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Lelechka [254]
3 years ago
8

A future use of space stations may be to provide hospitals for severely burned persons. It is very painful for a badly burned pe

rson on Earth to lie in bed. In a space station, the effect of gravity can be reduced or even eliminated. How long should each rotation take for a doughnut-shaped hospital of 200-m radius so that persons on the outer perimeter would experience 1/10 the normal gravity of Earth? A) 91 min B) 8.7 min C) 4.6 min D) 1.5 min E) 0.011 min
Physics
1 answer:
inessss [21]3 years ago
6 0

Answer:

1.5min

Explanation:

To solve the problem it is necessary to take into account the concepts related to Period and Centripetal Acceleration.

By definition centripetal acceleration is given by

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

Where,

V = Tangencial velocity

r = radius

With our values we know that

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

\frac{V^2}{r} = \frac{1}{10}g

Therefore solving to find V, we have:

V = \sqrt{\frac{1}{10}g*r}

V = \sqrt{\frac{9.81*200}{10}}

V = 14m/s

For definition we know that the Time to complete are revolution is given by

t = \frac{Perimeter}{Speed}

t = \frac{2\pi R}{V}

t = \frac{2\pi * 200}{14}

t = 1.5min

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2 years ago
A parallel plate capacitor with air between the plates has a potential difference of 71.0 V. Determine the potential difference
Gwar [14]

Answer:

15.8 V

Explanation:

The relationship between capacitance and potential difference across a capacitor is:

q=CV

where

q is the charge stored on the capacitor

C is the capacitance

V is the potential difference

Here we call C and V the initial capacitance and potential difference across the capacitor, so that the initial charge stored is q.

Later, a dielectric material is inserted between the two plates, so the capacitance changes according to

C'=kC

where k is the dielectric constant of the material. As a result, the potential difference will change (V'). Since the charge stored by the capacitor remains constant,

q=C'V'

So we can combine the two equations:

CV=CV'\\CV=(kC)V'\\V'=\frac{V}{k}

and since we have

V = 71.0 V

k = 4.50

We find the new potential difference:

V'=\frac{71.0}{4.50}=15.8 V

6 0
2 years ago
Why echo sound cannot be heard in small room
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7 0
3 years ago
A ball is dropped from rest at point O. After falling for some time, it passes by a window of height 3.3 m and it does so in 0.2
stiv31 [10]

Answer:

Speed at which the ball passes the window’s top = 10.89 m/s

Explanation:

Height of window = 3.3 m

Time took to cover window = 0.27 s

Initial velocity, u = 0m/s

We have equation of motion s = ut + 0.5at²

For the top of window (position A)

                     s_A=0\times t_A+0.5\times 9.81t_A^2\\\\s_A=4.905t_A^2

For the bottom of window (position B)

                     s_B=0\times t_B+0.5\times 9.81t_B^2\\\\s_A=4.905t_B^2

\texttt{Height of window=}s_B-s_A=3.3\\\\4.905t_B^2-4.905t_A^2=3.3\\\\t_B^2-t_A^2=0.673

We also have

                 t_B-t_A=0.27

Solving

         t_B=0.27+t_A\\\\(0.27+t_A)^2-t_A^2=0.673\\\\t_A^2+0.54t_A+0.0729-t_A^2=0.673\\\\t_A=1.11s\\\\t_B=0.27+1.11=1.38s

So after 1.11 seconds ball reaches at top of window,

       We have equation of motion v = u + at

                                     v_A=0+9.81\times 1.11=10.89m/s

Speed at which the ball passes the window’s top = 10.89 m/s                

7 0
3 years ago
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