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2 years ago

(e) For photons of energy 7.10 eV, what stopping potential would be required to arrest the current of photoelectrons

Physics

1 answer:

Masja [62]2 years ago

8 0

2.37eV stopping potential would be required to arrest the current of photoelectrons.

<h3 /><h3>What is stopping potential ?</h3>

The minimal negative voltage that must be provided to the anode to halt the photocurrent is known as stopping potential. When expressed in electron volts, the maximal kinetic energy of the electrons is equal to the stopping voltage.

Kmax = eV₀

2.37eV = eV₀

V₀ = 2.37eV

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Find expressions for the force needed to bring an object of mass m from rest to speed v in time t. express your answer in terms

VARVARA [1.3K]

Good morning.

We have that:

$\mathsf{V = a\cdot t}$ , since we have rest in the inicial time.

The acceleration can be found with Newton's Law:

$\mathsf{F = m\cdot a\iff a = \dfrac{F}{m}}$

Now we put the acceleratin in the velocity equation:

$\mathsf{V = \dfrac{F}{m} \cdot t}$

We want the force, so, let's isolate F:

$\mathsf{V\cdot m = F\cdot t}\\ \\ \\ \boxed{\mathsf{F = \dfrac{V\cdot m}{t}}}$

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A cricket starts at the 5 meter mark on a ruler. It jumps to the 10 meter mark. Then, it jumps back to the 6 meter mark. Calcula

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We have that the cricket’s displacement is

$D_t=9meters$

From the Question we are told that

Start position 5 meter mark

It jumps to the 10 meter mark.

it jumps back to the 6 meter mark.

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$D_t=Initial +final(displacement)\\\\D_t=(10-5)+(10-6)$

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3 years ago

A bicycle takes 8.0 seconds to accelerate at a constant rate from rest to a speed of 4.0 m/s. If the mass of the bicycle and rid

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3 years ago

Read 2 more answers

A rod extending between x = 0 and x = 13.0 cm has uniform cross-sectional area A = 8.00 cm2. Its density increases steadily betw

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Answer:

The mass is $m = 3.45408 kg$

Explanation:

From the question we are told that

The extension of the rod is $from , \ x_1 = 0 \to x_2 = 13.0$

The area is $A = 8.0 cm^2$

The density increase as follows $from \ \rho_1 =2.5 g/cm^2 \to \rho_2= 19.0 g/cm^3$

The equation $\rho = B + Cx$

at $x_1= 0$ $\rho_1 =2.5 g/cm^2$

$2.5 = B + 0$

=> $B =2.5$

So at $x_2 = 13.0$ , $\rho_2= 19.0 g/cm^3$

$19.0 = 2.5 + C(13)$

=> $C = 1.27$

Now

$m = 8 \int\limits^{13}_{0} {2.5 + 1.27x} \, dx$

$m = 8 [{2.5 +\frac{ 1.27x^2}{2} } ]\left | 13} \atop {0}} \right.$

$m = 8 [{2.5 +\frac{ 1.27(13)^2}{2} } ]$

$m = 3454.08 g$

$m = 3.45408 kg$

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