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Anon25 [30]
4 years ago
11

NASA scientists suggest using rotating cylindrical spacecraft to replicate gravity while in a weightless environment. Consider s

uch a spacecraft that has a diameter of d = 148 m. What is the speed v, in meters per second, the spacecraft must rotate at its outer edge to replicate the force of gravity on earth?
Physics
1 answer:
dusya [7]4 years ago
8 0

Answer:

Explanation:

Given

diameter of spacecraft d=148\ m

radius r=74\ m

Force of gravity F_g=mg

where m =mass of object

g=acceleration due to  gravity on earth

Suppose v is the speed at which spacecraft is rotating so a net centripetal  acceleration is acting on spacecraft which is given by

F_c=\frac{mv^2}{r}

F_c=F_g

\frac{mv^2}{r}=mg

\frac{v^2}{r}=g

v=\sqrt{gr}

v=\sqrt{1450.4}

v=38.08\ m/s    

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Answer:

The acceleration increases.

Explanation:

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3 years ago
Which is NOT part of the equation for angular momentum?
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8 0
3 years ago
Read 2 more answers
At one instant a bicyclist is 21.0 m due east of a park's flagpole, going due south with a speed of 13.0 m/s. Then 21.0 s later,
andre [41]

Answer:

Part a)

d = 21\sqrt2 = 29.7 m

Part b)

Direction is 45 degree North of West

Part c)

v_{avg} = 1.41 m/s

Part d)

direction of velocity will be 45 Degree North of West

Part e)

a = 0.875 m/s^2

Part f)

\theta = 45 degree North of East

Explanation:

Initial position of the cyclist is given as

r_1 = 21.0 m due East

final position of the cyclist after t = 21.0 s

r_2 = 21.0 m due North

Part a)

for displacement we can find the change in the position of the cyclist

so we have

d = r_2 - r_1

d = 21\hat j - 21\hat i

so magnitude of the displacement is given as

d = 21\sqrt2 = 29.7 m

Part b)

direction of the displacement is given as

\theta = tan^{-1}\frac{y}{x}

\theta = tan^{-1}\frac{21}{-21}

so it is 45 degree North of West

Part c)

For average velocity we know that it is defined as the ratio of displacement and time

so here the magnitude of average velocity is defined as

v_{avg} = \frac{\Delta x}{t}

v_{avg} = \frac{29.7}{21}

v_{avg} = 1.41 m/s

Part d)

As we know that average velocity direction is always same as that of average displacement direction

so here direction of displacement will be 45 Degree North of West

Part e)

Here we also know that initial velocity of the cyclist is 13 m/s due South while after t = 21 s its velocity is 13 m/s due East

So we have

change in velocity of the cyclist is given as

\Delta v = v_f - v_i

\Delta v = 13\hat i - (-13\hat j)

now average acceleration is given as

a = \frac{\Delta v}{\Delta t}

a = \frac{13\hat i + 13\hat j}{21}

so the magnitude of acceleration is given as

a = \frac{13\sqrt2}{21} = 0.875 m/s^2

Part f)

direction of acceleration is given as

\theta = tan^{-1}\frac{y}{x}

\theta = tan^{-1}\frac{13}{13}

\theta = 45 degree North of East

4 0
3 years ago
A 20 kg cart moving at 4 m/s has how much momentum?
masha68 [24]

Answer:

We conclude that the required momentum is 80 kgm/s.

Explanation:

Given

  • Mass m = 20 kg
  • Velocity v = 4 m/s

To determine

The momentum p = ?

Important Tip:

  • The momentum of an object is the product of mass and velocity.

We can determine the momentum using the formula

p = mv

where

  • p is the momentum
  • m is the mass
  • v is the velocity

now substituting m = 20 and v = 4 using the equation

p = mv

p = (20) (4)

p = 80 kg m/s

Therefore, we conclude that the required momentum is 80 kgm/s.

6 0
3 years ago
Why is. A paper clip attracted to a magnet, while a pencil is not?
Akimi4 [234]
The paper clip did because it is a conductor. A pencil is made with wood and that is an insulator.
7 0
3 years ago
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