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3 years ago

How much kinetic energy does a baseball with a mass of 0.143 kg have it it is traveling at a velocity of 41.1 m/s?

Physics

1 answer:

vampirchik [111]3 years ago

3 0

Answer:

E = 120.77 J

Explanation:

Given that,

Mass of a baseball, m = 0.143 kg

Velocity of a baseball, v = 41.1 m/s

We need to find the kinetic energy of the baseball. We know that the kinetic energy of an object is associated with its motion. It can be given by the formula as follow :

$E=\dfrac{1}{2}mv^2\\\\E=\dfrac{1}{2}\times 0.143\ kg\times (41.1\ m/s)^2\\E=120.77\ J$

So, the kinetic energy of the baseball is 120.77 J.

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What must be the magnitude of a uniform electric field if it is to have the same energy density as that possessed by a 0.50 T ma

Sindrei [870]

Answer:

E = 1.50 × $10^{8}$ V/m

Explanation:

given data

B = 0.50 T

solution

we know that energy density by the magnetic field is express as

$\mu _b = \frac{B}{2\mu _o}$ ...............1

and

energy density due to electric filed is

$\mu _e = \frac{\epsilon _o E^2}{2}$ ...............2

and here $\mu _b = \mu _ e$

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put here value and we get

$E = \frac{0.50}{\sqrt{4\pi \times 10^{-7} \times 8.852 \times 10^{-12}}}$

E = 3 × $10^{8}$ × 0.50

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Since the electric field between the plates is constant, If the two plates are brought closer together, the potential difference between the two plates decreases

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Since the electric field is maintained constant, the potential difference is directly inversely proportional to the distance between the plates.

The potential difference between the plates will therefore likewise decrease if the distance between the plates is reduced, we will state in this case.

The energy required to move a unit charge, or one coulomb, from one point to the other in a circuit is measured as the potential difference between the two points. Potential difference is measured in volts or joules per coulomb.

Refer to more about the potential difference here

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Explanation:

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Answer:

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