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STALIN [3.7K]
3 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]3 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]3 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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prohojiy [21]

Answer:

v= 449.8 m/s

Explanation:

Given data

Frequency=  346Hz

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The expression below is used to find the speed

v= f \lambda\\\\

substitute

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Hence the speed is v= 449.8 m/s

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Can a falling object reach terminal velocity in outer space? Explain.
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ou have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 660.0 kg and was trav
Montano1993 [528]

Answer:

    vₐ₀ = 29.56 m / s

Explanation:

In this exercise the initial velocity of car A is asked, to solve it we must work in parts

* The first with the conservation of the moment

* the second using energy conservation

let's start with the second part

we must use the relationship between work and kinetic energy

             W = ΔK                             (1)

for this part the mass is

             M = mₐ + m_b

the final velocity is zero, the initial velocity is v

friction force work is

              W = - fr x

the negative sign e because the friction forces always oppose the movement

we write Newton's second law for the y-axis

              N -W = 0

              N = W = Mg

friction forces have the expression

              fr =μ N

              fr = μ M g

we substitute in 1

               -μ M g x = 0 - ½ M v²

             v² = 2 μ g x

let's calculate

              v² = 2  0.750  9.8  6.00

              v = ra 88.5

              v = 9.39 m / s

Now we can work on the conservation of the moment, for this part we define a system formed by the two cars, so that the forces during the collision are internal and therefore the tsunami is preserved.

Initial instant. Before the crash

         p₀ = + mₐ vₐ₀ - m_b v_{bo}

instant fianl. Right after the crash, but the cars are still not moving

         p_f = (mₐ + m_b) v

         p₀ = p_f

         + mₐ vₐ₀ - m_b v_{bo} = (mₐ + m_b) v

           

         mₐ vₐ₀ = (mₐ + m_b) v + m_b v_{bo}

let's reduce to the SI system

          v_{bo} = 64.0 km / h (1000m / 1km) (1h / 3600s) = 17.778 m / s

let's calculate

         660 vₐ₀ = (660 +490) 9.39 + 490 17.778

         vₐ₀ = 19509.72 / 660

         vₐ₀ = 29.56 m / s

we can see that car A goes much faster than vehicle B

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PLzzz helpppp
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Answer: Charging.

Explanation: ...

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2 years ago
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a car headed north at 15 m/s accelerates for 4.25 s to reach a velocity of 28.3 m/s. What is the acceleration of the car?
Liono4ka [1.6K]

<u>Answer:</u>

The acceleration of the car is 3.13 m/s^2

<u>Explanation:</u>

In the question it is given that car initially heads north with a velocity 15 m/s. It then accelerates for 4.25 s and in the end its velocity is 28.3 m/s.

initial velocity u = 15 m/s

time t=4.25 s

final velocity v=28.3 m/s

The equation of acceleration is

a= \frac{(v-u)}{t}

= \frac{(28.3-15)}{4.25} =  \frac {13.3}{4.25} =3.13m/s^2

The value of acceleration is positive, here since the car is speeding up. If it was slowing down the value of acceleration would be negative.

7 0
3 years ago
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