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3 years ago

5

Find the Gravitational Potential at a point on the earth’s surface. Take mass of earth as 5.98 X 10 24 kg, its radius as 6.38 X

10 6 m and G = 6.67 X 10 – 11 Nm2 kg – 2.

1 answer:

kherson [118]3 years ago

5 0

Answer:

$-6.25 x 10^7 J/Kg\\\\$

Explanation:

The final answer is $-6.25 x 10^7 J/Kg\\\\$

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bonded pairs of electrons, lone pairs of electrons.

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3 years ago

Friction provides the force needed for a car to travel around a flat, circular race track. What is the maximum speed at which a

ad-work [718]

Answer:

Maximum speed of the car is 17.37 m/s.

Explanation:

Given that,

Radius of the circular track, r = 79 m

The coefficient of friction, $\mu=0.39$

To find,

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Solution,

Let v is the maximum speed of the car at which it can safely travel. It can be calculated by balancing the centripetal force and the gravitational force acting on it as :

$v=\sqrt{\mu rg}$

$v=\sqrt{0.39\times 79\times 9.8}$

v = 17.37 m/s

So, the maximum speed of the car is 17.37 m/s.

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3 years ago

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Dahasolnce [82]

Answer:

a is the correct choice

Explanation:

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3 years ago

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If a bar of copper is brought near a magnet, the copper bar will be

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3 years ago

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When tightening a bolt, you push perpendicularly on a wrench with a force of 165 N at a distance of 0.140 m from the center of t

choli [55]

Answer:

Part a)

$\tau = 23.1 Nm$

Part b)

$\tau = 17.05 Foot pound force$

Explanation:

As we know that torque is defined as the product of force and its perpendicular distance from reference point

so here we have

$\tau = \vec r \times \vec F$

now we have

$\tau = (0.140)(165)$

$\tau = 23.1 Nm$

Part b)

Now we know the conversion as

1 meter = 3.28 foot

1 N = 0.225 Lb force

now we have

$\tau = 23.1 Nm$

$\tau = 23.1 (0.225 Lb)(3.28 foot)$

$\tau = 17.05 Foot pound force$

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3 years ago