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MrMuchimi
3 years ago
6

How does height affects the potential energy of a body?​

Physics
1 answer:
ivann1987 [24]3 years ago
3 0

Answer:

The higher the height above the more PE (Potential Energy), the body posses as seen in the formula PE = mass*gravity*height. m and g are constant hence only h affect PE

Explanation:

Potential Energy (PE) is the energy the body posses by virtue of its vertical height above the ground because, once the body is released, it will release or create some energy eg hydroelectricity power generation or a car going downhill does not need to be accelerated by the driver since it can roll over on its own due to PE

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If i am changing velocity, i must also have _______ and a net _________
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if i am changing velocity, i must also have <u>acceleration</u> and a net <u>force</u>

<h2><u>Newton's</u><u> </u><u>first</u><u> </u><u>law</u><u> </u><u>of</u><u> </u><u>motio</u><u>n</u></h2>

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brainly.com/question/4931057

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An astronaut 100m from the spaceship observes a 200kg meteoroid that drifts toward the shop at 10m/sec. If the astronaut can gai
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8 0
3 years ago
a wire of a certain material has resistance r and diameter d a second wire of the same material and length is found to have resi
777dan777 [17]

Answer:

<h2>d₂ = 3d</h2><h2>The diameter of the second wire is 3 times that of the initial wire.</h2>

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Using the formula for calculating the resistivity of an object to find the diameter.

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R is the resistance of the material

A is the cross sectional area

L is the length of the material

Since A = πd²/4

P = R( πd²/4)/L

P = Rπd²/4L ... 1

If the second wire of the same material and length is found to have resistance R/9, the resistivity of the second material will be;

P₂ = (R/9)A₂/L₂

P₂ = (R/9)(πd₂²/4)/L₂

P₂ = (Rπd₂²/36)/L₂

P₂ = (Rπd₂²)/36L₂

Since the length and resistivity are the same;

P = P₂  and L =L₂

Equating 1 and 2;

Rπd²/4L =  (Rπd₂²)/36L₂

Rπd²/4L =  (Rπd₂²)/36L

d² = d₂²/9

d₂² = 9d²

Taking the square root of both sides;

√d₂² = √9d²

d₂ = 3d

Therefore the diameter of the second wire is 3 times that of the initial wire

5 0
3 years ago
the maximum range of a projectile is 2÷√3 times its actual range what is the angle of the projection for the actual range​
Murrr4er [49]

Answer:

The actual angle is 30°

Explanation:

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v_y(t)=vo*sin(A)-g*t

the velocity is Zero when the projectile reach in the maximum altitude:

0=vo-gt\\t=\frac{vo}{g}

When the time is vo/g the projectile are in the middle of the range.

<h2>x axis:</h2>

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**sin(2A)=2sin(A)cos(A)

<h2>The maximum range occurs when A=45°(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>

Let B the actual angle of projectile

\frac{vo^{2} }{g} =(\frac{2}{\sqrt{3} }) \frac{vo^{2} *sin(2B)}{g}\\\\1= \frac{2 }{\sqrt{3}} *sin(2B)\\\\sin(2B)=\frac{\sqrt{3}}{2}\\\\

2B=60°

B=30°

7 0
3 years ago
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