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Hatshy [7]
2 years ago
9

Who invented the telephone?

Physics
2 answers:
krok68 [10]2 years ago
5 0
<h2>Answer:</h2><h3><em><u>Alexander Graham Bell</u></em></h3><h2>Explanation:</h2>

Alexander Graham Bell is often credited as the inventor of the telephone since he was awarded the first successful patent.

tia_tia [17]2 years ago
3 0
Alexander graham bell
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You are pulling a child in a wagon. The rope handle is inclined upward at a 60∘ angle. The tension in the handle is 20 N.
dem82 [27]
  • Angle (θ) = 60°
  • Force (F) = 20 N
  • Distance (s) = 200 m
  • Therefore, work done
  • = Fs Cos θ
  • = (20 × 200 × Cos 60°) J
  • = (20 × 200 × 1/2) J
  • = (20 × 100) J
  • = 2000 J

<u>Answer</u><u>:</u>

<u>2</u><u>0</u><u>0</u><u>0</u><u> </u><u>J</u>

Hope you could get an idea from here.

Doubt clarification - use comment section.

6 0
2 years ago
Read 2 more answers
Questions 15 out of 20
Olin [163]

Answer:

D

Explanation:

7 0
3 years ago
lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A
qaws [65]

2.6×10^6\:\text{m}

Explanation:

The acceleration due to gravity g is defined as

g = G\dfrac{M}{R^2}

and solving for R, we find that

R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration F_c experienced by the satellite is equal to the gravitational force F_G or

F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)

The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as

v = \dfrac{2\pi r}{T}

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}

Solving for <em>M</em>, we get

M = \dfrac{4\pi^2 r^3}{GT^2}

Putting this expression back into Eqn(1), we get

R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}

\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}

\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}

\:\:\:\:= 2.6×10^6\:\text{m}

5 0
3 years ago
The escape speed from the moon is much smaller than from earth. True or False
Lisa [10]

Answer:

True

The escape speed from the Moon is much smaller than from Earth.

Explanation:

The escape speed is defined as:

v_{e} = \sqrt{\frac{2GM}{r}}  (1)

Where G is the gravitational constant, M is the mass and r is the radius.

The mass of the Earth is 5.972x10^{24}kg and its radius is 6371000m

Then, replacing those values in equation 1 it is gotten.

v_{e} = \sqrt{\frac{(2)(6.67x10^{-11}N.m^{2}/kg^{2})(5.972x10^{24}kg)}{(6371000m)}}  

v_{e} = 11.18m/s

For the case of the Moon:

v_{e} = \sqrt{\frac{(2)(6.67x10^{-11}N.m^{2}/kg^{2})(7.347x10^{22}Kg)}{(1737000m)}}  

v_{e} = 2.38m/s

Hence, the escape speed from the Moon is much smaller than from Earth.

Since it has a smaller mass and smaller radius compared to that from the Earth.

4 0
3 years ago
Someone help with this also
Mumz [18]
I’m pretty sure it’s average speed= total distance and total time which is A.
3 0
3 years ago
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