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3 years ago

Hii please help i’ll give brainliest if you give a correct answer please

Physics

1 answer:

elena55 [62]3 years ago

7 0

Answer:

they are equal and act in opposite directions

Explanation:

newton's third law of motion states that if object A exerts a force on object B, then object B will exert an equal but opposite in direction force.

hope it helps, I think my response was late,

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The electric force between two charged objects of charge +Q0 that are separated by a distance R0 is F0. The charge of one of the

vovikov84 [41]

Answer:

$F=3F_o(\frac{R_o}{R_{o2}})^2$

Explanation:

This problem is approached using Coulomb's law of electrostatic attraction which states that the force F of attraction or repulsion between two point charges, $Q_1$ and $Q_2$ is directly proportional to the product of the charges and inversely proportional to the square of their distance of separation R.

$F=\frac{kQ_1Q_2}{R^2}..................(1)$

where k is the electrostatic constant.

We can make k the subject of formula as follows;

$k=\frac{FR^2}{Q_1Q_2}...........(2)$

Since k is a constant, equation (2) implies that the ratio of the product of the of the force and the distance between two charges to the product of charges is a constant. Hence if we alter the charges or their distance of separation and take the same ratio as stated in equation(2) we will get the same result, which is k.

According to the problem, one of the two identical charges was altered from $Q_o$ to $3Q_o$ and their distance of separation from $R_o$ to $R_{o2}$ , this also made the force between them to change from $F_o$ to $F_{o2}$ . Therefore as stated by equation (2), we can write the following;

$\frac{F_oR_o^2}{Q_o*Q_o}=\frac{F_{o2}R_{o2}^2}{3Q_o*Q_o}.............(3)$

Therefore;

$\frac{F_oR_o^2}{Q_o^2}=\frac{F_{o2}R_{o2}^2}{3Q_o^2}.............(4)$

From equation (4) we now make the new force $F_{o2}$ the subject of formula as follows;

${F_oR_o^2}*{3Q_o^2}=F_{o2}R_{o2}^2*{Q_o^2}$

$Q_o$ then cancels out from both side of the equation, hence we obtain the following;

$3{F_oR_o^2}=F_{o2}R_{o2}^2.............(4)$

From equation (4) we can now write the following;

$F_{o2}=\frac{3F_oR_o^2}{R_{o2}^2}$

This could also be expressed as follows;

$F_{o2}=3F_o(\frac{R_o}{R_{o2}})^2$

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4 years ago

You can not exert force on a wall unless

ludmilkaskok [199]

unless...the wall simulaneiously exerts the same amount of force on you.

5 0

4 years ago

Which solid-state component can be used as a switch to turn current on or off?

Papessa [141]

The answer is Transistor. Its a semiconductor device used to amplify or switch electronic signals and electrical power. It is composed of semiconductor material usually with at least three terminals for connection to an external circuit.

4 0

3 years ago

Read 2 more answers

What happens to the charge on the conductive sphere when it is connected to a source of charge such as the electrostatic voltage

hichkok12 [17]

Answer and Explanation:

The charge on the conductive sphere spreads out non-uniformly over the surface of the sphere.

Normally, the charge on such spherical surface stay on this surface uniformly, but the presence of a voltage source tampers with that dynamic.

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3 years ago

A 1.23 x 10^3 kilogram car is traveling east at 25 meters per second. The brakes are applied and the car is brought to rest in 5

ahrayia [7]

Answer:

$\text{Magnitude: }30,000\:\text{Ns},\\\text{Direction: opposite direction of car's movement}}$

Explanation:

*Edit: The original question states a mass of $1.23\cdot 10^3\:\text{kg}$ . Since the, the poster has corrected it to $1.20\cdot 10^3\:\text{kg}$ and therefore the answers have been change to account for the typo.

The impulse-momentum theorem states that the impulse on a object is equal to the change in momentum of the object.

Therefore, we have the following equation:

$F\Delta t=\Delta p$ , where $F\Delta t$ is impulse (another way to find impulse) and $\Delta p$ is change in momentum.

Because the car is being slowed to a rest, its final velocity will be zero, and therefore its final momentum will also be zero. Since momentum is given as $p=mv$ , the car's change in momentum is $1.20\cdot 10^3\cdot 25-0=1230\cdot 25=30,000\:\text{kgm/s}$ .

As we wrote earlier, this is also equal to the magnitude of impulse on the object. The time it takes to stop the car is actually irrelevant to finding the total impulse. However, if we were to calculated the average applied force on the car, we would need how long it takes to bring it to rest (refer to $F\Delta t$ ).

The direction of the impulse must be exactly opposite to the car's direction, since we are slowing it to a stop.

Therefore, the impulse on the car is $\boxed{30,750\:\text{Ns, opposite direction of car's movement}}$ .

3 0

3 years ago