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Montano1993 [528]
3 years ago
13

Calculate the location xcm of the center of mass of the Earth-Moon system. Use a coordinate system in which the center of the Ea

rth is at x
Physics
1 answer:
Jet001 [13]3 years ago
8 0

Answer:

The center of mass of the Earth-Moon system is 4.673 kilometers away from center of Earth.

Explanation:

Let suppose that planet and satellite can be treated as particles. The masses of Earth and Moon (m_{E}, m_{M}) are 5.972\times 10^{24}\,kg and 7.349\times 10^{22}\,kg, respectively. The distance between centers is 384,403 kilometers. The location of the center of mass can be found by using weighted averages:

\bar x = \frac{x_{E}\cdot m_{E}+x_{M}\cdot m_{M}}{m_{E}+m_{M}}

If x_{E} = 0\,km and x_{M} = 384,403\,km, then:

\bar x = \frac{(0\,km)\cdot (5.972\times 10^{24}\,kg)+(384,403\,km)\cdot (7.349\times 10^{22}\,kg)}{5.972\times 10^{24}\,kg+7.349\times 10^{22}\,kg}

\bar x = 4.673\,km

The center of mass of the Earth-Moon system is 4.673 kilometers away from center of Earth.

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A cell phone uses a 3. 0 V battery. The circuit board it uses needs a 0. 05 A current. What size resistor is needed to generate
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The value of resistance will be 60 ohms. Resistance is the opposition force offered to the flow of electric current. The ratio of the voltage to the electric current passing through the wire.

<h3>What is a resistor?</h3>

A resistor is a two-terminal passive electrical component that acts as a circuit element by implementing electrical resistance.

Resistors are used to limit current flow, alter signal levels, divide voltages, bias active devices, and complete transmission lines in electronic circuits.

The given data in the problem is;

V is the voltage =  3. 0 V

I is value of electric current=  0. 05 A

R is the resistance=?

According to ohm's law, the potential difference between the two ends of the conductor is equal to the product of electric current and the resistance generated.

So from the ohm law;

V= IR

\rm R = \frac{V}{I} \\\\ \rm R = \frac{3}{0.05} \\\\ \rm R =60 \ ohm

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To learn more about the resistor refer to the link;

brainly.com/question/1322424

3 0
2 years ago
Estimate the electric field at a point 2.40 cm perpendicular to the midpoint of a uniformly charged 2.00-m-long thin wire carryi
nadya68 [22]

Answer:

E = 1.85*10^{12}\frac{N}{C}

Explanation:

Hi!

The perpendicular distance 2.4cm, is much less than the distance to both endpoints of the wire, which is aprox 1m. Then the edge effect is negligible at this field point, and we can aproximate the wire as infinitely long.

The electric filed of an infinitely long wire is easy to calculate. Let's call z the axis along the wire. Because of its simmetry (translational and rotational), the electric field E must point in the radial direction,  and it cannot depende on coordinate z. To calculate the field Gauss law is used, as seen in the image, with a cylindrical gaussian surface. The result is:

E = \frac{\lambda}{2\pi \epsilon_0 r}\\\lambda=\text{charge per unit length}=\frac{4.95 \mu C}{2 m} = 2.475 \frac{C}{m}\\r=\text{perpendicular distance to wire}\\\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}

Then the electric field at the point of interest is estimated as:

E = \frac{\22.475}{2\pi*( 8.85*10^{-12})*(2.4*10^{-2})}\frac{N}{C}=1.85*10^{12}\frac{N}{C}

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3 years ago
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