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Alex
3 years ago
12

A 70.0 kg astronaut is training for accelerations that he will experience upon reentry. He is placed in a centrifuge (r = 10.0 m

) and spun at a constant angular velocity of 16.3 rpm. Answer the following:
What is the angular velocity of the centrifuge in ?

What is the linear velocity of the astronaut at the outer edge of the centrifuge?

What is the centripetal acceleration of the astronaut at the end of the centrifuge?

How many g’s does the astronaut experience?

What is the centripetal force and net torque experienced by the astronaut? Give magnitudes and directions.
Physics
1 answer:
vovangra [49]3 years ago
7 0

Answer:

1.3823 rad/s

20.7345 m/s

28.66129935 m/s²

2006.29095 N radially outward

Explanation:

r = Radius = 15 m

m = Mass of person = 70 kg

g = Acceleration due to gravity = 9.81 m/s²

Angular velocity is given by

Angular velocity is 1.3823 rad/s

Linear velocity is given by

The linear velocity is 20.7345 m/s

Centripetal acceleration is given by

The centripetal acceleration is 28.66129935 m/s²

Acceleration in terms of g

Centripetal force is given by

The centripetal force is 2006.29095 N radially outward

The torque will be experienced when the centrifuge is speeding up of slowing down i.e., when it is accelerating and decelerating.

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Light from a laser (lambda= 406.192 nm) is used to illuminate two narrow slits. The interference pattern is observed on a screen
dsp73

Answer:

The spacing between the slits is    d = 0.00145m                

Explanation:

From the question we are told that

  The wavelength of the light is \lambda = 406.192nm = 406.192*10^{-9} m

   The distance of the slit from the screen is D = 5.937 \ m

    The number of bright fringe is n = 24

     The  length the fringes span is   L = 39.835 mm = \frac{39.835 }{1000} = 0.0398 m

The fringe width (i.e the distance of between two successive bright or dark fringe) is mathematically represented as

             \beta  = \frac{\lambda D}{d}

Where d is  the distance between the  slits

            \beta is the fringe width which can also be evaluated as

                         \beta = \frac{L}{n}

Substituting values

                        \beta = \frac{0.0398}{24}

                          \beta = 1.660 *10^{-3}

Making d the subject of formula in the above equation

                d = \frac{\lambda D}{\beta }

Substituting values

                d = \frac{406.192 *10^{-9} * 5.937 }{1.660 *10^{-3}}

                    d = 0.00145m                

           

4 0
4 years ago
Please can someone answer this question ​
EastWind [94]
B????????? I thinkkkkk
7 0
3 years ago
Read 2 more answers
Which nucleus complete the following equation
rosijanka [135]

(C) ^{208}_{84}\text{Po}

Explanation:

^{212}_{86}\text{Rn} \rightarrow \:^4_2\text{He} + \:^{208}_{84}\text{Po}

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3 years ago
An object is 10 cm from the mirror, its height is 1 cm and the focal length is 5 cm. What is the image height? (Indicate the obj
boyakko [2]

The image height is -10 cm (the image is upside down)

Explanation:

We can solve the problem by using the mirror equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where:

f = 5 cm is the focal length of the mirror

p = 10 cm is the distance of the object from the mirror

q is the distance of the image from the mirror

Solving for q, we find:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{5}-\frac{1}{10}=\frac{1}{10}\\\rightarrow q= 10 cm

So, the distance of the image from the mirror is 10 cm.

Now we can find the image height by using the magnification equation:

\frac{y'}{y}=-\frac{q}{p}

where

y' is the height of the image

y = 1 cm is the height of the object

and using

p = 10 cm

q = 10 cm

We find the size of the image:

y' = -\frac{qy}{p}=-\frac{(10)(1)}{10}=-10 cm

where the negative sign indicates that the image is upside down.

#LearnwithBrainly

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3 years ago
Examples of impact printers​
kaheart [24]

Answer:

Dot matrix printers, Daisy - wheel printers, ball printers.

7 0
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