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love history [14]
3 years ago
5

By experiment, determine what makes a force attractive or repulsive. Describe your experiments and observations with some exampl

es. 4. What evidence do you see that Newton's third law applies to electrostatic forces
Physics
1 answer:
kumpel [21]3 years ago
7 0

Answer:

he combs are brought closer, each one has to move, therefore a force must be applied to each one, and the only force is electrostatic. force must have the same magnitude in each one and in the opposite direction; this description is equivalent to Newton's third law.

Explanation:

A very simple experiment that we can do is to rub a brush or plastic comb with a piece of paper, this creates some electrical charges in the comb, we hang the comb by a thread.

We take another comb and we form it with the same paper, when we bring the combs closer, we see that they repel each other, this is proof that the electrical charges created are of the same type.  

If now we rub the comb with another material, for example plastic, and bring it closer to the comb that is hanging, we see that they are approaching, therefore, it is proven that the load created from another type

With these two simple experiments it can be proved that electric charges are of two types, which are called positive and negative.  

4. In the previous experiment we see that when the combs are brought closer, each one has to move, therefore a force must be applied to each one, and the only force is electrostatic. If we observe carefully we see that the breaking of the combs is the same, therefore the force must have the same magnitude in each one and in the opposite direction; this description is equivalent to Newton's third law.

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A particle moving along a straight line is subjected to a deceleration ???? = (−2???? 3 ) m/???? 2 , where v is in m/s. If it ha
dlinn [17]

Answer:

(a). The velocity is 0.099 m/s.

(b). The position is 19.75 m.

Explanation:

Given that,

The deceleration is

a=(-2v^3)\ m/s^2

We need to calculate the velocity at t = 25 s

The acceleration is the first derivative of velocity of the particle.

a=\dfrac{dv}{dt}

\dfrac{dv}{dt}=-2v^3

\dfrac{dv}{-v^3}=2dt

On integrating

int{\dfrac{dv}{-v^3}}=\int{2dt}

\dfrac{1}{2v^2}=2t+C

v^2=\dfrac{1}{4t+2C}....(I)

At t = 0, v = 10 m/s

10^2=\dfrac{1}{4\times0+2C}

C=\dfrac{1}{200}

Put the value of C in equation (I)

v^2=\dfrac{1}{4\times25+2\times\dfrac{1}{200}}

v=\sqrt{\dfrac{1}{4\times25+2\times\dfrac{1}{200}}}

v=0.099\ m/s

The velocity is 0.099 m/s.

(b). We need to calculate the position at t = 25 sec

The velocity is the first derivative of position of the particle.

\dfrac{ds}{dt}=v

On integrating

\int{ds}=\int(\sqrt{\dfrac{200}{800t+1}})dt

s=\dfrac{\sqrt{200}\times2\sqrt{800t+1}}{800}+C'

At t = 0, s = 15 m

15=\dfrac{200}{800}+C'

C'=15-\dfrac{200}{800}

C'=14.75

Put the value in the equation

s=\dfrac{\sqrt{200}\times2\sqrt{800\times25+1}}{800}+14.75

s=19.75\ m

The position is 19.75 m.

Hence, (a). The velocity is 0.099 m/s.

(b). The position is 19.75 m.

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Answer:

0.0195 m

Explanation:

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d = depth of puck below surface of mercury

According to Archimedes principle, the weight of puck is balanced by the weight of mercury displaced by puck

Weight of mercury displaced = Weight of puck

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