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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 confined aquifer with a porosity of 0.15 is 30 m thick. The potentiometric surface elevation at two observation wells 1000 m a
AlekseyPX

Answer:

Part (a) The flow rate per unit width of the aquifer is 1.0875 m³/day

Part (b) The specific discharge of the flow is 0.0363 m/day

Part (c) The average linear velocity of the flow is 0.242 m/day

Part (d) The time taken for a tracer to travel the distance between the observation wells is 4132.23 days = 99173.52 hours

Explanation:

Part (a) the flow rate per unit width of the aquifer

From Darcy's law;

q = -Kb\frac{dh}{dl}

where;

q is the flow rate

K is the permeability or conductivity of the aquifer = 25  m/day

b is the aquifer thickness

dh is the change in th vertical hight = 50.9m - 52.35m = -1.45 m

dl is the change in the horizontal hight = 1000 m

q = -(25*30)*(-1.45/1000)

q = 1.0875 m³/day

Part (b) the specific discharge of the flow

V = \frac{Q}{A} = \frac{q}{b} = -K\frac{dh}{dl}\\\\V = -(25 m/d).(\frac{-1.45 m}{1000 m}) = 0.0363 m/day

V = 0.0363 m/day

Part (c) the average linear velocity of the flow assuming steady unidirectional flow

Va = V/Φ

Φ is the porosity = 0.15

Va = 0.0363 / 0.15

Va = 0.242 m/day

Part (d) the time taken for a tracer to travel the distance between the observation wells

The distance between the two wells = 1000 m

average linear velocity = 0.242 m/day

Time = distance / speed

Time = (1000 m) / (0.242 m/day)

Time = 4132.23 days

        = 4132.23 days *\frac{24 .hrs}{1.day} = 99173.52, hours

4 0
3 years ago
Please help!!! 50 POINTS
taurus [48]

Answer:

The Energy Involved In A Reaction That Changes Methane Gas And Oxygen Into Carbon Dioxide And Water.

Explanation:

6 0
3 years ago
Read 2 more answers
A cart loaded with bricks has a total mass of 22.2 kg and is pulled at constant speed by a rope. The rope is inclined at 27.5 ◦
musickatia [10]

Answer:

W = 1.432 KJ

Explanation:

given,

mass = 22.2 Kg

angle of the rope = 27.5°

distance on the ground = 24 m

kinetic friction= μ = 0.32

acceleration due to gravity, g = 9.8 m/s²

Work done = ?

W = F d cosθ

a = 0 because it is moving with constant speed

equating all the forces acting  in x direction

F cosθ = F friction = μN  

equating all the forces acting  in y direction

F sinθ + N -mg =0

now,

N = mg - F sinθ

putting value of N

F cosθ = μ mg -μ F sinθ

F (cosθ + μsinθ ) = μ mg

F = \dfrac{\mu mg}{cos\theta + \mu sin\theta}

F = \dfrac{0.32 \times 22.2 \times 9.8}{cos 27.5^0+0.32 \times sin27.5^0}

F =67.28 N

now,

W=F d cosθ

W =67.28 x 24 x cos(27.5)

W =1432.27 J

W = 1.432 KJ

7 0
3 years ago
Read 2 more answers
A stone on ground is zero energy​
NNADVOKAT [17]

Answer:

A stone on the ground does not have zero energy…there is an internal potential in every object. Aldo is not in action or in any mechanical motion it is being acted upon by gravity and also molecular forces and energy.

<em>Hope</em><em> </em><em>this</em><em> helps</em><em> </em><em>!</em>

7 0
3 years ago
The speed of sound in air is around 330 m/s. If a bat emits a single high-pitched ‘click’ of sound in a cave that is 25m wide, c
Bingel [31]

Answer:

0.15 s

Explanation:

From the question given above, the following data were obtained:

Speed of sound (v) = 330 m/s

Distance (x) = 25 m

Time (t) =?

The time taken for the echo of the sound to the bat can be obtained as follow:

v = 2x / t

330 = 2 × 25 / t

330 = 50 / t

Cross multiply

330 × t = 50

Divide both side by 330

t = 50 / 330

t = 0.15 s

Thus, it will take 0.15 s for the echo of the sound to the bat

4 0
3 years ago
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