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3 years ago

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Two charged objects attract each other with a force of F. What happens to the force between them if one charge is doubled, the o

ther charge is tripled, and the separation distance between their centers is reduced to ¼ its original value?

1 answer:

Ray Of Light [21]3 years ago

3 0

Answer:

The force is increased by a factor of 96. (F1 = 96F)

Explanation:

Let the first charge be Q1

Let the second charge be Q2

Let the distance between their centers be r

The electrostatic force, F, between them is:

F = (k*Q1*Q2) / r²

If the first charge is doubled = 2Q1

If the second charge is tripled = 3Q2

The separation between their centers is reduced to ¼ = ¼r

The force between them becomes:

F1 = (k* 2Q1 * 3Q2) / (¼r)²

F1 = (6 * k * Q1 * Q2) / (r²/16)

F1 = 96(k * Q1 * Q2) /r²

F1 = 96F

The force is increased by a factor of 96.

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Answer:

The moment of inertial of the wheel, $I = 8(\frac{1}{3}M_sL^2 ) + M_rL^2$

Explanation:

Given;

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Starting at 1.0 m/s, a cheetah runs with a constant acceleration for 4.8 s reaching a speed of 28 m/s. What is the acceleration

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Answer:

c. $5.6m/s^2$

Explanation:

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Time taken by cheetah =4.8 s

Final velocity of cheetah,v=28 m/s

We have to find the acceleration of this cheetah.

We know that

Acceleration, $a=\frac{v-u}{t}$

Where v=Final velocity of object

u=Initial velocity of object

t=Time taken by object

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Then, we get

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An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. What is the minimum coef

navik [9.2K]

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. The minimum coefficient of static friction needed for this claim to be possible is 0.7

In an inclined plane, the coefficient of static friction is the angle at which an object slide over another.

As the angle rises, the gravitational force component surpasses the static friction force, as such, the object begins to slide.

Using the Newton second law;

$\sum F_x = \sum F_y = 0$

$\mathbf{mg sin \theta -f_s= N-mgcos \theta = 0 }$

So; On the L.H.S

$\mathbf{mg sin \theta =f_s}$

$\mathbf{mg sin \theta =\mu_s N}$

On the R.H.S

N = mg cos θ

Equating both force component together, we have:

$\mathbf{mg sin \theta =\mu_s \ mg \ cos \theta}$

$\mathbf{sin \theta =\mu_s \ \ cos \theta}$

$\mathbf{\mu_s = \dfrac{sin \theta }{ cos \theta}}$

From trigonometry rule:

$\mathbf{tan \theta= \dfrac{sin \theta }{ cos \theta}}$

∴

$\mathbf{\mu_s =\tan \theta}}$

$\mathbf{\mu_s =\tan 35^0}}$

$\mathbf{\mu_s = 0.700}}$

Therefore, we can conclude that the minimum coefficient of static friction needed for this claim to be possible is 0.7

Learn more about static friction here:

brainly.com/question/24882156?referrer=searchResults

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3 years ago