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ladessa [460]
3 years ago
10

Define reversible change​

Physics
1 answer:
inysia [295]3 years ago
8 0

A reversible change is a change that can be undone or reversed

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Find the wavelength of light given off by a hydrogen atom when its electron drops from the n = 4 to n = 1 energy level
4vir4ik [10]

Answer: 1.55 x 10 -19J on top

Explanation:

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A football player kicks a ball at a 30o angle from the ground with an initial velocity of 15 m/s. What is the final velocity of
navik [9.2K]

Given that,

Angle = 30°

Initial velocity = 15 m/s

We need to calculate the time of flight

Using formula of time of flight

T=\dfrac{2u\sin\theta}{g}

Where, u = initial velocity

g = acceleration due to gravity

Put the value into the formula

T=\dfrac{2\times15\sin30}{9.8}

T=1.5\ sec

We need to calculate the final velocity of the ball

Using equation of motion

v=u+gt

v=15+9.8\times1.5

v=29.7\ m/s

Hence, The final velocity of the ball is 29.7 m/s.

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What are two ways in which friction can be useful?
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If something is going down a hill it can help slow it down

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3 years ago
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Which other subatomic particle has the same mass as a neutron
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Calculate the speed of a proton after it accelerates from rest through a potential difference of 350 V.
AVprozaik [17]

The speed of a proton after it accelerates from rest through a potential difference of 350 V is 25.86 \times 10^4 ~m/s.

Initial velocity of the proton u = 0

Given potential difference \Delta V = 350V

let's assume that the speed of the proton is v,

Since the proton is accelerating through a potential difference, proton's potential energy will change with time. The potential energy of a particle of charge q when accelerated with a potential difference \Delta V is,

    U = q \Delta V

Due to Work-Energy Theorem and Conservation of Energy - <em>If there is no non-conservative force acting on a particle then loss in Potential energy  P.E must be equal to gain in Kinetic Energy K.E</em> i.e

\Delta K = \Delta V

If the initial and final velocity of the proton is u and v respectively then,

change in Kinetic Energy  \implies  \Delta K = \frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}mv^2 - 0

change in Potential Energy \implies \Delta U = q\Delta V

from conservation of energy,

             v= \sqrt{\frac{2q\Delta V}{m}}

so,         v = \sqrt{\frac{2\times 350 \times 1.6\times 10^{-19}}{1.67 \times 10^{-27}}

                = 25.86 \times 10^4 ~m/s

To read more about the conservation of energy, please go to brainly.com/question/14668053

7 0
1 year ago
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