An astronaut rides in a circular training accelerator at 16 m/s. If the radius of the accelerator is 8.0 meters, what is his centripetal acceleration? m/s2 Enter the answer Submit Answer I don't know this mprove this question​

Answers

Answer 1

Answer:

[tex]a=32\ m/s^2[/tex]

Explanation:

Given that,

The velocity of an astronaut in a circular path, v = 16 m/s

The radius of the accelerator, r = 8 m

We need to find his centripetal acceleration. The formula that is used to find the centripetal acceleration is given by :

[tex]a=\dfrac{v^2}{r}\\\\a=\dfrac{(16)^2}{8}\\\\a=32\ m/s^2[/tex]

So, the required centripetal acceleration is [tex]32\ m/s^2[/tex].


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i looked it up an thats all i can find

As we know, the moon is a satellite of our earth, what is the
theoretical period of the moon? The average radius of the
moon's orbit is 3.84 108 m and the mass of the earth is 5.97 x
1024 kg (in hours, G = 6.67 x 10-9 N (m/kg) 3).

Answers

Answer:c

Explanation:c

This question involves the concepts of the time period, orbital radius, and gravitational constant.

The theoretical period of the moon is "658 hr".

The theoretical time period of the moon around the earth can be found using the following formula:

[tex]\frac{T^2}{R^3}=\frac{4\pi^2}{GM}[/tex]

where,

T = Time Period of Moon = ?

R = Orbital Radius = 3.84 x 10⁸ m

G = Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of Earth = 5.97 x 10²⁴ kg

Therefore,

[tex]\frac{T^2}{(3.84\ x\ 10^8\ m)^3}=\frac{4\pi^2}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(5.97\ x\ 10^{24}\ kg)}\\\\T^2=(9.91\ x\ 10^{-14}\ s^2/m^3)(56.62\ x\ 10^{24}\ m^3)\\\\T=\sqrt{561.34\ x\ 10^{10}\ s^2}[/tex]

T = 2.37 x 10⁶ s[tex](\frac{1\ h}{3600\ s})[/tex]

T = 658 hr

Learn more about the orbital time period here:

https://brainly.com/question/14494804?referrer=searchResults

The attached picture shows the derivation of the formula for orbital speed.

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