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Mariana [72]
3 years ago
6

Two satellites revolve around the Earth. Satellite A has mass m and has an orbit of radius r. Satellite B has mass 6m and an orb

it of unknown radius rb. The forces of gravitational attraction between each satellite and the Earth is the same. Find rb.
Physics
1 answer:
melomori [17]3 years ago
8 0

Answer:

aaaaa

Explanation:

M = Mass of the Earth

m = Mass of satellite

r = Radius of satellite

G = Gravitational constant

F=G\frac{Mm}{r^2}

F=G\frac{M6m}{r_b^2}

G\frac{Mm}{r^2}=G\frac{M6m}{r_b^2}\\\Rightarrow \frac{1}{r^2}=\frac{6}{r_b^2}\\\Rightarrow \frac{r_b^2}{r^2}=6\\\Rightarrow \frac{r_b}{r}=\sqrt{6}\\\Rightarrow r_b=2.44948r

r_b=2.44948r

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Your electric drill rotates initially at 5.21 rad/s. You slide the speed control and cause the drill to undergo constant angular
RUDIKE [14]

Answer:

The drill's angular displacement during that time interval is 24.17 rad.

Explanation:

Given;

initial angular velocity of the electric drill, \omega _i = 5.21 rad/s

angular acceleration of the electric drill, α = 0.311 rad/s²

time of motion of the electric drill, t = 4.13 s

The angular displacement of the electric drill at the given time interval is calculated as;

\theta = \omega _i t \ + \ \frac{1}{2}\alpha t^2\\\\\theta = (5.21 \ \times \ 4.13) \ + \ \frac{1}{2}(0.311)(4.13)^2\\\\\theta = (21.5173 ) \ + \ (2.6524)\\\\\theta =24.17 \ rad

Therefore, the drill's angular displacement during that time interval is 24.17 rad.

6 0
3 years ago
when an object is placed near a concave mirror, at what position does it forms a magnified and erect image.​
alexira [117]

Answer:

Between the principal focus and the pole of the mirror

4 0
3 years ago
A wave front has the form of a
podryga [215]
Your answer is "<span>surface of a sphere"

Hope this helps.</span>
3 0
3 years ago
A student witnesses a flash of lightning and then t=2.5s later the student hears assiciated clap of thunder. (please show work)
frosja888 [35]

Answer:

857.5 m

2.8583×10⁻⁶ seconds

Explanation:

Time taken by the sound of the thunder to reach the student = 2.5 s

Speed of sound in air is 343 m/s

Speed of light is 3×10⁸ m/s

Distance travelled by the sound = Time taken by the sound × Speed of sound in air

⇒Distance travelled by the sound = 2.5×343 = 857.5 m

⇒Distance travelled by the sound = 857.5 m

Time taken by light = Distance the light travelled / Speed of light

\text{Time taken by light}=\frac{857.5}{3\times 10^8}\\\Rightarrow  \text{Time taken by light}=2.8583\times 10^{-6}

Time taken by light = 2.8583×10⁻⁶ seconds

3 0
3 years ago
A dumbell has a mass of 95 kg. What force must be applied to accelerate it upward at 2.2 m/s2?
Sveta_85 [38]
A :-) F = ma
Given - m = 95 kg
a = 2.2 m/s^2
Solution -
F = ma
F = 95 x 2.2
F = 209

.:. The force is 209 N
5 0
3 years ago
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