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Dima020 [189]
3 years ago
5

A 1.0 kilogram ball is thrown into the air with an initial velocity of 30m/s. How much kinetic energy does the ball have? (B) ho

w much potential energy does the ball have when it reaches the top of its accent? (C) how high into the air did the ball travel?
Physics
1 answer:
Gala2k [10]3 years ago
7 0

Answer:

450 joules ; 450 joules ; 45.9 m

Explanation:

Given that :

Initial Velocity, u = 30m/s

Mass, m = 1 kg

Kinetic Energy of ball (KE) = 0.5mu²

K. E = 0.5 * 1 * 30^2

K.E = 0.5 * 900

K.E = 450 Joules

B.) Potential Energy (P. E)

P. E = mgh

At the highest point, all kinetic energy has would have become potential energy, hence

K. E = P. E = 450 Joules

C) Height of the ball :

From ; P. E = mgh

Where ; g = acceleration due to gravity = 9.8m/s² ; h = height

450 = 1 * 9.8 * h

450 = 9.8h

h = 450 / 9.8

h = 45.918

h = 45.9 m

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A car possesses 20,000 units of momentum. what would be the car's new momentum if ... its velocity was doubled?
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7 0
3 years ago
A photon of wavelength 7.33 pm scatters at an angle of 157° from an initially stationary, unbound electron. What is the de Brogl
Ann [662]

Answer:

4.63 p.m.

Explanation:

The problem given here can be solved by the 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=157^{\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-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42(1-cos157^\circ ) p.m.

Therfore,

\lambda^{'}-\lambda=4.64 p.m.

Here, the photon's incident wavelength is \lamda=7.33pm

So,

\lambda^{'}=7.33+4.64=11.97 p.m

From the conservation of momentum,

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

here, \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}}

Therfore,

\lambda_{e}=\frac{7.33\times 11.97}{\sqrt{7.33^{2}+11.97^{2}-2\times 7.33\times 11.97\times cos157^\circ }} p.m.\\\lambda_{e}=\frac{87.7401}{18.935} = 4.63 p.m.

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

8 0
3 years ago
Josh tests an unknown liquid substance using litmus paper. When he compared the used litmus paper to a pH scale, the color match
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Answer:

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because ph below 7 is acid

8 0
2 years ago
Imagine that the ball on the left is given a nonzero initial velocity in the horizontal direction, while the ball on the right c
masya89 [10]

Answer:

vₓ = xg/2y

Explanation:

In this question, let us  find the time it takes for the ball on the right that has zero initial velocity to reach the ground.

By newton equation of motion we know that

y = v₀ t - ½ g t²

t = 2y / g

This is the time it takes for the ball on the right to reach the ground; at this time the ball on the left travels a distance

vₓ = x/t

vₓ = xg/2y

vₓ = xg/2y

Where we assume that x and y are known.

7 0
3 years ago
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