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Hitman42 [59]
3 years ago
5

What is simple machine

Physics
2 answers:
const2013 [10]3 years ago
6 0
Any of the basic mechanical devices for applying a force, such as an inclined plane, wedge, or lever.
vodka [1.7K]3 years ago
6 0

Hello!

Explanation:

↓↓↓↓↓↓↓↓↓↓↓↓

Simple machines have few or no moving parts. They make work easier.

  1. Inclined plane
  2. Wedge
  3. Screw
  4. Lever
  5. Wheel and axle
  6. Pulley

Inclined plane is a sloping surface, such as a ramp. Inclined plane make it easier to lift objects. The distance is greater, but the amount of force used is less. The amount of work done is the same. Wedge is used to split things apart. Force is applied to the large area of the wedge. The force is concentrated on a smaller area at the pointed end. The wedge looks like two inclined planes placed back to back. A screw can hold things together. The threading grips materials. The threading on a screw looks like an inclined plane spiraling around the outside of the screw. Large screw are also used to lift things. A car jack lifts a car by using a large screw. A lever is a strong bar that turns about a fulcrum, or fixed point. It can be used to lift heavy objects. Force applied at one end is transmitted to the other end. A lever can change the direction of a force. Moving the position of the fulcrum affects how much force is needed. A wheel and axle consists of two circular objects of different sizes that are connected. The smaller-diameter object, the axle, is a rod that goes through the wheel and lets the wheel turn. When the axle is turned, the wheel moves a greater distance, but less force is needed to move it. A pulley changes the direction of a force. A pulley consists of a grooved wheel and a rope. Pulling down on the rope lifts objects. A construction crane uses a series of pulleys. A flagpole has a pulley at the top.

Hope this helps!

Thank you for posting your question at here on brainly.

-Charlie

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Why isn't pluto considered a planet anymore?
aivan3 [116]

The answer is. It did not meet the three criteria the IAU uses to define a full-sized planet.

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3 0
3 years ago
A charge of 32.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of 4.30x 104 V/
hodyreva [135]

A) The work done by the electric field is zero

B) The work done by the electric field is 9.1\cdot 10^{-4} J

C) The work done by the electric field is -2.4\cdot 10^{-3} J

Explanation:

A)

The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).

The work done by a force is given by the equation

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement of the particle

\theta is the angle between the direction of the force and the direction of the displacement

In this problem, we have:

  • The force is directed vertically upward (because the field is directed vertically upward)
  • The charge moves to the right, so its displacement is to the right

This means that force and displacement are perpendicular to each other, so

\theta=90^{\circ}

and cos 90^{\circ}=0: therefore, the work done on the charge by the electric field is zero.

B)

In this case, the charge move upward (same direction as the electric field), so

\theta=0^{\circ}

and

cos 0^{\circ}=1

Therefore, the work done by the electric force is

W=Fd

and we have:

F=qE is the magnitude of the electric force. Since

E=4.30\cdot 10^4 V/m is the magnitude of the electric field

q=32.0 nC = 32.0\cdot 10^{-9}C is the charge

The electric force is

F=(32.0\cdot 10^{-9})(4.30\cdot 10^4)=1.38\cdot 10^{-3} N

The displacement of the particle is

d = 0.660 m

Therefore, the work done is

W=Fd=(1.38\cdot 10^{-3})(0.660)=9.1\cdot 10^{-4} J

C)

In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is

\theta=90^{\circ}+45^{\circ}=135^{\circ}

Moreover, we have:

F=1.38\cdot 10^{-3} N (electric force calculated in part b)

While the displacement of the charge is

d = 2.50 m

Therefore, we can now calculate the work done by the electric force:

W=Fdcos \theta = (1.38\cdot 10^{-3})(2.50)(cos 135.0^{\circ})=-2.4\cdot 10^{-3} J

And the work is negative because the electric force is opposite direction to the displacement of the charge.

Learn more about work and electric force:

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kaheart [24]
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