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Verdich [7]
3 years ago
14

If the mass of the Moon were suddenly to double, and the distance to the moon cut in half, how would the gravitational force bet

ween the moon and the Earth Change?
Physics
2 answers:
Andrei [34K]3 years ago
6 0

Isaac Newton compared the acceleration of the moon to the acceleration of objects on earth. Believing that gravitational forces were responsible for each, Newton was able to draw an important conclusion about the dependence of gravity upon distance. This comparison led him to conclude that the force of gravitational attraction between the Earth and other objects is inversely proportional to the distance separating the earth's center from the object's center. But distance is not the only variable affecting the magnitude of a gravitational force. Consider Newton's famous equation

Fnet = m • a

Newton knew that the force that caused the apple's acceleration (gravity) must be dependent upon the mass of the apple. And since the force acting to cause the apple's downward acceleration also causes the earth's upward acceleration (Newton's third law), that force must also depend upon the mass of the earth. So for Newton, the force of gravity acting between the earth and any other object is directly proportional to the mass of the earth, directly proportional to the mass of the object, and inversely proportional to the square of the distance that separates the centers of the earth and the object.

Alenkinab [10]3 years ago
6 0

Isaac Newton compared the acceleration of the moon to the acceleration of objects on earth. Believing that gravitational forces were responsible for each, Newton was able to draw an important conclusion about the dependence of gravity upon distance. This comparison led him to conclude that the force of gravitational attraction between the Earth and other objects is inversely proportional to the distance separating the earth's center from the object's center. But distance is not the only variable affecting the magnitude of a gravitational force. Consider Newton's famous equation

\

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A coin 15.0 mm in diameter is placed 15.0 cm from a spherical mirror. The coin's image is 5.0 mm in diameter and is erect. Is th
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Answer:

15 cm

Explanation:

h_{o} = Diameter of the coin = 15 mm

h_{i} = Diameter of the image of coin =  5 mm

d_{o} = distance of the coin from mirror = 15 cm

d_{i} = distance of the image of coin from mirror = ?

Using the equation

\frac{d_{i}}{d_{o}} = \frac{- h_{i}}{h_{o}}

\frac{d_{i}}{15} = \frac{- (5)}{15}

d_{i} = - 5 cm

R = radius of curvature

Using the mirror equation

\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{2}{R}

\frac{1}{15} + \frac{1}{- 5} = \frac{2}{R}

R = - 15 cm

6 0
3 years ago
An astronaut lands on an alien planet. He places a pendulum (L = 0.200 m) on the surface and sets it in simple harmonic motion,
Ne4ueva [31]
A)  f = 1.8 rev/s = 2 Hz 
<span>T = 1 / f = 0.55s

B)  not really sure..srry

C)  </span><span>T = 2 pi √ ( L / g ) </span>
<span>0.57 = 2 x 3.14 x √ ( 0.2 / g )
</span><span>
g = 25.5 m/s²
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Hope this helps a little at least.. :)

5 0
3 years ago
Based on the graph, which data point is most likely to have experimental
Len [333]

Answer:

B. 59 kg

Explanation:

From the graph you notice that a linear relation in indicated by the line joining the points such that the points on the line represent the data that show a correct relationship in the experiment.

This means that the point outside the line has an error .

This point is the value 59 kg that does not align with other values which are included in the graph.

8 0
2 years ago
What is a neutral atom?
IgorC [24]
A neutral atom is simply an atom that has no charge.
8 0
3 years ago
Read 2 more answers
Unpolarized light with intensity S is incident on a series of polarizing sheets. The first sheet has its transmission axis orien
jeka94

Answer:

Explanation:

Given

Initial Intensity of light is S

when an un-polarized light is Passed through a Polarizer then its intensity reduced to half.

When it is passed through a second Polarizer with its transmission axis \theat =45^{\circ}

S_1=S_0\cos ^2\theta

here S_0=\frac{S}{2}

S_1=\frac{S}{2}\times \frac{1}{(\sqrt{2})^2}

S_1=\frac{S}{4}

When it is passed through third Polarizer with its axis 90^{\circ} to first but \theta =45^{\circ} to second thus S_2

S_2=S_0\cos ^2\theta

S_2=\frac{S}{4}\times \frac{1}{2}

S_2=\frac{S}{8}

When middle sheet is absent then Final Intensity will be zero                    

3 0
3 years ago
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