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2 years ago

It takes 500W of power to move an object 96 m in 12 s. What force is being applied to the object?

Physics

1 answer:

Marianna [84]2 years ago

5 0

The force being applied to the object is calculated to be 62.5N.

<h3>POWER:</h3>

Power is the rate at which work is done. It is measured in watts (W). Power can be calculated by multiplying the force applied on a given object by its speed.

Power (P) = Force (F) × speed (V)

According to this question, it takes 500W of power to move an object 96 m in 12 s. The speed of the object in m/s is calculated as: 96m/12s = 8m/s.

The force applied on the object = 500/8

Force = 62.5N

Therefore, the force being applied to the object is calculated to be 62.5N.

Learn more about force at: brainly.com/question/1436187

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3 years ago

Read 2 more answers

A plastic rod 1.3 m long is rubbed all over with wool, and acquires a charge of -3e-08 coulombs. we choose the center of the rod

Anestetic [448]

(a) The plastic rod has a length of L=1.3m. If we divide by 8, we get the length of each piece:
$L/8=1.3m/8=0.1625 m$

(b) The center of the rod is located at x=0. This means we have 4 pieces of the rod on the negative side of x-axis, and 4 pieces on the positive side. So, starting from x=0 and going towards positive direction, we have: piece 5, piece 6, piece 7 and piece 8. Each piece is 0.1625 m long. Therefore, the center of piece 5 is at 0.1625m/2=0.0812 m. And the center of piece 6 will be shifted by 0.1625m with respect to this:
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(c) The total charge is $Q=-3 \cdot 10^{-8}C$ . To get the charge on each piece, we should divide this value by 8, the number of pieces:
$Q/8=-3\cdot 10^{-8}C/8=-3.75\cdot 10^{-9}C$

(d) We have to calculate the electric field at x=0.7 generated by piece 6. The charge on piece 6 is the value calculated at point (c):
$q= -3.75\cdot 10^{-9}C$
If we approximate piece 6 as a single charge, the electric field is given by
$E=k_e \frac{q}{d^2}$
where $k_e=8.99\cdot 10^9Nm^2C^{-2}$ and d is the distance between the charge (center of piece 6, located at 0.2437m) and point a (located at x=0.7m). Therefore we have
$E= 8.99\cdot 10^9 Nm^2C^{-2} \frac{-3.75\cdot 10^9 C}{(0.2437m-0.7m)^2} =-161.9 V/m$
poiting towards the center of piece 6, since the charge is negative.

(e) missing details on this question.

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To explain it in more detail:

Right in the center of the Earth there is an internal core of solid iron at a very high temperature, this iron is as hot as the surface of the Sun, however, the pressure caused by the gravity force prevents it from becoming liquid. Around this inner core is the outer core, a layer of approximately 2000 km thick, composed of iron, nickel and small amounts of other metals in liquid state. Due to the lower pressure of the outer core, in relation to the inner core, the metal here is fluid (molten).

The differences in temperature, pressure and composition within the outer core cause convection currents in the molten metal, as the cold, dense matter sinks, and the warm, less dense matter rises. In addition, the Coriolis force, resulting from the spin of the Earth, also causes eddies in this mixture of molten metals.

This is how the movement of liquid iron inside the Earth generates electric currents, which in turn produce magnetic fields, and the spiral caused by the Coriolis force causes the magnetic fields to align in approximately the same direction, being this combined effect added to produce a vast magnetic field that envelops the entire planet.

This field extends from the terrestrial nucleus, progressively attenuating itself in outer space. However, it should be noted that unlike the field of a bar magnet, the Earth's magnetic field changes over time because it is actually generated by the movement of the iron alloys in the Earth's outer core, which causes the poles to move.

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