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klemol [59]
3 years ago
15

At t=10 ~\text{s}t=10 s, a particle is moving from left to right with a speed of 5.0 ~\text{m/s}5.0 m/s. At t=20 ~\text{s}t=20 s

, the particle is moving right to left with a speed of 8.0 ~\text{m/s}8.0 m/s. Assuming the particle’s acceleration is constant, determine its average acceleration. Assume positive is on the right.
Physics
1 answer:
EastWind [94]3 years ago
8 0

Answer:

a=6.5m/s^2 to the left.

Explanation:

We can use the equation v=v_0+a(t-t_0) where v is the velocity at time t and v_0 the velocity at t_0. Since we want the acceleration we write this equation as:

a=\frac{v-v_0}{t-t_0}

Considering the <em>direction to the right as the positive one</em>, we have v_0=+5m/s at t_0=10s, and v=-8m/s at t_0=8s, so we substitute:

a=\frac{v-v_0}{t-t_0}=\frac{(-8m/s)-(5m/s)}{(10s)-(8s)}=\frac{-13m/s}{2s}=-6.5m/s^2

Where the minus sign indicates it is directed to the left.

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When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a
sergij07 [2.7K]

Answer:

The right solution is:

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Explanation:

Given:

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Now,

(a)

The energy of photon will be:

E = \frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}

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or,

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(b)

As we know,

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By substituting the values, we get

⇒ 1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0

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or,

⇒                            =\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}

⇒                            =1.4375 \ eV

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anyanavicka [17]

Answer:

I think the answer is

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7m is the answer

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