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2 years ago

The moon Phobos orbits Mars

Physics

1 answer:

podryga [215]2 years ago

3 0

Explanation:

For a circular orbit v= $\sqrt{\frac{G.m}{r} }$ with G = 6.6742 × $10^{-11}$

Given m = 6.42 x 10^23 kg and r=9.38 x 10^6 m

=> v = 2137.3 m/s

I hope this is the correct way to solve

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A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is

aniked [119]

1) At the moment of being at the top, the piston will not only tend to push the penny up but will also descend at a faster rate at which the penny can reach in 'free fall', in that short distance. Therefore, at the highest point, the penny will lose contact with the piston. Therefore the correct answer is C.

2) To solve this problem we will apply the equations related to the simple harmonic movement, hence we have that the acceleration can be defined as

$a = -\omega^2 A$

Where,

a = Acceleration

A = Amplitude

$\omega$ = Angular velocity

From a reference system in which the downward acceleration is negative due to the force of gravity we will have to

$a = -g$

$-\omega^2 A = -g$

$\omega = \sqrt{\frac{g}{A}}$

From the definition of frequency and angular velocity we have to

$\omega = 2\pi f$

$f = \frac{1}{2\pi} \sqrt{\frac{g}{A}}$

$f = \frac{1}{2\pi} \sqrt{\frac{9.8}{4*10^{-2}}}$

$f = 2.5Hz$

Therefore the maximum frequency for which the penny just barely remains in place for the full cycle is 2.5Hz

6 0

3 years ago

A viscous fluid is flowing through two horizontal pipes. They have the same length, although the radius of one pipe is three tim

Rudik [331]

The conservation of the mass of fluid through two sections (be they A1 and A2) of a conduit (pipe) or current tube establishes that the mass that enters is equal to the mass that exits. Mathematically the input flow must be the same as the output flow,

$Q_1 = Q_2$

The definition of flow is given by

$Q =VA$

Where

V = Velocity

A = Area

The units of the flow of flow are cubic meters per second, that is to say that if there is a continuity, the volume of input must be the same as that of output, what changes if the sections are modified are the proportions of speed.

In this way

$Q_A = Q_B$