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vichka [17]
3 years ago
9

An astronaut in the International Space Station cannot stand on a weighing scale. But an astronaut inside a rotating space stati

on (not yet built) can stand on a weighing scale. Explain. Your answer
Physics
1 answer:
VLD [36.1K]3 years ago
6 0

The astronaut when positioned on a scale of an international space station that is static in rotational motion, and turning around the earth, will have no effect, since there is no external force acting on it. This is in perpetual free fall around the earth.

When an astronaut stands on a scale within a rotating space station, he will have the Rotational Centripetal Force acting on it, said Force, from the inside acting as 'Artificial Gravity' which will generate a given measurement on the scale.

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It takes you 5 min to walk with an average velocity of .75 m/s to the north from the parking lot to the entrance of the amusemen
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A displacement is a vector quantity that takes into account the shortest distance from the starting point to the endpoint. 

The given above gave a time interval in minutes which needs to be converted to seconds. Given that each minute is 60 seconds, 5 minutes equal 300 seconds. To determine the distance, multiply time with speed. The product is 225 m. 

Thus, the displacement is 225 m. 
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3 years ago
Define linear expansivity
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Linear expansivity is a type of thermal expansion. It is described by a fraction that represents the fractional increase in length of a thin beam of a material exposed to a temperature increase of one degree Celsius. ... Linear expansivity is used in many real world applications.

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3 years ago
Read 2 more answers
A car travels around a level, circular track that is 750m across. What coefficient of friction is required to ensure the car can
Crank

The coefficient of friction must be 0.196

Explanation:

For a car moving on a circular track, the frictional force provides the centripetal force needed to keep the car in circular motion. Therefore, we can write:

\mu mg = m\frac{v^2}{r}

where the term on the left is the frictional force acting between the tires of the car and the road, while the term on the right is the centripetal force. The various terms are:

\mu is the coefficient of friction between the tires and the road

m is the mass of the car

g=9.8 m/s^2 is the acceleration of gravity

v is the speed of the car

r is the radius of the curve

In this problem,

r = 750 m is the radius

v=85 mph \cdot \frac{1609}{3600}=38.0 m/s is the speed

And solving for \mu, we find the coefficient of friction required to keep the car in circular motion:

\mu = \frac{v^2}{rg}=\frac{38.0^2}{(750)(9.8)}=0.196

Learn more about circular motion:

brainly.com/question/2562955  

brainly.com/question/6372960  

#LearnwithBrainly

8 0
3 years ago
A record player turntable initially rotating at 3313 rev/min is braked to a stop at a constant rotational acceleration. The turn
Rus_ich [418]

Answer:

(A) It will take 22 sec to come in rest

(b) Work done for coming in rest will be 0.2131 J              

Explanation:

We have given the player turntable initially rotating at speed of 33\frac{1}{3}rpm=33.333rpm=\frac{2\times 3.14\times 33.333}{60}=3.49rad/sec

Now speed is reduced by 75 %

So final speed \frac{3.49\times 75}{100}=2.6175rad/sec

Time t = 5.5 sec

From first equation of motion we know that '

\alpha =\frac{\omega -\omega _0}{t}=\frac{2.6175-3.49}{4}=-0.158rad/sec^2

(a) Now final velocity \omega =0rad/sec

So time t to come in rest  t=\frac{0-3.49}{-0.158}=22sec

(b) The work done in coming rest is given by

\frac{1}{2}I\left ( \omega ^2-\omega _0^2 \right )=\frac{1}{2}\times 0.035\times (0^2-3.49^2)=0.2131J

4 0
3 years ago
Explain why hitting/catching a baseball are great examples of each type of collision
NeX [460]
When you hit a ball it collides with the bat. When you catch a ball it collides with your hand.
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3 years ago
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