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cluponka [151]
2 years ago
14

In this problem, you will calculate the location of the center of mass for the Earth-Moon system, and then you will calculate th

e center of mass of the Earth-Moon-Sun system. The mass of the Moon is 7.35×1022 kg , the mass of the Earth is 6.00×1024 kg , and the mass of the sun is 2.00×1030 kg . The distance between the Moon and the Earth is 3.80×105 km . The distance between the Earth and the Sun is 1.50×108 km
A.) Where is the center of mass of the Earth-Moon system? The radius of the Earth is 6378 km and the radius of the Moon is 1737 km. Select one of the answers below:

a. The center of mass is exactly in the center between the Earth and the Moon.
b. The center of mass is nearer to the Moon than the Earth, but outside the radius of the Moon.
c. The center of mass is nearer to the Earth than the Moon, but outside the radius of the Earth.
d. The center of mass is inside the Earth.
e. The center of mass is inside the Moon.
f.) Calculate the location of the center of mass of the Earth-Moon-Sun system during a full Moon. A full Moon occurs when the Earth, Moon, and Sun are lined up as shown in the figure. (Figure 2) Use a coordinate system in which the center of the sun is at x=0 and the Earth and Moon both lie along the positive x direction.
Physics
1 answer:
Radda [10]2 years ago
8 0

Answer:

a) Option D is correct.

The center of mass between the Eartg and the moon is inside the Earth.

Explanation:

Given,

Mass of the moon = (7.35×10²²) kg

Mass of the Earth = (6.00×10²⁴) kg

Mass of the Sun = (2.00×10³⁰) kg

Distance between the Earth and the moon = (3.80×10⁵) km

Distance between the Earth and the Sun = (1.50×10⁸) km

With the assumption that all.of the bodies being considered are on the same straight line on the x-axis,

Note that Centre of mass is given as

C.M = (Σmx)/(Σm)

For the Earth-moon system, let the earth be x=0, then the moon is at x = (3.80 × 10 5) km away.

C.M = (Σmx)/(Σm)

Σmx = (6.00×10²⁴)) × (0) + (7.35×10²²) × (3.80×10⁵) = (2.793 × 10²⁸) kg.km

Σm = (6.00×10²⁴) + (7.35×10²²) = (6.0735 × 10²⁴) kg

CM = (2.793 × 10²⁸) ÷ (6.0735 × 10²⁴)

CM = (4.60 × 10³) km = 4600 km

This means the centre of mass is 4600 km from the Earth.

The Earth's radius = 6378 km

Hence, the centre if mass is inside the Earth.

Hope this Helps!!!

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Psi communication refers to the transfer of information through a/an ________ process.
shtirl [24]

Answer:

unknown

Explanation:

Psi communication: The term "Psi communication" was initially originated in an article named 'Mass Communication and Para-Social Interaction' during 1956 by Donald Horton, and is also denoted as "parasocial interaction". It is described as a phenomenon that involves "one-sided interaction" between people, encompassing the viewer knowing a specific celebrity well whereas the viewer himself or herself is being completely unknown for the celebrity in return.

In the question above, the given statement represents that the Psi communication is responsible for transforming via an unknown process as one of its members are considered as unknown of the transfer, so the correct answer would be "unknown process".

4 0
2 years ago
An overhead projector lens is 32.0 cm from a slide (the object) and has a focal length of 30.1 cm. What is the magnification of
puteri [66]

Answer: 15.8

Explanation:

You are given that the

Object distance U = 32 cm

Focal length F = 30.1 cm

First calculate the image distance V by using the formula

1/F = 1/U + 1/V

Substitute F and V into the formula

1/30.1 = 1/32 + 1/V

1/V = 1/30.1 - 1/32

1/V = 0.00197259

Reciprocate both sides

V = 506.94 cm

Magnification M is the ratio of image distance to object distance.

