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3 years ago

A third class lever has a mechanical advantage of <1. What is an example of a third class lever and why use it?

Physics

2 answers:

AVprozaik [17]3 years ago

7 0

B.) Baseball bat; increases velocity

All third class levers have a mechanical advantage less than 1. Since the output end has a longer distance from the fulcrum than the input point, the output end moves at a greater velocity than the input point. Because of this, third class levers are commonly used when trying to hit an object with as much velocity as possible.

34kurt3 years ago

5 0

Baseball bat. The handle of the bat is the fulcrum. Exerting a force from the handle supplies the input force just near the middle, while the other end of the baseball bat pushes the ball with the output forces. The input force is greater than the output force but the output load is able to move farther, and this increases the ball's velocity.

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Which statements describe a situation in which work is being done? Check all that apply.

Sphinxa [80]

Answer: A, D, E

Explanation:

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3 years ago

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(a) If a proton with a kinetic energy of 6.2 MeV is traveling in a particle accelerator in a circular orbit with a radius of 0.5

Tju [1.3M]

Answer:

The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

Explanation:

Suppose Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$

Given that,

Kinetic energy = 6.2 MeV

Radius = 0.500 m

We need to calculate the acceleration

Using formula of acceleration

$a=\dfrac{v^2}{r}$

Put the value into the formula

$a=\dfrac{\dfrac{1}{2}mv^2}{\dfrac{1}{2}mr}$

Put the value into the formula

$a=\dfrac{6.2\times10^{6}\times1.6\times10^{-19}}{\dfrac{1}{2}\times1.67\times10^{-27}\times0.51}$

$a=2.32\times10^{15}\ m/s^2$

We need to calculate the rate at which it emits energy because of its acceleration is

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$

Put the value into the formula

$\dfrac{dE}{dt}=\dfrac{(1.6\times10^{-19})^2\times(2.3\times10^{15})^2}{6\pi\times8.85\times10^{-12}\times(3\times10^{8})^3}$

$\dfrac{dE}{dt}=3.00\times10^{-23}\ J/s$

The energy in ev/s

$\dfrac{dE}{dt}=\dfrac{3.00\times10^{-23}}{1.6\times10^{-19}}\ J/s$

$\dfrac{dE}{dt}=1.875\times10^{-4}\ ev/s$

We need to calculate the fraction of its energy that it radiates every second

$\dfrac{\dfrac{dE}{dt}}{E}=\dfrac{1.875\times10^{-4}}{6.2\times10^{6}}$

$\dfrac{\dfrac{dE}{dt}}{E}=3.02\times10^{-11}$

Hence, The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

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3 years ago

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Law Incorporation [45]

1.from low potential to high potential
2. A circuit is an unbroken loop of conductive material through which electrons flow continuously.
3. There are two, a series circuit and a parallel circuit.
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3 years ago

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A charged box (m=445 g, ????=+2.50 μC) is placed on a frictionless incline plane. Another charged box (????=+75.0 μC) is fixed i

victus00 [196]

The concept required to perform this exercise is given by the coulomb law.

The force expressed according to this law is given by

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