The point in the orbit of a planet, asteroid, or comet at which it is closest to the sun.

Q=l^2 times R times t

Where Q - heat, I -current, R - resistance and t is time

If you increase I twice (and it's squared), than Q gonna went up 4 times (2 squared).

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The new **force** between the **charges **when the distance become **2d** will be **F'=4F**

<h3>What is electrostatic force?</h3>

**When **two charged particles are **separated** by the distance d then the force of **attraction **or repultion acts on the **charged** particle depending upon the nature of the **charge **whether it is positive or **negative**.

The **formula **for the **electrostatic **force is given by

Now if the **value **of **d** becomes** 2d** then the **formula **will become:

Hence the new **force** between the **charges **when the distance become **2d** will be **F'=4F**

To know more about **electrostatic force** follow

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The question here is that if sneezy hands from a similar rope while delivering presents at the earth's equator, what will be the tension in the rope be. Here is the solution:The tension on the rope when it is at pole, T= 455 NTo find, the tension, t= mgTo solve for mass, m= t/g. Substituting this we have, m=455/9.8. m=46.43 kgAssume that the downwards acceleration is, a= -46.43 m/s^2.T = mg + maT = (46.43 kg) ( 9.8 m/s^2) - (46.43 kg) (-46.43 m/s^2)T = 455.01 kg-m/s^2 - -2155.74 kg-m/s^2T = 2610.75 kg-m/s^2 = 2610.75 N

Answer: 330.88 J

Explanation:

Given

Linear velocity of the ball, v = 17.1 m/s

Distance from the joint, d = 0.47 m

Moment of inertia, I = 0.5 kgm²

The rotational kinetic energy, KE(rot) of an object is given by

KE(rot) = 1/2Iw²

Also, the angular velocity is given

w = v/r

Firstly, we calculate the angular velocity. Since it's needed in calculating the Kinetic Energy

w = v/r

w = 17.1 / 0.47

w = 36.38 rad/s

Now, substituting the value of w, with the already given value of I in the equation, we have

KE(rot) = 1/2Iw²

KE(rot) = 1/2 * 0.5 * 36.38²

KE(rot) = 0.25 * 1323.5

KE(rot) = 330.88 J