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SVEN [57.7K]
3 years ago
7

A centrifuge is used to test space pilots. The centrifuge spins with a centripetal acceleration of 3.04g. If the length of the a

rm of the centrifuge is 21m what is the speed of the centrifuge?
Physics
1 answer:
uranmaximum [27]3 years ago
3 0

Answer:

approx.= 25\frac{m}{s}

Explanation:

Centripetal acceleration (a) is defined as the square of an object's velocity (V^2) divided by the distance of the object from it's point/axis of revolution (r). So:

a=\frac{V^{2} }{r}

which allows us to solve for the velocity:

V=\sqrt{ar}\\ Since: a=3.04g=(3.04)(9.81),r=21;\\V=\sqrt{(3.04)(9.81)(21)} =25.02...

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The flagpole is held vertical by two ropes. The first of these ropes has a tension in it of 100 N and is at an angle of 60° to t
KatRina [158]

Answer:

T₂ = 123.9 N,  θ = 66.2º

Explanation:

To solve this exercise we use the law of equilibrium, since the diaphragm does not appear, let's use the adjoint to see the forces in the system.

The tension T1 = 100 N, we create a reference frame centered on the pole

X axis

       T₁ₓ - T_{2x} = 0

        T_{2x}= T₁ₓ

Y axis y

      T_{1y} + T_{2y} - 200N = 0

      T_{2y} = 200 -T_{1y}

let's use trigonometry to find the component of the stresses

         sin 60 = T_{1y} / T₁

         cos 60 = t₁ₓ / T₁

         T_{1y} = T₁ sin 60

         T1x = T₁ cos 60

         T_{1y}y = 100 sin 60 = 86.6 N

         T₁ₓ = 100 cos 60 = 50 N

for voltage 2 it is done in the same way

         T_{2y} = T₂ sin θ

         T₂ₓ = T₂ cos θ

we substitute

         

           T₂ sin θ= 200 - 86.6 = 113.4

           T₂ cos θ = 50              (1)

to solve the system we divide the two equations

           tan θ = 113.4 / 50

           θ = tan⁻¹ 2,268

           θ = 66.2º

we caption in equation 1

           T₂ cos 66.2 = 50

           T₂ = 50 / cos 66.2

           T₂ = 123.9 N

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The power of a purely resistive lead is always positive although the current and voltage are sometimes negative. explain​
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Answer:

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