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ehidna [41]
3 years ago
7

gayle cooks a roast in her microwave oven. the klystron tube in the oven emits photons whose energy is 1.20 x 10^-3 ev. what are

the wavelengths of these photons
Physics
1 answer:
AveGali [126]3 years ago
7 0

Answer:

\lambda=1.03\times 10^{-3}\ m

Explanation:

Given that,

The energy of the microwave oven is 1.2\times 10^{-3}\ eV.

We need to find the wavelength of these photons.

1.2\times 10^{-3}\ eV=1.2\times 10^{-3}\times 1.6\times 10^{-19}\\\\=1.92\times 10^{-22}\ J

The energy of a wave is given by :

E=\dfrac{hc}{\lambda}\\\\\lambda=\dfrac{hc}{E}

Put all the values,

\lambda=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{1.92\times 10^{-22}}\\\\\lambda=1.03\times 10^{-3}\ m

So, the wavelength of these photon is 1.03\times 10^{-3}\ m.

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A double concave lens is a ________ lens, and a double convex lens is a ______ lens.
forsale [732]

Answer:

b. diverging, converging

4 0
2 years ago
If a range hood removes contaminants at a flow rate of F liters of air per​ second, then the percent P of contaminants that are
podryga [215]

Explanation:

The given value of P is as follows.

           P = 1.06F + 22.18,           10 \leq F \leq 70

or,       P' = 1.06

As p' is defined and non-zero. Hence, only critical points are boundary points.

For F = 10, the value of P will be calculated as follows.

         P = 1.06 \times 10 + 22.18

            = 32.78

For F = 70, the value of P will be calculated as follows.

        P = 1.06 \times 70 + 22.18

           = 96.38

Therefore, the minimum value of P is 32.78 and maximum value of P is 96.38.

3 0
3 years ago
Suppose that during any period of ¼ second there is one instant at which the crests or troughs of component waves are exactly in
kolezko [41]
<h3><u>Answer;</u></h3>

A. 4

<h3><u>Explanation;</u></h3>
  • <em><u>The period of a wave or periodic time is the time taken for a complete oscillation to occur. </u></em>For example its is the time taken between two successive crests or troughs.
  • <em><u>The beats or oscillation that occur in one second represents the frequency. Frequency is the number of complete oscillations or beats in one second in a wave.</u></em>
  • Frequency, measured in Hertz is given by the reciprocal of the periodic time.
  • Thus; <u><em>Frequency or beats per second = 1/(1/4) = 4</em></u>
  • <u><em>Hence , 4 beats per second</em></u>

6 0
3 years ago
A mass of 1 kg is moving in a circle of radius 3.3 meters. What is the liner velocity v m/s that would give a Centripetal Force
Ghella [55]

Answer:

The linear velocity of the object is 8.71 m/s.

Explanation:

Given;

mass of the object, m = 1 kg

radius of the circle, r = 3.3 meters

centripetal force, F = 23 N

Centripetal force is given by;

F_c = \frac{mv^2}{r}\\\\

where;

v is the linear velocity of the object

F_c = \frac{mv^2}{r}\\\\mv^2 = F_cr\\\\v^2 = \frac{F_cr}{m}\\\\v= \sqrt{\frac{F_cr}{m}} \\\\v= \sqrt{\frac{23*3.3}{1}}\\\\v = 8.71 \ m/s

Therefore, the linear velocity of the object is 8.71 m/s.

4 0
3 years ago
(a) Consider a fusion generator built to create 3.00 GW of power. Determine the rate of fuel burning in grams per hour if the DT
Stels [109]

The rate of fuel burning in grams per hour if the DT reaction is used is 1.08 ×10^1^3 J/g per hour

<h3>How is the rate of fuel burning in grams per hour calculated when the D-T reaction is used?</h3>
  • D + T → He + n
  • The D-T fusion reaction results in a Helium (He) and  neutron (n)

E = 17.59 MeV

Mass = 2.014u + 3.016u

= 5.030u

Energy per Kg = (17.59×10^6×1.6×10^-^1^9) ÷ ( 5.030×1.66×10^-^2^7)

= 3.37×10^1^4 J/Kg

= 3.0× 10^9 J/g

Rate of fuel burning in grams per hour = 3.0× 10^9 ×  3600

= 3.6×3.0×10^1^2

= 1.08 ×10^1^3 J/g per hour

To learn more about fusion reactor and energy production, refer

brainly.com/question/13399644

#SPJ4

4 0
2 years ago
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