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navik [9.2K]
3 years ago
14

A ball with an initial velocity of 12 m/s is rolling up a hill. The ball accelerates at a rate of -2.6 m/s/s.

Physics
1 answer:
Anon25 [30]3 years ago
4 0

Answer:

4.61 seconds

Explanation:

Given data

Initial velocity= 12m/s

acceleration= -2.6m/s^2

From the given data

we can find the time t

we know that

Acceleration= velocity/time

time= velocity/acceleration

time= 12/2.6

time= 4.61 seconds

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Se lanza una pelota de béisbol desde la azotea de un edificio de 25 m de altura con velocidad inicial de magnitud 10 m/s y dirig
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 v_f = 24.3 m / s

Explanation:

A) In this exercise there is no friction so energy is conserved.

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Final point. On the floor

         Em_f = K = ½ m v_f²

         Emo = Em_g

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6 0
2 years ago
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Answer:

\lambda=1.37\times 10^{-11}\ m

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Give that,

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Using the conservation of energy,

2meV=\dfrac{h^2}{\lambda^2}\\\\\lambda=\sqrt{\dfrac{h^2}{2meV}}

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\lambda=\sqrt{\dfrac{(6.63\times 10^{-34})^2}{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 8\times 10^3}}\\\\\lambda=1.37\times 10^{-11}\ m

So, the wavelength of the electrons is 1.37\times 10^{-11}\ m.

7 0
2 years ago
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