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s2008m [1.1K]
3 years ago
10

A 213.7 kg satellite is in a circular orbit of 22,236 miles (35,768 km) in radius. The force keeping the satellite in orbit is 4

4.1 N. What is the velocity (speed) of the satellite
Physics
2 answers:
nikdorinn [45]3 years ago
5 0

Answer:

<em>v = 3.08 k ms^-1</em>

Explanation:

As we know that, v=\sqrt{GM/r}

where r = R + h = 35768 + 6400 = 42168 km = 4.2168 x 10^7 m

G = 6.673 x 10^-11 N m^-2 kg^-2 and M = 6.0 x 10^24 kg

By substituting the required values, we get...

v = \sqrt{}6.673 x 10^-11 x 6.0 x 10^24 / 4.2168 x 10^7 = \sqrt{}9.494 x 10^6 ms^-1 = 3081.2 ms^-1 = 3.08 k ms^-1

julia-pushkina [17]3 years ago
5 0

Answer:

2716.84 m/s

Explanation:

Using,

F = mv²/r................. Equation 1

Where F = The force keeping the satellite in the orbit, m = mass of the satellite, v = velocity of the satellite, r = radius of the circular orbit.

make v the subject of equation 1

v = √(Fr/m).................. Equation 2

Given: F = 44.1 N, r = 35768 km = 35768000 m, m = 213.7 kg.

Substitute into equation 2

v = √(44.1×35768000/213.7)

v = √(7381229.761)

v = 2716.84 m/s

Hence the velocity of the satellite = 2716.84 m/s

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Answer:

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Explanation:

maybe it will cause the eletric cause a fire that will mean it will spread till the fire men should come

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Obi Wan uses his Jedi mind tricks to compel people to do his will. The words he uses begin in his vocal cords which _____ the ai
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A spherically spreading EM wave comes from an 1800-W source. At a distance of 5.0 m, what is the intensity, and what is the rms
Aleonysh [2.5K]

Explanation:

It is given that,

Power of EM waves, P = 1800 W

We need to find the intensity at a distance of 5 m. Also, the rms value of the electric field.

Intensity,

I=\dfrac{P}{4\pi r^2}\\\\I=\dfrac{1800}{4\pi\times (5)^2}\\\\I=5.72\ W/m^2

The formula that is used to find the rms value of the electric field is as follows :

I=\epsilon_o cE^2_{rms}

c is speed of light and \epsilon_o is permittivity of free space

So,

E_{rms}=\sqrt{\dfrac{I}{\epsilon_o c}}\\\\E_{rms}=\sqrt{\dfrac{5.72}{8.85\times 10^{-12}\times 3\times 10^8}}\\\\E_{rms}=46.41\ V/m

Hence, this is the required solution.

4 0
3 years ago
One string of a certain musical instrument is 70.0 cm long and has a mass of 8.79 g . It is being played in a room where the spe
Svetach [21]

To solve this problem we will apply the concepts of linear mass density, and the expression of the wavelength with which we can find the frequency of the string. With these values it will be possible to find the voltage value. Later we will apply concepts related to harmonic waves in order to find the fundamental frequency.

The linear mass density is given as,

\mu = \frac{m}{l}

\mu = \frac{8.79*10^{-3}}{70*10^{-2}}

\mu = 0.01255kg/m

The expression for the wavelength of the standing wave for the second overtone is

\lambda = \frac{2}{3} l

Replacing we have

\lambda = \frac{2}{3} (70*10^{-2})

\lambda = 0.466m

The frequency of the sound wave is

f_s = \frac{v}{\lambda_s}

f_s = \frac{344}{0.768}

f_s = 448Hz

Now the velocity of the wave would be

v = f_s \lambda

v = (448)(0.466)

v = 208.768m/s

The expression that relates the velocity of the wave, tension on the string and linear mass density is

v = \sqrt{\frac{T}{\mu}}

v^2 = \frac{T}{\mu}

T= \mu v^2

T = (0.01255kg/m)(208.768m/s)^2

T = 547N

The tension in the string is 547N

PART B) The relation between the fundamental frequency and the n^{th} harmonic frequency is

f_n = nf_1

Overtone is the resonant frequency above the fundamental frequency. The second overtone is the second resonant frequency after the fundamental frequency. Therefore

n=3

Then,

f_3 = 3f_1

Rearranging to find the fundamental frequency

f_1 = \frac{f_3}{3}

f_1 = \frac{448Hz}{3}

f_1 = 149.9Hz

7 0
3 years ago
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