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aleksklad [387]
4 years ago
7

When a certain rubber band is stretched a distance x, it exerts a restoring force of magnitude f = ax, where a is a constant. th

e work done by a person in stretching this rubber band from x = 0 to x = l is:?
Physics
1 answer:
Veseljchak [2.6K]4 years ago
4 0
Given:
F = ax
where
x = distance by which the rubber band is stretched
a =  constant

The work done in stretching the rubber band from x = 0 to x = L is
W=\int_{0}^{L} Fdx = \int_{0}^{L}ax \, dx = \frac{a}{2}  [x^{2} ]_{0}^{L} =  \frac{aL^{2}}{2}

Answer:  \frac{aL^{2}}{2}

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If the star Alpha Centauri were moved to a distance 10 times farther than it is now, its parallax angle would
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Answer:

decrease by a factor 10

Explanation:

The parallax angle of a close star is given by

p=\frac{1}{d}

where

p is the parallax angle

d is the distance of the star from Earth, in parsecs

From the formula we see that the parallax angle is inversely proportional to the distance.

In this problem, the distance of the star is increased by a factor 10:

d' = 10 d

so the new parallax angle would be

p'=\frac{1}{10 d}=\frac{1}{10}\frac{1}{d}=\frac{p}{10}

So, the parallax angle would decrease by a factor 10.

4 0
3 years ago
An asteroid of mass 5.7×104 kg carrying a negative charge of 14.0 μC is 190 m from a second asteroid of mass 5.5×104 kg carrying
Finger [1]

Answer:

repulsion force es  R = 5,224 10⁻³ N

Explanation:

We use Newton's second law to add forces, the elective force is repulsive because the asteroids have the same type of charge and gravitational force is always attractive.

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      Fg = G m1 m2 / r²

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Electric force

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The resulting force is

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     R = 5.2296 10⁻³ - 5.795 10⁻⁶=  5229.6 10⁻⁶ - 5,795 10⁻⁶

     R = 5,224 10⁻³ N

this is a repulsive force and the asteroids move away from each other

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3 years ago
I need help finding the solution
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Answer:

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3 0
3 years ago
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Answer:

Laterite would be your answer.

Explanation:

I hope this helps!

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3 years ago
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