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kirill [66]
3 years ago
15

An eccentric inventor attempts to levitate by first placing a large negative charge on himself and then putting a large positive

charge on the ceiling of his workshop. Instead, while attempting to place a large negative charge on himself, his clothes fly off. Explain.
Physics
1 answer:
GuDViN [60]3 years ago
3 0

Answer:

shirt is little attached to the body, it can come off and fly away

Explanation:

In electrostatics, charges of different signs attract and charges of the same sign repel.

In this case, when a negative charge is placed on it, both the inventor and the shirt are charged, therefore there is a repulsive force, also there is an attraction between the positive charge of the roof attracts the negative charge, such as the shirt. of weak the two forces not greater than the resistance of the walk.

As the shirt is little attached to the body, it can come off and fly away

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A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl
Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)

here, \lambda is the initial photon wavelength, \lambda^{'} is the scattered photon wavelength, h is he Planck's constant, m_e is the free electron mass, c is the velocity of light, \theta is the angle of scattering.

Given that, the scattering angle is, \theta=147^{\circ}

Putting the respective values, we get

\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.

Here, the photon's incident wavelength is \lamda=2.78pm

Therefore,

\lambda^{'}=2.78+4.45=7.23 pm

From the conservation of momentum,

\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}

where,\vec{P_\lambda} is the initial photon momentum, \vec{P_{\lambda^{'}}} is the final photon momentum and \vec{P_e} is the scattered electron momentum.

Expanding the vector sum, we get

P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta

Now expressing the momentum in terms of De-Broglie wavelength

P=h/\lambda,

and putting it in the above equation we get,

\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}

Therefore,

\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm

This is the de Broglie wavelength of the electron after scattering.

6 0
4 years ago
A tennis ball, 0.314kg, is accelerated at a rate of 164m/s2 when hit by a professional tennis player. What force does the player
Colt1911 [192]

Newton's 2nd law of motion:

Force = (mass) x (acceleration)

= (0.314 kg) x (164 m/s²)

=  51.5 newtons

(about 11.6 pounds).

Notice that the ball is only accelerating while it's in contact with the racket. The instant the ball loses contact with the racket, it stops accelerating, and sails off in a straight line at whatever speed it had when it left the strings.

~ I hope this helped, and I would appreciate Brainliest. ♡ ~ ( I request this to all the lengthy answers I give ! )

5 0
3 years ago
Kepler-62e is an exoplanet that orbits within the habitable zone around its parent star. The planet has a mass that is 3.57 time
skad [1K]

Answer:

g' = 13.5 m/s²

Explanation:

The acceleration due to gravity on surface of earth is given by the formula:

g = GMe/Re²   --------------- euation 1

where,

g = acceleration due to gravity on surface of earth

G = Universal Gravitational Constant

Me = Mass of Earth

Re = Radius of Earth

Now, the the acceleration due to gravity on the surface of Kepler-62e is:

g' = GM'/R'²   --------------- euation 1

where,

g' = acceleration due to gravity on surface of Kepler-62e

G = Universal Gravitational Constant

M' = Mass of Kepler-62e = 3.57 Me

R' = Radius of Kepler-62e = 1.61 Re

Therefore,

g' = G(3.57 Me)/(1.61 Re)²

g' = 1.38 GMe/Re²

using equation 1:

g' = 1.38 g

where,

g = 9.8 m/s²

Therefore,

g' = 1.38(9.8 m/s²)

<u>g' = 13.5 m/s²</u>

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