M = V/U

substitute the values of V and U into the formula

M = 506.94/32

M = 15.8

Therefore, the magnification of the image is 15.8 or approximately 16.

6 0
2 years ago
Think of something from everyday life that follows a two-dimensional path. It could be a kicked football, a bus that's turning a
OLEGan [10]

Answer:

Let us consider the case of a bus turning around a corner with a constant velocity, as the bus approaches the corner, the velocity at say point A is Va, and is tangential to the curve with direction pointing away from the curve. Also, the velocity at another point say point B is Vb and is also tangential to the curve with direction pointing away from the curve.<em> </em><em>Although the velocity at point A and the velocity at point B have the same magnitude, their directions are different (velocity is a vector quantity), and hence we have a change in velocity. By definition, an acceleration occurs when we have a change in velocity, so the bus experiences an acceleration at the corner whose direction is away from the center of the corner</em>.

The acceleration is not aligned with the direction of travel because<em> the change in velocity is at a tangent (directed away) to the direction of travel of the bus.</em>

4 0
2 years ago
The starter armature is rubbing on the field coils. technician a says the bushings need to be replaced. technician b says the br
Viefleur [7K]
In a situation in which the smarter armiture is rubbing on the field the most commin reason are defective bushings. So, the ststement the techician A ssys that the brushings shojld be replaced is correct. Techinican A is right.
6 0
2 years ago
Read 2 more answers
A trebuchet was a hurling machine built to attack the walls of a castle under siege. A large stone could be hurled against a wal
Studentka2010 [4]

(a) 18.9 m/s

The motion of the stone consists of two independent motions:

- A horizontal motion at constant speed

- A vertical motion with constant acceleration (g=9.8 m/s^2) downward

We can calculate the components of the initial velocity of the stone as it is launched from the ground:

u_x = v_0 cos \theta = (25.0)(cos 41.0^{\circ})=18.9 m/s\\u_y = v_0 sin \theta = (25.0)(sin 41.0^{\circ})=16.4 m/s

The horizontal velocity remains constant, while the vertical velocity changes due to the acceleration along the vertical direction.

When the stone reaches the top of its parabolic path, the vertical velocity has became zero (because it is changing direction): so the speed of the stone is simply equal to the horizontal velocity, therefore

v=18.9 m/s

(b) 22.2 m/s

We can solve this part by analyzing the vertical motion only first. In fact, the vertical velocity at any height h during the motion is given by

v_y^2 - u_y^2 = 2ah (1)

where

u_y = 16.4 m/s is the initial vertical velocity

v_y is the vertical velocity at height h

a=g=-9.8 m/s^2 is the acceleration due to gravity (negative because it is downward)

At the top of the parabolic path, v_y = 0, so we can use the equation to find the maximum height

h_{max} = \frac{-u_y^2}{2a}=\frac{-(16.4)^2}{2(-9.8)}=13.7 m

So, at half of the maximum height,

h = \frac{13.7}{2}=6.9 m

And so we can use again eq(1) to find the vertical velocity at h = 6.9 m:

v_y = \sqrt{u_y^2 + 2ah}=\sqrt{(16.4)^2+2(-9.8)(6.9)}=11.6 m/s

And so, the speed of the stone at half of the maximum height is

v=\sqrt{v_x^2+v_y^2}=\sqrt{18.9^2+11.6^2}=22.2 m/s

(c) 17.4% faster

We said that the speed at the top of the trajectory (part a) is

v_1 = 18.9 m/s

while the speed at half of the maximum height (part b) is

v_2 = 22.2 m/s

So the difference is

\Delta v = v_2 - v_2 = 22.2 - 18.9 = 3.3 m/s

And so, in percentage,

\frac{\Delta v}{v_1} \cdot 100 = \frac{3.3}{18.9}\cdot 100=17.4\%

So, the stone in part (b) is moving 17.4% faster than in part (a).

4 0
2 years ago
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