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oksano4ka [1.4K]
3 years ago
7

A common, though incorrect, statement is, "The Moon orbits the Earth." That creates an image of the Moon?s orbit that looks like

that shown in the figure. (Figure 1) The Earth's gravity pulls on the Moon, causing it to orbit. However, by Newton?s third law, it is known that the Moon exerts a force back on the Earth. Therefore, the Earth should move in response to the Moon. Thus a more accurate statement is, "The Moon and the Earth both orbit the center of mass of the Earth-Moon system." In this problem, you will calculate the location of the center of mass for the Earth-Moon system, and then you will calculate the center of mass of the Earth-Moon-Sun system. The mass of the Moon is 7.35?1022kg , the mass of the Earth is 6.00?1024kg , and the mass of the sun is 2.00?1030kg . The distance between the Moon and the Earth is 3.80?105km . The distance between the Earth and the Sun is 1.50?108km . Part A Calculate the location x_cm of the center of mass of the Earth-Moon system. Use a coordinate system in which the center of the Earth is at x=0 and the Moon is located in the positive x direction.
Physics
1 answer:
pychu [463]3 years ago
8 0

Answer:

x_{cm} = 4.6 10⁶ m

Explanation:

The definition of mass center is

    x_{cm} = 1/M ∑ xi mi

Where M is the total mass, mi and xi the mass and position of each body

Let us apply this equation to our case, as they indicate that we take the center of reference to Earth, its distance is zero; let's write the data they give us

Earth

    M = 6.00 10²⁴ kg

    r1 = 0

Moon

    m = 7.35 10²² kg

    r = 3.80 10⁵ km (1000m / 1km) = 3.80 10⁸ m

Let's calculate

        x_{cm} = 1 /(m + M) (0 + m r)

        x_{cm} = m / (m + M) r

        x_{cm} = 7.35 10²² / (7.35 10²² + 600 10²²) 3.80 10⁸

        x_{cm} = 4.6 10⁶ m

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Stars form from an accumulation of gas and dust, which collapses due to gravity and starts to form stars. Stars are typically classified by their spectrum in what is known as the Morgan-Keenan or MK system.
6 0
3 years ago
Three point charges are arranged on a line. Charge q3 = 5 nC and is at the origin. Charge q2 = - 3 nC and is at x = 4 cm. Charge
Taya2010 [7]

Answer:

q₁ = + 1.25 nC

Explanation:

Theory of electrical forces

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Known data

q₃=5 nC

q₂=- 3 nC

d₁₃=  2 cm

d₂₃ = 4 cm

Graphic attached

The directions of the individual forces exerted by q1 and q₂ on q₃ are shown in the attached figure.

For the net force on q3 to be zero F₁₃ and F₂₃ must have the same magnitude and opposite direction, So,  the charge q₁ must be positive(q₁+).

The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs ,then. F₁₃ is directed to the left (-x).

The force (F₂₃) of q₂ on q₃ is attractive because the charges have opposite signs.  F₂₃ is directed to the right (+x)

Calculation of q1

F₁₃ = F₂₃

\frac{k*q_{1}*q_3 }{(d_{13})^{2}  } = \frac{k*q_{2}*q_3 }{(d_{23})^{2}  }

We divide by (k * q3) on both sides of the equation

\frac{q_{1} }{(d_{13})^{2} } = \frac{q_{2} }{(d_{23})^{2} }

q_{1} = \frac{q_{2}*(d_{13})^{2}   }{(d_{23} )^{2}  }

q_{1} = \frac{5*(2)^{2} }{(4 )^{2}  }

q₁ = + 1.25 nC

3 0
3 years ago
How long does it take someone to run 8 miles if they are running at a speed of 7 km/hr?
ZanzabumX [31]

Answer:

1.84 hours or 110.4 minutes

Explanation:

8 miles             1.61 km                 1 hr

------------    x   -----------------  x   -------------- = 1.84 hours or 110.4 minutes

                          1 mile                7  km    

7 0
3 years ago
London lives 1200 meters south of Chick-fil-A and is thinking about going there for lunch. When she is ready to leave, she reali
saul85 [17]

Answer:

  1750 m

Explanation:

The distance traveled is the 750 meters to the Aunt's house plus the 1000 m from there to Chick-fil-A.

  750 +1000 = 1750 . . . meters traveled

6 0
3 years ago
You drive a car east on the highway at 26 m/s. Another car passes you moving east traveling at 32 m/s. How fast do you view the
tamaranim1 [39]

The speed of the car passing you is 6 m/s while car is moving 6 m/s behind the car.

<h3>Relative velocity of the car</h3>

The speed of the car passing you is determined by applying relative velocity principle as shown below;

Vr = Va - Vb

Vr = 26 m/s - 32 m/s

Vr = -6 m/s

Thus, the speed of the car passing you is 6 m/s while car is moving 6 m/s behind the car.

Learn more about relative velocity here: brainly.com/question/17228388

#SPJ1

3 0
2 years ago
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