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3 years ago

Alice kicks a 0.25 kg soccer ball with 0.5 N of force. What force does the ball exert on Alice’s foot as she kicks it

2 answers:

7 0

You do not doubt it. The third Law of Newton really works. I would say it is the most reliable law of the Universe. Action and reaction. It is not subject to special conditions, it works always. If an object exerts a force over other object, the second object exerts a force of equal magnitude but in the opposed direction over the first.

So, the answer, undoubtedly, is that the ball exerts a force of 0.5 N over Alices's foot as she kicks it.

mamaluj [8]3 years ago

4 0

Answer:0.5

Explanation:I did exam

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The laST ONE I THINK

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Force F acts between a pair of charges, q1 and q2, separated by a distance d. For each of the statements, use the drop-down menu

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The initial force between the two charges is given by:

$F=k \frac{q_1 q_2}{d^2}$

where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:

1. F

In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.

So, we have:

$q_1' = \frac{q_1}{2}\\q_2' = 2 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(\frac{q_1}{2})(2q_2)}{d^2}=k \frac{q_1 q_2}{d^2}=F$

So the force has not changed.

2. F/4

In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.

So, we have:

$q_1' = q_1\\q_2' = q_2\\d' = 2d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{q_1 q_2)}{(2d)^2}=\frac{1}{4} k \frac{q_1 q_2}{d^2}=\frac{F}{4}$

So the force has decreased by a factor 4.

3. 6F

In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.

So, we have:

$q_1' = 2 q_1\\q_2' = 3 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(2 q_1)(3 q_2)}{d^2}=6 k \frac{q_1 q_2}{d^2}=6F$

So the force has increased by a factor 6.

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A 50.0-kg wolf is running at 10.0 m/sec. What is the wolfs kinetic energy

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Kinetic energy = mass time squared speed divided by 2
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3 years ago

A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V. What are (a) the total energy stored i

Debora [2.8K]

Answer:

(A) Total energy will be equal to $0.044\times 10^{-5}J$

(b) Energy density will be equal to $0.0175J/m^3$

Explanation:

We have given diameter of the plate d = 2 cm = 0.02 m

So area of the plate $A=\pi r^2=3.14\times 0.02^2=0.001256m^2$

Distance between the plates d = 0.50 mm = $0.50\times 10^{-3}m$

Permitivity of free space $\epsilon _0=8.85\times 10^{-12}F/m$

Potential difference V =200 volt

Capacitance between the plate is equal to $C=\frac{\epsilon _0A}{d}=\frac{8.85\times 10^{-12}\times 0.001256}{0.50\times 10^{-3}}=0.022\times 10^{-9}F$

(a) Total energy stored in the capacitor is equal to

$E=\frac{1}{2}CV^2$

$E=\frac{1}{2}\times 0.022\times 10^{-9}\times 200^2=0.044\times 10^{-5}J$

(b) Volume will be equal to $V=Ad$ , here A is area and d is distance between plates

$V=0.001256\times 0.02=2.512\times 10^{-5}m^3$

So energy density $=\frac{Energy}{volume}=\frac{0.044\times 10^{-5}}{2.512\times 10^{-5}}=0.0175J/m^3$

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The elephant and the mouse having zero weight in a gravity free space will not bump into you at the same effect.

<u>Explanation: </u>

When both are in a gravity free space, the weights are zero, as we know that the $\text {weight of the body}=\text {mass of the body} \times \text {acceleration due to gravity}$

$\text {here, the weight of elephant}=\text {mass of elephant } \times \text {zero gravti} y=zero$

$\text {similarly,weight of mouse}=\text {mass of mouse } \times \text {zero gravity}=zero$

But when they will acquire the speed of same magnitude, say v, their different masses will acquire different momentum, which will make the difference in effect while bumping.

$\text { momentum of elephant }=\text { mass of elephant } \times v$ $\text { momentum of mouse = mass of mouse } \times v$

And as we know $\text { mass of elephant }>\text { mass of mouse }$ Therefore, effect of impact by elephant will be more than that of mouse . An elephant breaking into you will take you back faster than a mouse in space hits you.

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