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nasty-shy [4]
3 years ago
8

A 65-kg ice skater stands facing a wall with his arms bent and then pushes away from the wall by straightening his arms. At the

instant at which his fingers lose contact with the wall, his center of mass has moved 0.55 m , and at this instant he is traveling at 3.5 m/s .
Part A: What is the average force exerted by the wall on him? Express your answer with the appropriate units. FavF a v = 720 N

Part B: What is the work done by the wall on him?
Physics
1 answer:
Marrrta [24]3 years ago
6 0

Our values can be defined like this,

m = 65kg

v = 3.5m / s

d = 0.55m

The problem can be solved for part A, through the Work Theorem that says the following,

W = \Delta KE

Where

KE = Kinetic energy,

Given things like that and replacing we have that the work is given by

W = Fd

and kinetic energy by

\frac {1} {2} mv ^ 2

So,

Fd = \frac {1} {2} m ^ 2

Clearing F,

F = \frac {mv ^ 2} {2d}

Replacing the values

F = \frac {(65) (3.5)} {2 * 0.55}

F = 723.9N

B) The work done by the wall is zero since there was no displacement of the wall, that is d = 0.

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Using an unmanned rocket to visit the space station requires 85.2 trillion BTU of energy. The best fuel for the mission will hav
Illusion [34]
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1 lbs/454 ml x 3785.41 ml/gal → 3785.41 lbs/454gal
Conversion of g/ml = 8.34 lbs/gal
Looking at each fuel:

Kerosene:
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The best fuel in terms of energy to volume ratio is Gasoline.
Gallons required:
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5 0
3 years ago
Using a simple machine, a student is able to lift a 500N weight by applying only 100N.
pishuonlain [190]
Hey there!

a \ student  \ is \ able \ to \ lift \ a \ 500(n) \\ weight \ by \ applying \ only  \ 100(n) \\ \\ so, \ this \ info \ here, \ we \ simply \ divide \ by \\ how \ much \ this \ kid \ lifted, \ by \ the \ weight \ he/she \\ \ is \ applying. \\ \\   \left[\begin{array}{ccc}\boxed{\boxed{500(n)/100(n)}}\end{array}\right] \\ \\ and \ from \ this,\ your \ answer \ would \ \\ conclude \ to \ be \ (5) \\ \\ \boxed{5} \ would \ be \ your \ answer!

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7 0
3 years ago
If we were to illuminate them only with light from the Balmer transition considered above, would the solar panels produce a curr
Ugo [173]

Answer:

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The Balmer series consists of lines in the visible spectrum. It corresponds to emission of a photon of light when electrons descend from higher energy levels to the n=2 level in the hydrogen spectrum. The various wavelengths in the Balmer series can be separated by a prism since they are all in the visible region of the electromagnetic spectrum.

In solar panels, light corresponding to the wavelengths in the Balmer series is merely reflected by the panel and not absorbed. Since light is not absorbed, no current can be produced when the panel is irradiated with light corresponding to the wavelengths in the Balmer series.

6 0
3 years ago
You have 1.2kg of (AU)gold (that's quite a bit of money at $1839/ounce). Using a heat source you apply 3096 J to the gold and re
melamori03 [73]

Answer:

129 J/Kg°C

Explanation:

Given :

Mass of gold, m = 1.2kg

Quantity of heat applied, Q = 3096 J

Temperature, t2 = 40°C

Temperature, t1 = 20°C

Change in temperature, dt = (40-20)°C = 20°C

Using the relation :

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3096 J = 24kg°C * C

C = 3096 J / 24 kg°C

C = 129 J/Kg°C

7 0
3 years ago
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