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iVinArrow [24]
3 years ago
13

Find the gravitational potential at a point on the earth surface. Take mass as of earth as 5.98×10^24kg,it's radius as6.38×10^6n

and G=6.67×10^-11Nm^2kg^-2​
Physics
1 answer:
andrew-mc [135]3 years ago
5 0

Answer:

-6.25\cdot 10^7 J

Explanation:

The gravitational potential at a point on the Earth surface is given by:

U=-\frac{GM}{R^2}

where

G=6.67×10^-11Nm^2kg^-2 is the gravitational constant

M=5.98×10^24kg is the Earth's mass

R=6.38×10^6 m is the Earth's radius

Substituting the numbers into the equation, we find

U=-\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})}{(6.38\cdot 10^6)}=-6.25\cdot 10^7 J

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Answer:

Force = 186 N

Explanation:

Torque is the rotational equivalent of linear force. It can be easely calculated using the formula :

Torque = \vec{r} \times \vec{F}

Where \vec{r} is a vector that from the origin of the coordinate system to the point at which the force is applied (the position vector), \vec{F} is the applied force.

The easiest way of computing the force is by setting the origin of the coordinate system to the lowest point of the torque wrench.  By doing this we have that r (the magnitud of the position vector) is 35cm.

Before computing the force we need to set all our values to the international system of units (SI). The torque is already in SI. The one missing is the length of the torque wrench (it is in centimeters and we need it in meters). So :

35cm * \dfrac{1m}{100cm} = 0.35m

Now using the torque formula:

Torque = \vec{r} \times \vec{F}

Torque = rFsin(\theta)}

Where \theta is the smaller angle between the force and the position vector. Because the force is applied perpendiculary to the position vector  \theta = 90°, thus :

Torque = rFsin(\theta)}

65 N m = (0.35m)Fsin(90°)}

F = \dfrac{65N m} {(0.35m)sin(90°)}

F = \dfrac{65N m} {(0.35m)sin(90°)}

F = \dfrac{1300} {7}N

so the force is approximately 186 N.

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Answer:

88 m/s

Explanation:

To solve the problem, we can use the following SUVAT equation:

v^2-u^2=2ad

where

v is the final velocity

u is the initial velocity

a is the acceleration

d is the distance covered

For the car in this problem, we have

d = 484 m is the stopping distance

v = 0 is the final velocity

a=-8.0 m/s^2 is the acceleration

Solving for u, we find the initial velocity:

u=\sqrt{v^2-2ad}=\sqrt{-2(8.0)(484)}=88 m/s

6 0
3 years ago
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Ivan

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a crate is being lifted into a truck. if it is moved with a 2470n force and 3650 j of work is done , then how far is the crate b
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Answer:

The crate was being lifted by a height of 1.48 meters.

Explanation:

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distance = 3650/2470

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3 years ago
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