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iren2701 [21]
3 years ago
7

Earth is subject not only to the gravitational force of the Moon but also to the gravitational pull of the Sun. However, Earth i

s 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
1 answer:
Margaret [11]3 years ago
4 0

Answer:

option (B)

Explanation:

The gravitational force on earth depends on the mass of sun and the distance of sun from earth.

As the mass of sun is much more than the mass of moon, so the gravitational force of sun on earth is much more than the force on earth due to the moon.

The distance does not matter because the mass of sun is much more.

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Cerrena [4.2K]

Answer:

H2O is also known as for water

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2 years ago
Which will a positively charged objects attract
Vitek1552 [10]
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3 years ago
Read 2 more answers
During a hurricane in 2008, the Westin Hotel in downtownNew Orleans suffered damage. Suppose a piece of glass dropped near the t
tangare [24]

Answer:

  • <u>77.8 m/s, downward</u>

Explanation:

For uniform acceleration motion, the average speed is equal to half the soum of the initial velocity, Vi, and the final velocity, Vf

  • Average speed = (Vf + Vi)/2

Also, by definition, the average speed is the distance divided by the time:

  • Average speed = distance / time

Then:

  • (Vf + Vi)/2 = 300m/6.62s

Other kinematic equation for uniform acceleration is:

  • Vf = Vi + a×t

Since the window is falling and the air resistance is ignored, a = g (gravitational acceleration ≈ 9.8m/s²)

Replacing the known values we can set a system of two equations:

From (Vf + Vi)/2 = 300m/6.62s

(Vf + Vi) = 2 × 300m/6.62s

  • Vf + Vi = 90.634      equation 1

From Vf = Vi + a×t

Vf - Vi = 9.8 (6.62)

  • Vf - Vi = 64.876     equation 2

Adding the two equations:

  • 2Vf = 155.510

  • Vf = 77.8 m/s downward (velocities must be reported with their directions)
8 0
3 years ago
What is the torque τa about axis a due to the force f⃗ ? express the torque about axis a at cartesian coordinates (0,0)?
pentagon [3]

Answer:

\tau_a = F a sin \theta

Explanation:

The torque of a force is given by:

\tau = F d sin \theta

where

F is the magnitude of the force

d is the distance between the point of application of the force and the centre of rotation of the system

\theta is the angle between the direction of the force and d

In this problem, we have:

F, the force

a, the distance of application of the force from the centre (0,0)

\theta, the angle between the direction of the force and a

Therefore, the torque is

\tau_a = F a sin \theta

5 0
3 years ago
A box is moving along the x-axis and its position varies in time according to the expression:
Colt1911 [192]

Answer:

38.4 m/s

Explanation:

a) at t = 3.2s. x = 6 * 3.2^2 = 61.44 m

b) at t = 3.2 + Δt. x = 6*(3.2 + \Delta t)^2

c) As Δt approaches 0. We can find the velocity by the ratio of Δx/Δt

v = \frac{\Delta x}{\Delta t} = \frac{x_2 - x_1}{\Delta t}

v = \frac{6*(3.2 + \Delta t)^2 - 61.44}{\Delta t}

v = \frac{6(3.2^2 + 6.4\Delta t + \Delta t^2) - 61.44}{\Delta t}

v = \frac{61.44 + 38.4\Delta t + \Delta t^2 - 61.44}{\Delta t}

v = \frac{\Delta t(38.4 + \Delta t)}{\Delta t}

v = 38.4 + \Delta t

As Δt approaches 0, v = 38.4 + 0 = 38.4 m/s

3 0
3 years ago
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