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Nutka1998 [239]
3 years ago
15

Find the wavelength of a photon which has 500.0 eV of energy.

Physics
1 answer:
jekas [21]3 years ago
5 0

Answer:

2.5 x 10⁻⁹ m

Explanation:

E = Energy of photon = 500 eV = 500 x 1.6 x 10⁻¹⁹ J

c = speed of photon = 3 x 10⁸ m/s

λ = wavelength of photon = ?

Energy of photon is given as

E = \frac{hc}{\lambda }

inserting the values

500\times 1.6\times 10^{-19} = \frac{(6.63\times 10^{-34})(3\times 10^{8})}{\lambda }

λ = 2.5 x 10⁻⁹ m

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A 3.50-meter length of wire with a cross-sectional area of 3.14 × 10-6 meter2 is at 20° Celsius. If the wire has a resistance of
maks197457 [2]

Answer:

5.6\times 10^{-8}\ Ohm.m

Explanation:

Resistivity is given by \rho=\frac {AR}{L} where A is cross-sectional area, R is resistance, L is the length and \rho is the reistivity. Substituting 0.0625 for R, 3.14 × 10-6 for A and 3.5 m for L then the resistivity is equivalent to

\rho=\frac {3.14\times 10^{-6}\times 0.0625}{3.5}=5.60714285714285714285714285714285714285\times 10^{-8}\approx 5.6\times 10^{-8}\ Ohm.m

8 0
3 years ago
Please help!!
sukhopar [10]

Answer:

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The force exerted by the thrusters caused the pod to change direction.

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This pod moved differently because it was more massive.

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3 0
2 years ago
Hans Langseth's beard measured 5.33 m in 1927. Consider two charges, q1 = 2.42 nC and an unspecified charge, q2, are separated 5
schepotkina [342]

Answer:

-7.89 * 10^(-9) C

Explanation:

Parameters given:

q1 = 2.42 nC = 2.42 * 10^(-9) C

Distance between q1 and q2 = 5.33 m

q3 = 1.0 nC = 1 * 10^(-9) C

Distance between q1 and q3 = 1.9 m

Distance between q2 and q3 = 5.33 - 1.9 = 3.43 m

The net force acting on q3 is:

F = F(q1, q3) + F(q2, q3)

F = (k*q1*q3)/1.9² + (k*q2*q3)/3.43²

F = (9 * 10^(9) * 2.42 * 10^(-9) * 1 * 10^(-9))/3.61 + (9 * 10^(9) * q2 * 1 * 10^(-9))/11.7649

F = 6.033 * 10^(-9) + 0.765*q2

If the net force is zero:

0 = 6.033 * 10^(-9) + 0.765*q2

-0.765*q2 = 6.033 * 10^(-9)

=> q2 = -[6.033 * 10^(-9)]/0.765

q2 = -7.89 * 10^(-9) C

3 0
2 years ago
A weight lifter lifts a dumbbell a certain height in 2.0 s, while a competitor does the same workin 1.0 s. Compared to the power
Aloiza [94]

Answer:

a. one-half as great

Explanation:

The power developed by the first lifter is one-half as great as that of the second person.

  Power is defined as the rate at which work is done;

          Power  = \frac{workdone}{time}

Since the two lifters do the same work at different time, let us estimate their power;

       P₁ = \frac{workdone}{2}                     P₂ = \frac{workdone }{1}

   We see that for P₁, power is half of the work done whereas in P₂ power is the same as the work done.

Therefore,

           The power of the first weight lifter is one-half the second lifter.

4 0
2 years ago
What's the minimum Out PUT WORK<br> required to raise 14,0m3 of water 26.0m?
BartSMP [9]

Answer:

3.57 MJ

Explanation:

ASSUMING it's fresh water with density of 1000 kg/m³

W = ΔPE = mgΔh = 14.0(1000)(9.81)(26.0) = 3,570,840 J

Salt water would require more.

3 0
2 years ago
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