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Vladimir [108]
3 years ago
12

If the work function for a certain metal is 1.8eV, what is the stopping potential for electrons ejected from the metal when ligh

t of wavelength 400nm shines on the metal? And what is the maximum speed of the ejected electrons? ...?
Physics
1 answer:
taurus [48]3 years ago
8 0
First of all the equation you need is this one
<span>E=<span>Ek</span>+ϕ</span> <span><span>Ek</span>=e<span>V0
</span></span>Please remember that the energy of a photon of wavelength lambda is given by <span>E=<span><span>hc</span>λ

</span></span>you can then calculate the maximum kinetic energy since you know the work function of the metal (1.8eV), and the wavelength of the photons (400 nm) (remembering to convert to SI units of joules for the work function). You can then find the maximum velocity of the electrons easy enough.You can obtain the stopping potential usign the next formula
Ek=eV0
<span>Because it is a stopping voltage, V_{0}  and remember it will have the units of volts.
</span>I think with that I can help you a lot. 
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A grating whose slits are 3.2x10^-4 cm apart produces a third-order fringe at a 25.°0 angle. What is the wavelength of light tha
Ber [7]

Answer:

The light used has a wavelenght of 4.51×10^-7 m.

Explanation:

let:

n be the order fringe

Ф be the angle that the light makes

d is the slit spacing of the grating

λ be the wavelength of the light

then, by Bragg's law:

n×λ = d×sin(Ф)

λ = d×sin(Ф)/n

λ = (3.2×10^-4 cm)×sin(25.0°)/3

  = 4.51×10^-5 cm

  ≈ 4.51×10^-7 m

Therefore, the light used has a wavelenght of 4.51×10^-7 m.

7 0
3 years ago
The net force (unbalanced force) applied on an object is the product of_________and___________​
RSB [31]

Answer:

of its <u>mass</u> and its <u>acceleration</u>

<u>Explanation:</u>

based on Newton's second law of motion

5 0
2 years ago
The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in
OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

T_1+V_1=T_2+V_2

0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2}  \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}

Also;

\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}

Thus;

k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}

where;

\delta = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}

k = 104.82\ \  N/m

Thus; the spring constant = 104.82 N/m

b

The angular velocity can be calculated by also using the conservation of energy;

T_1+V_1 = T_3 +V_3  \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3}  \omega_3^2 +  \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0

\frac{m(a+b)^2}{3} \omega_3^2  + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0

\frac{1.53(0.6+0.6)^2}{3} \omega_3^2  + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0

0.7344 \omega_3^2 = 2.128

\omega _3 = \sqrt{\frac{2.128}{0.7344} }

\omega _3 =1.70 \ rad/s

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s

7 0
3 years ago
Adog has a mass of 12 kg. What is its weight? Round your answer to the nearest whole number
love history [14]

Answer:

26.5

Explanation:

3 0
3 years ago
True or False:
bekas [8.4K]
I think it false. Sorry if i'm wrong.

5 0
3 years ago
Read 2 more answers
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