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STALIN [3.7K]
4 years ago
11

Tidal forces are gravitational forces exerted on different parts of a body by a second body. Their effects are particularly visi

ble on the earth's surface in the form of tides. To understand the origin of tidal forces, consider the earth-moon system to consist of two spherical bodies, each with a spherical mass distribution. Let re be the radius of the earth, m be the mass of the moon, and G be the gravitational constant.
1) Earth is subject not only to the gravitational force of the Moon but also to the gravitational pull of the Sun. However, Earth is much farther away from the Sun than it is from the Moon. In fact, the center of Earth is at an average distance of 1.5×1011m from the center of the Sun. Given that the mass of the Sun is 1.99×1030kg, which of the following statements is correct?

A) The force exerted on Earth by the Sun is weaker than the corresponding force exerted by the Moon.
B) The force exerted on Earth by the Sun is stronger than the corresponding force exerted by the Moon.
C) The force exerted on Earth by the Sun is of the same order of magnitude of the corresponding force exerted by the Moon.
Physics
2 answers:
Nastasia [14]4 years ago
4 0

Answer:

B) True.

Explanation:

In this exercise to answer the statements we must calculate the force on the earth by the two bodies.

In both cases, the law of universal gravitation describes the process.

    F = G m₁ m₂ / r²

Let's calculate each force

- bodies Moon - Earth

Let's call m₁ the mass of the earth (m₁ = me), m₂ the mass of the moon (m₂ = m), the distance from the earth to the moon is r₁ = 3.84 10⁸m and the radius of the earth is re. The force on the tide that is a body on the surface of the Earth have a distance

      R₁ = r₁ -re

      R₁ = 3.84 10⁸ - 6.37 10⁶

      R₁ = 3.77 10⁸ m

Let's calculate

    F₁ = G me m / R₁²

    F₁ = (G me) 7.36 10²² / (3.77 10⁸)²

    F₁ = (G me) 5.2 10⁵

- bodies Earth -Sun

Let's call the mass of the sun M (m2 = M) the distance from the sun to the earth is 1.5 10¹¹ m, so the distance to the surface of the earth

    R₂ = r₂ - re

    R₂ = 1.5 10¹¹ - 6.37 10⁶

    R₂ = 1.5 10¹¹ m

The radius of the earth is too small compared to the earth-sun distance

Let's calculate

    F₂ = G me M / R₂²

    F₂ = (G me) 1.99 10³⁰ / (1.5 10¹¹)²

    F₂ = (G me) 8.8 10⁷

Let's see the statements:

A) False. It´s oppsote

B) True. In the previous part it has a differentiated 10² orders of magnitude

C) False. We saw that they are very different

andrew11 [14]4 years ago
4 0

Answer:

B. The force exerted on Earth by the Sun is stronger than the corresponding force exerted by the Moon.

Explanation:

The gravitational force between two planets is given by the following formula:

F = G \cdot \frac{m\cdot M}{r^{2}}

Where:

G - Gravitational constant.

m, M - Masses of planets.

r - Distance between planets.

The following is the description of the forces exerted on Earth as both as by the Sun and by the Moon:

Earth - Moon

F = \left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot \left[\frac{(5.972\times 10^{24}\,kg)\cdot (7.348\times 10^{22}\,kg)}{(384000000\,m)^{2}} \right]

F = 1.986\times 10^{20}\,N

Earth - Sun

F = \left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot \left[\frac{(5.972\times 10^{24}\,kg)\cdot (1.99\times 10^{30}\,kg)}{(1.5\times 10^{11}\,m)^{2}} \right]

F = 3.525\times 10^{22}\,N

The force exerted on Earth by the Sun is greater than the force by the Moon. Hence, the answer is B.

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What mathematical relationship between variables is suggested by a graph showing a diagonal line from the lower left to the uppe
Tresset [83]

Answer:

Direct proportionality

Explanation:

The graph of variables that are directly proportional such as the temperature and volume of a gas has a graph consisting of a diagonal line that from the lower left of the graph to the upper right of the graph

According to Charles law, the volume of a given mass of gas is directly proportional to its temperature in Kelvin at constant pressure

Charles law can be represented mathematically as V ∝ T

From which we have;

V₁/T₁ = V₂/T₂, therefore, the graph of V to T has a constant slope, ΔV/ΔT.

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3 years ago
How is a hypothesis different from a guess? A. It is different because it is a scientist making the guess. B. It is not differen
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Answer:

C

Explanation:

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Coquina is a sedimentary rock you looked at in the lab that consists of tiny chunks of sea shells all stuck together by other mi
HACTEHA [7]

Coquina rock is deposited on the water bodies of the Earth's surface and oceans.

<u>Explanation:</u>

Coquina is nothing but a type of sedimentary rock generally limestone that is formed by the chunks of shells, trilobites, brachiopods, or other invertebrates.

Coquina is deposited on the oceans and water bodies of the Earth's surface. Coquina looks like a soft white rock; therefore it can be used in construction of buildings.

8 0
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How long is the photons journey from the milky way to earth
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3 years ago
Will binoculars work properly if their prisms (assume n = 1.50) are immersed in water (nwater= 1.33)? assume that binoculars wor
Amanda [17]

The binoculars will not work properly.

Please refer to the diagram attached.

Binoculars work on the principle of total internal reflection. The incident ray from air enters the prism through on face normally and it strikes the opposite face at an angle of 45°. The ray undergoes Total internal reflection if its critical angle is less than the angle of incidence,which , in this case is 45°.

The critical anglei_{c} is related to the refractive index of the material of glass μ as shown below.

\mu =\frac{1}{sini_c}

For a refractive index equal to 1.5, the critical angle is found as shown below.

sini_c=\frac{1}{\mu} \\ =\frac{1}{1.5} \\ =0.666

Take the inverse of the value 0.666 to determine the critical angle.

sin^{-1} (0.666)=41.8^o

The critical angle in air is less than the angle of incidence, and hence total internal reflection occurs in air.

When the binoculars are immersed in water, the ray passes into the glass through water. therefore, the refractive index of the prism when immersed in water is given by,

\mu_g_w=\frac{\mu_g}{\mu_w}

Therefore the critical angle <em>c</em> in this case is given by,

sinc=\frac{\mu_w}{\mu_g}

Substitute 1.33 for \mu_w and 1.5 for \mu_g.

sinc=\frac{\mu_w}{\mu_g} \\ =\frac{1.33}{1.5} \\ =0.8866

Take the inverse of the value 0.8866.

c=sin^{-1} (0.8866)\\ =62^o

Since the critical angle of the prism when immersed in water, is 62°, which is greater than the angle of incidence of 45° required for viewing the object, the binoculars which are set for Total reflection in air, will not function when immersed in water.

4 0
4 years ago
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