqrt 2a^2}

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