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Anon25 [30]
3 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]3 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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aivan3 [116]

1) Example of contact force: friction

2) Examples of non-contact forces: gravity and electromagnetic force

Explanation:

1)

Contact forces are forces that acts only when the objects involved are touching.

An example of contact force in the geosphere is friction. Friction is a force that acts when two objects slide past each other, and the surfaces of the two objects are in contact. Due to the presence of "microbumps" on the two surfaces, there is a resistive force opposing the motion of the two objects, and this force is called friction.

Friction also acts when an object is moving through a fluid, although it takes a different name: resistance. Also in this case, the resistance acts in the direction opposite to the motion of the object, slowing it down.

2)

Non-contact forces are forces that act from a distance, therefore they act even when the objects involved are not touching.

Examples of non-contact forces are:

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4 0
3 years ago
Boat A and Boat B have the same mass. Boat A's velocity is three times greater than that of Boat B. Compared to
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Answer:

nine times as much.

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K.E of A = 9 times K.E of B

7 0
3 years ago
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Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges
Nostrana [21]

Answer:

r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})

Explanation:

Here two charges are placed at distance "d" apart

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so we will have

electric field due to charge 1 = electric field due to charge 2

E_1 = E_2

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\frac{kq_1}{r^2} = \frac{kq_2}{(d-r)^2}

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\frac{(d - r)^2}{r^2} = \frac{q_2}{q_1}

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Well, to understand this in a better way, let's begin by explaining that water is special due to its properties, which makes this fluid useful for many purposes and for the existence of life.

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How is this possible?

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Explanation:

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