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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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A 5.00 kg crate is on a 21.0 degree hill. Using X-Y axes tilted down the plane, what is the y-component of the normal force?
Rama09 [41]

Answer:

The y-component of the normal force is 45.74 N.

Explanation:

Given that,

Mass of the crate, m = 5 kg

Angle with hill, \theta=21^{\circ}

We need to find the y component of the normal force. We know that the y component of the normal force is given by :

F_y=F\ \cos\theta\\\\F_y=mg\ \cos\theta\\\\F_y=5\times 9.8\ \cos(21)\\\\F_y=45.74\ N

So, the y-component of the normal force is 45.74 N. Hence, this is the required solution.

7 0
2 years ago
A battery is a source of:<br> a. current<br> b. voltage<br> c. resistance<br> d. all of these
kolezko [41]

The correct answer is b


3 0
3 years ago
A positively charged sphere with a charge of the of 8Q is separated from a negatively charged sphere - 2Q by a distance r. The s
professor190 [17]

Answer: The new force is:

F1 = -(9/16)*F0

where F0 is the initial force.

Explanation:

The charges of the spheres is q1 = 8Q and q2 = -2Q

and as you know, the force between charged objects is:

F = k*q1*q2/r^2

where k is a constant and r is the distance

F0 = -(k16Q^2)/r^2

where the negative sign means that the force is attractive.

Now, when the spheres touch each other, the charge must be distributed equally in both spheres. So the total charge is q1 + q2 = 8Q - 2Q = 6Q

then each sphere has now a charge of 3Q.

The new force is:

F1 = k*3Q*3Q/r^2 = (k*9*Q^2)/r^2 = -(9/16)*F0

You can see that F1 is positive, this means that the force now is repulsive.

4 0
3 years ago
Identify the arrows that represent the process of cooling.
elixir [45]
We need to the where the arrows are..
3 0
3 years ago
Someone help me answer this.
horrorfan [7]

The angular velocity of the ring when the bug is halfway around and the angular velocity of the ring when the bug is back at the pivot is [m₂v / {(2m₁ +m₂)R}].

<h3>What is angular velocity?</h3>

The velocity of a particle when moving in the circular path.

Let speed of the bug with respect to ground is u.

Speed of bag with respect to ring will be

v = u - (- Rω) =

Then, u = v- Rω...............(1)

Angular momentum of ring and bug will remain conserved.

Initial  momentum: L ring + Lbug =0

Final  momentum:  -2m₁ R²ω + m₂uR =0...............(2)

Using equation (1) and (2), the angular velocity expression will be

ω =[m₂v / {(2m₁ +m₂)R}] in positive z direction

Thus, the angular velocity of the ring when the bug is halfway around and the angular velocity of the ring when the bug is back at the pivot is [m₂v / {(2m₁ +m₂)R}].

Learn more about angular velocity.

brainly.com/question/17592191

#SPJ1

8 0
2 years ago
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