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3 years ago

How does the input distance of a single fixed pulley compare to the out- put distance?

1 answer:

4 0

A pulley is another sort of basic machine in the lever family. We may have utilized a pulley to lift things, for example, a banner on a flagpole.

<u>Explanation:</u>

The point in a fixed pulley resembles the support of a lever. The remainder of the pulley behaves like the fixed arm of a first-class lever, since it rotates around a point. The distance from the fulcrum is the equivalent on the two sides of a fixed pulley. A fixed pulley has a mechanical advantage of one. Hence, a fixed pulley doesn't increase the force.

It essentially alters the direction of the force. A moveable pulley or a mix of pulleys can deliver a mechanical advantage of more than one. Moveable pulleys are appended to the item being moved. Fixed and moveable pulleys can be consolidated into a solitary unit to create a greater mechanical advantage.

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An electron accelerated from rest through a voltage of 780 v enters a region of constant magnetic field. part a part complete if

maxonik [38]

The electron is accelerated through a potential difference of $\Delta V=780 V$ , so the kinetic energy gained by the electron is equal to its variation of electrical potential energy:
$\frac{1}{2}mv^2 = e \Delta V$
where
m is the electron mass
v is the final speed of the electron
e is the electron charge
$\Delta V$ is the potential difference

Re-arranging this equation, we can find the speed of the electron before entering the magnetic field:
$v= \sqrt{ \frac{2 e \Delta V}{m} } = \sqrt{ \frac{2(1.6 \cdot 10^{-19}C)(780 V)}{9.1 \cdot 10^{-31} kg} }=1.66 \cdot 10^7 m/s$

Now the electron enters the magnetic field. The Lorentz force provides the centripetal force that keeps the electron in circular orbit:
$evB=m \frac{v^2}{r}$
where B is the intensity of the magnetic field and r is the orbital radius. Since the radius is r=25 cm=0.25 m, we can re-arrange this equation to find B:
$B= \frac{mv}{er}= \frac{(9.1 \cdot 10^{-31}kg)(1.66 \cdot 10^7 m/s)}{(1.6 \cdot 10^{-19}C)(0.25 m)} =3.8 \cdot 10^{-4} T$

3 0

3 years ago

POSSIBLE POINTS: 1.92

gogolik [260]

Answer:

jnfal4u4ryhfsbjls5

Explanation:

duehdakjweyedufkbshegygfr

7 0

3 years ago

Help!!

scZoUnD [109]

If Fg=mg=ma and, Fg(planetX)=1/5Fg(earth)

then the time would be 5x of the time as gravity is acceleration. So 3.9s*5=19.5s

As the force of gravity is less, then the acceleration of masses is also less, therefore it will take more time for the object to fall by the factor of the force of gravity difference

7 0

3 years ago

Read 2 more answers

4. Many of the teachers think that if a student sleeps more on the night before the test, then they

Effectus [21]

Answer:

a sleep doesnt get to depict whether you get a good grade or not its what you know.

8 0

1 year ago

A 0.68 kg squirrel is resting on a branch 8 meters above the ground. What is the gravitational potential energy of a squirrel? A

ANEK [815]

Answer:

The gravitational potential energy of a squirrel is 53.312 J.

Explanation:

We have,

Mass of a squirrel is 0.68 kg

It is placed at a height of 8 m above the ground.

It is required to find the gravitational potential energy of a squirrel. It is possessed by an object due to its position. Its formula is given by :

$E=mgh\\\\E=0.68\times 9.8\times 8\\\\E=53.312\ J$

So, the gravitational potential energy of a squirrel is 53.312 J.

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3 years ago