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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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What current flow (I) is associated with an input voltage of 5.0V and resistors R1 = 1.5 kiloohms and R2 = 0.5 kiloohms? Calcula
Y_Kistochka [10]

Answer:

current in series is 2.50 mA

current in parallel is 13.51 mA

Explanation:

given data

voltage = 5 V

resistors R1 = 1.5 kilo ohms

resistors R2 = 0.5 kilo ohms

to given data

current flow

solution

current flow in series is express as here

current = voltage / resistor    .................1

put here all value  in equation 1

current = 5 / (1.5 + 0.5)

current = 5 / 2.0

so current = 2.50 mA

and

current flow in parallel is express as

current = voltage / resistor   ....................2

put here all value in equation  2

current = 5 / (1/ (1/1.5 + 1/0.5))

current = 5 / 0.37

so current = 13.31 mA

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3 years ago
A 20 kg mass is dropped from a tall rooftop and accelerates at 9.8 m/s2. What is the weight of the dropped object?
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Suppose you want to determine the resistance of a resistor that is nominally 100 . You should be able to apply 10 V across the r
Butoxors [25]

Answer:

a) For y = 102 mA, R = 98.039 ohms

For y = 97 mA, R = 103.09 ohms

b) Check explanatios for b

Explanation:

Applied voltage, V = 10 V

For the first measurement, current y_{1} = 102 mA = 0.102 A

According to ohm's law, V = IR

R = V/I

Here, I = y_{1}

R = \frac{V}{y_{1} } \\R = \frac{10}{0.102} \\R = 98.039 ohms

For the second measurement, current y_{2} = 97 mA = 0.097 A

R = \frac{V}{y_{2} }

R = \frac{10}{0.097} \\R = 103 .09 ohms

b) y = \left[\begin{array}{ccc}y_{1} &y_{2} \end{array}\right] ^{T}

y = \left[\begin{array}{ccc}y_{1} \\y_{2} \end{array}\right]

y = \left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3}  \end{array}\right]

A linear equation is of the form y = Gx

The nominal value of the resistance = 100 ohms

x = \left[\begin{array}{ccc}100\end{array}\right]

\left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3}  \end{array}\right] =  \left[\begin{array}{ccc}G_{1} \\G_{2}  \end{array}\right] \left[\begin{array}{ccc}100\end{array}\right]\\\left[\begin{array}{ccc}G_{1} \\G_{2}  \end{array}\right] =  \left[\begin{array}{ccc}102*10^{-5} \\97*10^{-5}  \end{array}\right]

3 0
2 years ago
Now the elevator is moving downward with a velocity of v = -2.8 m/s but accelerating upward with an acceleration of a = 5.5 m/s2
borishaifa [10]

Answer:

160.75 N

Explanation:

The downward velocity has no effect on the force situation, it is only changes in velocity (plus, of course, gravity, which is always there) that require a force. At constant velocity, the bottom spring s_3 is supporting its mass m_3 to balance gravity.

As the elevator slows, though, it also ends up slowing down the spring arrangement, too. However, because the stretching takes time, it means that some damped harmonic motion will be set up in the spring chain.

When the motion has finally damped out, the net force the bottom spring s3 exerts on m3 has two components--that of gravity and of the deceleration of the elevator:

F_3net = m3 * (g + a) = 10.5×(9.81+5.5)= 10.5×15.31= 160.75 N

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3 years ago
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