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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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Nata [24]

Answer:

5 Km/h

Explanation:

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

Distance travelled = 10 Km

Time = 2 hours

Speed =?

Speed is simply defined as the distance travelled per unit time. Mathematically, it can be represented as:

Speed = distance travelled /time.

With the above formula, we can obtain the speed at which the duck is travelling as follow:

Distance travelled = 10 Km

Time = 2 hours

Speed =?

Speed = distance travelled /time.

Speed = 10 / 2

Speed = 5 Km/h

Thus, the duck is travelling at a speed of 5 Km/h

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Answer:

is sugar and water becouse sugar can solote the water

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the graph shows the cost of natural hazards in the us in 2011 (in billions of dollars)

Explanation:

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2 years ago
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In a pressure cooker, a lid is screwed tightly onto a pot so that the air pressure in the pot is increased. What effect, if any,
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A hotel elevator ascends 200m with maximum speed of 5m/s. Its acceleration and deceleration both have a magnitude of 1.0m/s2. Pa
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Answer:

45 s .

Explanation:

The accelerator will first accelerate , then move with uniform velocity and at last it will decelerate to rest .

displacement s = ?

acceleration a = 1 m /s²

Final speed v = 5 m/s

initial speed u = 0

v² = u² + 2as

5² = 0 + 2 x 1 x s

s = 12.5  m

B)  Let time of acceleration or deceleration be t

v = u + a t

5 = 0 + 1 t

t = 5 s

Similarly displacement during deceleration = 12.5 m

Total distance during uniform motion = 200 - ( 12.5 + 12.5 ) =  175 m .

velocity of uniform motion = 5 m /s

time during which there was uniform velocity = 175 / 5 = 35 s

Total time = 5 + 35 + 5 = 45 s .

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