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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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When designing a user interface, the most important information should be placed in the ______ of the screen.
musickatia [10]

Answer:

upper-left corner

Explanation:

Most vital information are positioned in a place where users can view them clearly and without obstruction.

3 0
3 years ago
A(n) 2602 kg van runs into the back of a(n)
forsale [732]

Answer:

8.5 m/s

Explanation:

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8 0
2 years ago
The question states: two large, parallel conducting plates are 12cm
AnnZ [28]

Answer:

1. 24375 N/C

2. 2925 V

Explanation:

d = 12 cm = 0.12 m

F = 3.9 x 10^-15 N

q = 1.6 x 10^-19 C

1. The relation between the electric field and the charge is given by

F = q E

So, E=\frac{F}{q}

E=\frac{3.9 \times 10^{-15}}{1.6 \times 10^{-19}}

E = 24375 N/C

2. The potential difference and the electric field is related by the given relation.

V = E x d

where, V be the potential difference, E be the electric field strength and d be the distance between the electrodes.

By substituting the values, we get

V = 24375 x 0.12 = 2925 Volt

6 0
3 years ago
The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi
Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which   describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}

Where,

B= Magnetic Field

l = length

\mu_0 = Vacuum permeability

\epsilon_0 = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}

Recall that the speed of light is equivalent to

c^2 = \frac{1}{\mu_0 \epsilon_0}

Then replacing,

B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}

B = \frac{r}{2C^2} \frac{dE}{dt}

Our values are given as

dE = 2150N/C

dt = 5s

C = 3*10^8m/s

D = 0.440m \rightarrow r = 0.220m

Replacing we have,

B = \frac{r}{2C^2} \frac{dE}{dt}

B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}

B =5.25*10^{-16}T

Therefore the magnetic field around this circular area is B =5.25*10^{-16}T

3 0
2 years ago
An electron of mass 9.11x10^-31 kg has an initial speed of 2.40x10^5 m/s. It travels in a straight line, and its speed increases
egoroff_w [7]

Answer: A) Force = 3.841*10^-18 N.

B) force (f) is 4.30* 10^12 times greater than the weight (Fg).

Explanation: mass of electronic charge = 9.11*10^-31kg

v = final velocity = 6.80*10^5 m/s

u = initial velocity = 2.40 * 10^5 m/s

S= distance covered = 4.8cm = 0.048m

a = acceleration

Since the acceleration of the electron is assumed to be constant, newton's laws of motion are valid.

Thus, recall that

v² = u² + 2aS

(6.80*10^5)² = ( 2.40*10^5)² + 2*a( 0.048)

46.24 * 10^10 = 5.76 * 10^10 + 0.096a

46.24 *10^10 - 5.76* 10^10 = 0.096a

40.48* 10^10 = 0.096a

a = 40.48 * 10^10/0.096

a = 4.2167*10^12m/s².

Force = mass * acceleration

Force = 9.11*10^-31 * 4.2167*10^12

Force = 3.841*10^-18 N.

Weight =Fg= mg where g = acceleration due gravity = 9.8m/s²

Fg= 9.11*10^-31 * 9.8

Fg = 8.9278* 10^-30 N

By comparing the force and the weight, we have that

F/Fg = 3.841 * 10^-18/8.9278 * 10^-30 = 4.30* 10^12.

This implies that the force (f) is 4.30* 10^12 times greater than the weight (Fg).

7 0
3 years ago
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