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mylen [45]
3 years ago
10

Which of the following renaissance scientists made improvements to the telescope?

Physics
1 answer:
LenaWriter [7]3 years ago
7 0
Michelaneglo DDDDDDDDDDDDDDDDDDDDDDDDD

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Can You Please Give Me Some Examples of Density-Independent and Density-Dependent Limiting Factors?
ivolga24 [154]
Density-Dependent: 
1<span><span><span><span>. </span>competition.</span><span>   
<span>2. </span>overcrowding.</span><span>   
3<span>. </span>predators.</span></span><span> 

(These are a few from a test I took, hopefully they help you a bit >.<)</span></span>
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
3
kramer

Answer:

60.2 J

Explanation:

Efficiency is the ratio of work out to work in.

e = Wout / Win

0.86 = Wout / 70 J

Wout = 60.2 J

5 0
3 years ago
Read 2 more answers
Sharon throws a 0.20 Kg with an acceleration of 10 m/s/s.
OlgaM077 [116]
Force=A×M
10m/s×0.20kg
=2Newton
4 0
3 years ago
Helpp me pleassee....
lilavasa [31]

Answer:

The fundamental wavelength of the vibrating string is 1.7 m.

Explanation:

We have,

Velocity of wave on a guitar string is 344 m/s

Length of the guitar string is 85 cm or 0.85 m

It is required to find the fundamental wavelength of the vibrating string. The fundamental frequency on the string is given by :

f=\dfrac{v}{2l}\\\\f=\dfrac{344}{2\times 0.85}\\\\f=202.35\ Hz

Now fundamental wavelength is :

\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{344}{202.35}\\\\\lambda=1.7\ m

So, the fundamental wavelength of the vibrating string is 1.7 m.

4 0
3 years ago
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