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2 years ago

Come in my free candy van located at Skellerupvej 43, 8600 Silkeborg, Denmark

Physics

1 answer:

Vsevolod [243]2 years ago

5 0

Answer:

Did you think you could pull that off-

I live in Japan anyway so.. ウーフ-

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What is an experience from everyday life that appears to support the geocentric model

fomenos

One everyday life experience that seems to support the geocentric model is the rising and setting of the Sun and Moon. The Moon rises and falls because it does revolve around the Earth and so it is easy to assume the same is true for the Sun.

3 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 v is the velocity of the satellite around the planet defined as

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

where r is the radius of the satellite's orbit in meters and T 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 M, 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

One way to measure g on another planet or moon by remote sensing is to measure how long it takes an object to fall a given dista

Jlenok [28]

Answer:

(a) 0.94 m/s²

(b) g (planet) = 0.096g

Explanation:

(a)

From Newton's equation of motion,

S = ut + 1/2gt²......................... equation 1

Making g the subject of equation 1

g =( S - ut)/t² ........................ equation 2

Where s = distance ( m), u = initial velocity (m/s), t = time (s), g = acceleration due to gravity (m/s²)

From the question, S = 12.02 m, t = 3.58 s, u= 0 ( at rest),

Substituting these values in equation 2

g = {12.02 -(0×3.58)}/3.58²

g = (12.02)/12.82

g = 0.94 m/s²

∴ The acceleration due to gravity on the planet = 0.94 m/s²

(b) g (planet)/g (earth) = 0.94/9.80

g (planet) = 0.096 g (earth).

The acceleration due to gravity of the planet in terms of the earth g is

g (planet) = 0.096g

5 0

3 years ago

Which telescopes must be placed in orbit around earth in order to observe short-wavelength radiation?.

son4ous [18]

Space telescopes must be placed in orbit around earth in order to observe short-wavelength radiation.

<h3>What is telescope?</h3>

A telescope is an optical instrument that uses lenses, curved mirrors, or a combination of both to watch distant objects.

When atoms in a gas reach this temperature, they travel so quickly that when they collide, they release X-ray photons with wavelengths smaller than 10 nanometers.

Because the Earth's atmosphere prevents all X-rays from space, these wavelengths must be seen using space telescopes.

To study short-wavelength radiation, space telescopes must be put in orbit around the Earth.

Hence, space telescope is the correct answer.

To learn more about the telescope, refer:

brainly.com/question/556195

#SPJ1

6 0

1 year ago

BigorU [14]

Could you maybe show us the diagram

3 0

3 years ago