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inna [77]
3 years ago
12

Suppose a skydiver (mass = 80kg) is falling toward the Earth. Calculate the skydiver’s gravitational potential energy at a point

when the skydiver is 100m above the Earth.
(Show the equation, show your work and answer with units.)
Physics
2 answers:
Artemon [7]3 years ago
8 0

Answer:

80,000 Joules or 80KJ

Explanation:

Gravitational potential energy is defined as the energy possessed by a body under the influence of gravitational force. The body will therefore possess acceleration due to gravity (g).

Mathematically,

Gravitational potential energy = mass (m) × acceleration due to gravity (g) × height (h)

Given mass of skydiver = 80kg

Height = 100m

taking g = 10m/s²

Substituting the values given in the formula to get the gravitational potential energy, we will have;

GPE = 80×100×10

GPE = 80,000Joules

GPE = 80KJ

Cerrena [4.2K]3 years ago
5 0

Answer:

The gravitational potential energy of the skydiver=79200 k g m^{2} / s^{2}

Given:

Mass of the skydiver=80kg

Distance of the skydiver from the earth=100m

To find:

Gravitational potential energy of the skydiver

<u>Step by Step Explanation:</u>

Solution:

According to the formula, Gravitational Potential Energy (GHE) is calculated as

\text {Gravitational Potential Energy(GHE)}=m g h

Where m= Mass of the skydiver

g=Acceleration due to gravity

h=Distance of the skydiver from the earth

Substitute the known values in the above equation we get

G H E=80 \times 9.8 \times 100

=79200 \mathrm{kg} \mathrm{m}^{2} / \mathrm{s}^{2}

Result:

Thus the gravitational potential energy of the skydiver is =79200 k g m^{2} / s^{2}

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The change in speed of this object is 3m/s

According to Newton's second law;

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Hence the change in speed of this object is 3m/s

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Answer:

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           tan θ = y / L

     

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therefore, by carefully measuring the zero intensity point, we can deduce the movement of the screen.

 

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Answer:

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For t = 3.0 s we have:

v(3.0s) = -13*3.0 =-39 \frac{m}{s}

b) v_{avg}= \frac{z_f - z_i}{t_f -t_i}

And we can find the positions for the two times required like this:

z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m

z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m

And now we can replace and we got:

V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}

Explanation:

The particle position is given by:

z(t) = -(6.5 \frac{m}{s^2}) t^2, t\geq 0

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In order to find the velocity we need to take the first derivate for the position function like this:

z'(t) =v(t) = -13t

Now we can replace the velocity for t=1.75 s

v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}

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Part b

For this case we can find the average velocity with the following formula:

v_{avg}= \frac{z_f - z_i}{t_f -t_i}

And we can find the positions for the two times required like this:

z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m

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And now we can replace and we got:

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8 0
3 years ago
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