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Bas_tet [7]
3 years ago
12

The moon Phobos orbits Mars

Physics
1 answer:
Soloha48 [4]3 years ago
8 0

Answer:

Explanation:

We are basically needing to solve for the time in the equation d = rt, where d is the distance around Mars (aka the circumference), r is the velocity, and t is time. We need to find the circumference and the velocity. We will begin with the velocity.

Because the gravitational attraction between Phobos and Mars provides the centripetal acceleration necessary to keep Phobos in its (sort of) circular path, the equation we use for this is:

F_g=F_c which says that Force supplied by gravity is equal to the centripetal force. Expanding that:

\frac{Gm_{Phobos}m_{Mars}}{r^2}=\frac{m_{Phobos}v^2}{r}

When we move that around mathematically to solve for the velocity value, what we end up with is:

v=\sqrt{\frac{Gm_{Mars}}{r} and filling in:

v=\sqrt{\frac{(6.67*10^{-11})(6.42*10^{23})}{9.38*10^6} } and we get that

v = 2100 m/s

Now for the circumference:

C = 2πr and

C = 2(3.1415)(9.38 × 10⁶) so

C = 5.9 × 10⁷

Putting that all together in the C = vT equation:

5.9 × 10⁷ = 2100T so

T = 2.8 × 10⁴ sec or 7.8 hours

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3 years ago
The cone cells of the human eye are sensitive to three wavelength ranges, which the eye interprets as blue (419nm), green (531 n
Sati [7]
Sorry, This is one of the questions i wouldn't know, Hope you find it though
8 0
3 years ago
(a) What is the tangential acceleration of a bug on the rim of a 13.0-in.-diameter disk if the disk accelerates uniformly from r
Oksi-84 [34.3K]

Answer:

0.321659377 m/s²

1.383138458 m/s

0.321659377 m/s²

0.62667 m/s²

0.7044 m/s² and 27.17°

Explanation:

d = Diameter of rim = 13 in = 13\times 0.0254=0.3302\ m

r = Radius = \frac{d}{2}=\frac{0.3302}{2}=0.1651\ m

\omega_f = Final angular velocity = 80\times\frac{2\pi}{60}=8.37758\ rad/s

\omega_i = Initial angular velocity = 0

\alpha = Angular acceleration

t = Time taken = 4.3 s

Equation of rotational motion

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{8.37758-0}{4.3}\\\Rightarrow \alpha=1.94827\ rad/s^2

Tangential acceleration is given by

a_t=r\alpha\\\Rightarrow a_t=0.1651\times 1.94827\\\Rightarrow a_t=0.321659377\ m/s^2

The tangential acceleration of the bug is 0.321659377 m/s²

Tangential velocity is given by

v=r\omega\\\Rightarrow v=0.1651\times 8.37758\\\Rightarrow v=1.383138458\ m/s

The tangential velocity of the bug is 1.383138458 m/s

The tangential acceleration is constant which is 0.321659377 m/s²

Centripetal acceleration is given by

a_c=\frac{a_tt^2}{r}\\\Rightarrow a_c=\frac{0.321659377^2\times 1}{0.1651}\\\Rightarrow a_c=0.62667\ m/s^2

The centripetal acceleration of the bug is 0.62667 m/s²

The resultant of the acceleration gives us total acceleration

a=\sqrt{a_t^2+a_c^2}\\\Rightarrow a=\sqrt{0.321659377^2+0.62667^2}\\\Rightarrow a=0.7044\ m/s^2

Direction is given by

\theta=cos^{-1}\frac{a_c}{a}\\\Rightarrow \theta=cos^{-1}\frac{0.62667}{0.7044}\\\Rightarrow \theta=27.17^{\circ}

The magnitude and direction of the acceleration is 0.7044 m/s² and 27.17°

8 0
4 years ago
A small sphere with mass 5.00×10−7 kg and charge +8.00 μC is released from rest a distance of 0.500 m above a large horizontal i
Damm [24]
<span>Work done on charge is W = Eqd = σ/(2ε₀) x q x d = {(8.00 x 10⁻¹²)/(2 x 8.854187 x 10⁻¹²)} x 3.00 x 10⁻⁶ x (0.650 - 0.250) = 5.42116402J. KE of sphere = 0.5mv² = 0.5 x 5.00 x 10⁻⁷v² = work done by E-field on charge during its fall = 5.42116402→ v = 4657 m/s.</span>
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3 years ago
Which of the following states that energy cannot be created or destroyed?
natali 33 [55]

Answer:

germany

Explanation:

5 0
3 years ago
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