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Elena-2011 [213]
3 years ago
12

What is the gravitational force on a 70kg that is 6.38x10^6m above the earths surface

Physics
1 answer:
NARA [144]3 years ago
5 0

Answer:

171.5 N

Explanation:

The gravitational force on an object due to the Earth is given by

F=mg

where

m is the mass of the object

g is the acceleration due to gravity

The acceleration due to gravity at a certain height h above the Earth is given by

g=\frac{GM}{(R+h)^2}

where:

G is the gravitational constant

M=5.98\cdot 10^{24} kg is the Earth's mass

R=6.37\cdot 10^6 m is the Earth's radius

Here,

h=6.38\cdot 10^6  m

So the acceleration due to gravity is

g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})}{(6.37\cdot 10^6 + 6.38\cdot 10^6)^2}=2.45 m/s^2

We know that the mass of the object is

m = 70 kg

So, the gravitational force on it is

F=mg=(70)(2.45)=171.5 N

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Explanation:

4 0
3 years ago
A block is released to slide down a frictionless incline of 15∘ and then it encounters a frictional surface with a coefficient o
Elodia [21]

The block's potential energy at the top of the incline (at a height h from the horizontal surface) is equal to its kinetic energy at the bottom of the incline, so that

mgh = 1/2 mv²

where v is its speed at the bottom of the incline. It follows that

v = √(2gh)

If the incline is 20.4 m long, that means the block has a starting height of

sin(15°) = h/(20.4 m)   ⇒   h = (20.4 m) sin(15°) ≈ 5.2799 m

and so the block attains a speed of

v = √(2gh) ≈ 10.1728 m/s

The block then slides to a rest over a distance d. Kinetic friction exerts a magnitude F over this distance and performs an amount of work equal to Fd. By the work-energy theorem, this quantity is equal to the block's change in kinetic energy, so that

Fd = 0 - 1/2 mv²   ⇒   d = (-1293.58 J)/F

By Newton's second law, the net vertical force on the block as it slides is

∑ F [vertical] = n - mg = 0

where n is the magnitude of the normal force, so that

n = mg = (25 kg) g = 245 N

and thus the magnitude of friction is

F = -0.16 (245 N) = -39.2 N

(negative since it opposes the block's motion)

Then the block slides a distance of

d = (-1293.58 J) / (-39.2 N) ≈ 32.9994 m ≈ 33 m

5 0
2 years ago
A molecular cloud fragments as it collapses because Choose one: A. the interstellar wind is stronger in some places than others.
kondaur [170]

The cause for a molecular cloud forming fragments when it collapses is indicated correctly by option D. density variations from place to place grow larger as the cloud collapses.

Molecular cloud:

A molecular cloud, also known as a stellar nursery, is a specific kind of interstellar cloud, whose density and size allow the development of molecules, absorption nebulae, and H II regions. In contrast, some regions of the interstellar medium mostly consist of ionized gas.

Molecular clouds are cold, dense areas of space where stars form. The cloud collapses into a proto-star when the gravitational force pulling it in outweighs the internal pressure pushing it in.

When a molecular cloud collapses, it is observed that the density varies from place to place with the variation increasing with collapse. As a result, the collapse is characterized by fragmentation of the cloud.

Thus the correct option is: D. density variations from place to place grow larger as the cloud collapses.

Learn more about molecular clouds,

brainly.com/question/3459894

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6 0
1 year ago
The kinetic energy of a book on a shelf is equal to the work done to lift the book to the shelf. t/f
Svet_ta [14]
The kinetic energy of a book on a shelf is equal to the work done to lift the book to the shelf is false. The kinetic energy on the shelf is zero because it is not in action.
7 0
3 years ago
Read 2 more answers
The most soaring vocal melody is in Johann Sebastian Bach's Mass in B minor. In one section, the basses, tenors, altos, and sopr
OlgaM077 [116]

Answer:

Detailed step wise solution is attached below

Explanation:

(a) wavelength of the initial note 2.34 meters

(b) wavelength of the final note 0.389 meters

(d) pressure amplitude of the final note 0.09 Pa

(e) displacement amplitude of the initial note 4.78*10^(-7) meters

(f) displacement amplitude of the final note 3.95*10^(-8) meters

6 0
3 years ago
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