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LiRa [457]
3 years ago
15

At its Ames Research Center, NASA uses its large "20-G" centrifuge to test the effects of very large accelerations ("hypergravit

y") on test pilots and astronauts. In this device, an arm 8.84 m long rotates about one end in a horizontal plane, and the astronaut is strapped in at the other end. Suppose that he is aligned along the arm with his head at the outermost end. The maximum sustained acceleration to which humans are subjected in this machine is typically 12.5 g. How fast must the astronaut's head be moving to experience the maximum acceleration?
Physics
1 answer:
Over [174]3 years ago
7 0

Answer:

v=32.9m/s

Explanation:

The acceleration needed to mantain a circular motion of radius r and speed v is given by the equation a=v^2/r

This is the centripetal acceleration. The person will feel what is called a centrifugal acceleration, of the same value, because he is not in an inertial frame (thus subject to fictitious forces, product of inertia).

We want to know the speed of his head when it is subject to 12.5 times the value of the acceleration of gravity while moving on a 8.84m radius circle, so we must do:

v=\sqrt{ar} = \sqrt{12.5gr}=\sqrt{(12.5)(9.8m/s)(8.84m)}=32.9m/s

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olga_2 [115]
The answer is:

V = d/t d = 86 km t = 1.3 hrs

V = 86 km/ 1.3 hrs

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I hope this helps!!
5 0
2 years ago
Which one of the following is a decomposition reaction?
BabaBlast [244]
You would know a decomposition reaction occurred if the reactants separated. For example from AB → A+B.

Now if you look at your options only 1 works out for that equation. Letter A. 
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It cannot be letter B because synthesis/combination occurred. The same goes for letter C. Letter D, single displacement occurred. 

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4 0
3 years ago
A 1090 kg car has four 12.7 kg wheels. When the car is moving, what fraction of the total kinetic energy of the car is due to ro
max2010maxim [7]

Answer:

\frac{KE_{Rotational}}{KE_{Total}} = 0.018

Explanation:

To develop this exercise we proceed to use the kinetic energy equations,

In the end we replace

KE_{Total}=KE_{Translational}+KE_{Rotational}

KE_{Total}=\frac{1}{2}m_{car}+4*\frac{1}{2}*I*(\frac{v}{r})^2

Here

I=\frac{1}{2}m_{wheels}*r^2 meaning the 4 wheels,

So replacing

KE_{Rotational}=4\frac{1}{2}*(\frac{1}{2}m_{wheels}*r^2)*(\frac{v}{r})^2=m*v^2

So,

\frac{KE_{Rotational}}{KE_{Total}} = \frac{m_{wheels}*v^2}{\frac{1}{2}m_{car}*v^2+m_{wheels}*v^2}

\frac{KE_{Rotational}}{KE_{Total}} = \frac{m_{wheels}}{\frac{1}{2}m_{car}+m_{wheels}}

\frac{KE_{Rotational}}{KE_{Total}} =  \frac{10}{545+10}

\frac{KE_{Rotational}}{KE_{Total}} = 0.018

3 0
3 years ago
A skateboarder is moving 5.25 m/s when he starts to roll up a frictionless hill. How much higher is he when his velocity has red
Taya2010 [7]
Mechanical energy E = mgh + 1/2mv²

When he starts, let h = 0 ⇒ E₁ = 1/2mv₁²
When he reaches height h ⇒ E₂ = mgh + 1/2mv₂²

Without friction, energy is conserved at all times.

E₁ = E₂
     ↓
1/2mv₁² = mgh + 1/2mv₂²
     ↓
1/2v₁² = gh + 1/2v₂²
     ↓
gh = 1/2(v₁² - v₂²)
    ↓
h = (v₁² - v₂²) / (2g)


5 0
3 years ago
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Answer:

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Explanation:

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