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chubhunter [2.5K]
3 years ago
10

Calculate the gravitational potential energy of the interacting pair of the Earth and a 4 kg block sitting on the surface of the

Earth. You would need to supply the absolute value of this result to move the block to a location very far from the Earth (actually, you would need to use even more energy than this due to the gravitational potential energy associated with the Sun-block interacting pair).
Ugrav = J
Physics
2 answers:
Airida [17]3 years ago
6 0

Answer:

U=-2.4896\times 10^{7} J

Explanation:

From the equation we know that the gravitational potential energy:

U= -G\frac{M.m}{r}.......................(1)

<em>The negative potential indicates a bound state.</em>

where:

M= mass of the earth

m= mass of the object

r= radial distance from the center of the earth

G= universal gracvitational constant= 6.67\times 10^{-11} N.m^{2}.kg^{-2}

Given:

m= 4kg

We, have

M=5.972\times 10^{24} kg

∵The object is on the earth surface, we have the radius of the earth:

r= 64000 km

Putting the values in the eq. (1)

U=-6.67\times 10^{-11}\times \frac{5.972\times 10^{24}\times 4}{64000\times 1000}

U=-2.4896\times 10^{7} J

STALIN [3.7K]3 years ago
5 0

Answer:

- 2.425 x 10^5 J

Explanation:

The gravitational potential energy between earth and the bock is given by

U=-G\frac{Mm}{r}

Where, G is the universal gravitational constant = 6.67 x 10^-11 Nm^2/kg^2

M is the mass of earth = 5.8 x 10^24 kg

m is the mass of block = 4 kg

r be the radius of earth = 6380 km = 6380 x 10^3 m

by substituting the values in the above expression, we get

U=-6.67\times10^{-11}\frac{5.8\times 10^{24}\times 4}{6380\times 10^{3}}

U = - 2.425 x 10^5 J

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Bumek [7]

Answer:443.1 s

Explanation:

Given

Engine of a locomotive exerts a force of 8.1\times 10^5 N

Mass of train=1.9\times 10^7

Final speed (v)=68 km/h \approx 18.88 m/s

F=ma

so acceleration(a) =\frac{F}{m}=\frac{8.1\times 10^5}{1.9\times 10^7}

a=0.042631 m/s^2

and acceleration is

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0.042631=\frac{18.88-0}{t}

t=443.089 \approx 443.1 s

8 0
3 years ago
Read 2 more answers
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Natalija [7]

Answer:

105 m/s

Explanation:

Given that the speed of train A, V_A = 45 m/s from west to east.

Speed of train B, V_B = 60 m/s from east to west.

Train B is moving in the opposite direction with respect to the speed of train A. Assuming that the speed from east to west direction is positive.

So, the speed of train A from east to west= - 45 m/s

The speed of train B w.r.t train A = V_B - V_A=60-(-45)=60+45=105 m/s

Hence, the speed of train B w.r.t train A is 105 m/s from east to west.

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3 years ago
What occurs when the moon blocks the view of the sun
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solar eclipse

Explanation:

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hope it helps :)

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Consider three force vectors Fi with magni- tude 53 N and direction 116º, F2 with mag- nitude 57 N and direction 217°, and F3 wi
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Answer:

a. Fnet =37.67N

b. The direction = 133.4 from the x axis counter clockwise.

c. Option 2

Explanation:

Given that F1 is 53N at 116°, then it will be at a direction of 116-90=26° in the second quadrant.

Given that F2 is 57N at 116°, then it will be at a direction of 217-180=37° in the third quadrant..

Given that F1 is 71N at 20°, then it is in the first quadrant.

a. Fnet= F1+F2+F3

Fnet= -F1sin26i+F2cos26j-F2cos37i-F2sin37j+F3cos20i+F3sin20j

Fnet= 53sin26i+53cos26j-57cos37i-57sin37j+71cos20i+71sin20j

Resolving the vectors into x and y components.

Fnet= -2.04i+37.62j

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Fnet= √((-2.04)^2+(37.62)^2)

Fnet= 37.67N

Fnet is approximately 38N.

b. Direction of the Fnet.

Angle=arctan(y/x)

Angle=arctan(-37.61/2.04)

Angle= -43.37°

The angle is in the negative x axis and positive y axis.

Then the direction becomes 180-43.37

Therefore, the direction of the net force is 133.37°.

c. The instantaneous velocity of a body is always in the direction of the net force at that instant. Option 2 is correct.

Fnet=ma

Fnet= mv/t

So the velocity is in the direction of the Fnet.

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The images output from your new color laser printer seem to be a little too blue. What can you do to fix this?
Svetradugi [14.3K]

Answer:

The images output from your new color laser printer seem to be a little too blue. to fix this problem we need to calibrate the printer.

Explanation:

This can be done by opening the toolbox, clicking in the device setting folder their you get print quality page click on it. Under the print quality option click on the calibrate next to calibrate now. Then click OK unless when the 'your request has been sent to the device' appears on the screen. When the calibration ends again try printing. calibrating is useful for managing the proper alignment of the inkjet cartridge nozzle to the paper and each other, without proper calibration the print quality deteriorates.

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3 years ago
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