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Advocard [28]
3 years ago
9

A person standing on the surface of the earth has a weight of 650 N. What is its weight at a distance of two earth radii from th

e surface? If the person is at a distance of two earth radii from the surface, he/she is at a distance of three earth radii from the earth’s center. Since we have an inverse square law, three times the distance means 1/9th of the weight.
Physics
1 answer:
snow_lady [41]3 years ago
4 0

Answer:

72.22 N

Explanation:

F = weight

m = mass of body

M = mass of earth

R = radius of earth

G = universal constant of gravitation

F_1= 650 N

F_1 = GMm/R^2

two earth radius above the surface of the earth:

F_2=  GMm/(3R)^2=  GMm/9R^2= F_1/9= 650/9

=72.22 N

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A ball is rolling along at speed v without slipping on a horizontal surface when it comes to a hill that rises at a constant ang
MissTica

Answer:

The ball has an initial linear kinetic energy and initial rotational kinetic energy which can both be converted into gravitational potential energy. Therefore the hill with friction will let the ball reach higher.

Explanation:

The ball has an initial linear kinetic energy and initial rotational kinetic energy which can both be converted into gravitational potential energy. Therefore the hill with friction will let the ball reach higher.

This is because:

If we consider the ball initially at rest on a frictionless surface and a force is exerted through the centre of mass of the ball, it will slide across the surface with no rotation, and thus, there will only be translational motion.

Now, if there is friction and force is again applied to the stationary ball, the frictional force will act in the opposite direction to the force but at the edge of the ball that rests on the ground. This friction generates a torque on the ball which starts the rotation.

Therefore, static friction is infact necessary for a ball to begin rolling.

Now, from the top of the ball, it will move at a speed 2v, while the centre of mass of the ball will move at a speed v and lastly, the bottom edge of the ball will instantaneously be at rest. So as the edge touching the ground is stationary, it experiences no friction.

So friction is necessary for a ball to start rolling but once the rolling condition has been met the ball experiences no friction.

6 0
3 years ago
It is easier to overcome load when the load is shifted towards the wheel in a wheelbarrow why ? give reason​
krok68 [10]

Answer:

The wheelbarrow's wheel and axle help the wheelbarrow to move without friction thus making it easier to push or pull. That's why it will be easier to lift a load in wheel barrow of the load is transferred towards the wheel.

4 0
2 years ago
Read 2 more answers
The velocity of a 48.0 g shell leaving a 2.95 kg rifle is 391. m/s. What is the recoil velocity of the rifle?
Monica [59]

Hi there!

\large\boxed{  -6.36m/s}

Use the equation:

v_{2} = -\frac{m_{1}}{m_{2}} v_{1}

Where m2 and v2 deal with the larger object, and m1 and v1 with the smaller object. Plug in the given values:

v2 = ?

m1 = 0.048 kg (converted)

m2 = 2.95

v1 = 391

v_{2} = -\frac{0.048}{2.95} *391

v_{2} = -6.36m/s

8 0
2 years ago
Car B starts at point X and moves clockwise around a circular track at a constant rate of 2 mph. Ten hours later, Car A leaves f
stealth61 [152]

Answer:

20.96 h

Explanation:

The perimeter of the track is 2*pi*r = 20pi miles

In 10 hours, car B would have moved 20miles. So, when Car A leaves from point X, car B is 20pi - 20 miles from point X counter-clockwise and car A.

From here, we can express the distance of A from X like this:

xa = 3t

And the distance of B would be:

xb = 20pi - 20 - 2t

The time t where they would passed each other and put  12 miles between them would be the one where xa - xb is equal to 12:

xa - xb = 12

3t - (20pi - 20 - 2t) = 12

5t = 20 pi - 8

t = (20pi - 8)/5 = 10.96 h

Remember to add this value to the 10 hours car B had already been racing:

t = 20.96h

4 0
3 years ago
An apple with a mass of 0.95 kilograms hangs from 3.0 meters above the ground. What is the potential energy of the apple
Vesnalui [34]

As Potential energy =mgh

m= 0.95kg

h=3 meter

g = 9.8 m/sec^2. ( acceleration due to gravity)

So P.E =(0.95)(9.8)(3)kgm^2/s^2

P.E =27.93 joules

3 0
3 years ago
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