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OverLord2011 [107]
3 years ago
6

How many photons per second are emitted by the antenna of a microwave oven if its power output is 1.00 kw at a frequency of 2515

mhz?
Physics
1 answer:
Norma-Jean [14]3 years ago
8 0
The power of the antenna is
P=1.00 kW=1000 W
and since the power is the emitted energy divided by the time, the energy emitted per second is
E=Pt = (1000 W)(1 s)=1000 J

The frequency of the emitted photons is
f=2515 MHz=2.515 \cdot 10^9 Hz
So the energy of a single photon is
E_1=hf=(6.6 \cdot 10^{-34} Js)(2.515 \cdot 10^9 Hz)=1.66 \cdot 10^{-24} J

Therefore, to find the number of emitted photons per second, we should divide the total energy emitted by the antenna in 1 second by the energy of a single photon:
N= \frac{E}{E_1}= \frac{1000 J}{1.66 \cdot 10^{-24} J}=  6.02 \cdot 10^{26}
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3 years ago
The component of the external magnetic field along the central axis of a 78-turn circular coil of radius 34.0 cm decreases from
grigory [225]

Answer:

Induced current, I = 18.88 A

Explanation:

It is given that,

Number of turns, N = 78

Radius of the circular coil, r = 34 cm = 0.34 m

Magnetic field changes from 2.4 T to 0.4 T in 2 s.

Resistance of the coil, R = 1.5 ohms

We need to find the magnitude of the induced current in the coil. The induced emf is given by :

\epsilon=-N\dfrac{d\phi}{dt}

Where

\dfrac{d\phi}{dt} is the rate of change of magnetic flux,

And \phi=BA

\epsilon=-NA\dfrac{dB}{dt}

\epsilon=-78\times \pi (0.34)^2\dfrac{(0.4-2.4)}{2}

\epsilon=28.32\ V

Using Ohm's law, \epsilon=I\times R

Induced current, I=\dfrac{\epsilon}{R}

I=\dfrac{28.32}{1.5}

I = 18.88 A

So, the magnitude of the induced current in the coil is 18.88 A. Hence, this is the required solution.

5 0
3 years ago
39 g aluminum spoon (specific heat 0.904 J/g·°C) at 24°C is placed in 166 mL (166 g) of coffee at 83°C and the temperature of th
tatuchka [14]

<u>Answer:</u> The final temperature of the solution is 80.14^oC

<u>Explanation:</u>

The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, \text{heat}_{absorbed}=\text{heat}_{released}

To calculate the amount of heat released or absorbed, we use the equation:  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

Also,

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]    ..........(1)

where,

q = heat absorbed or released

m_1 = mass of aluminium = 39 g

m_2 = mass of coffee = 166 g

T_{final} = final temperature = ?

T_1 = temperature of aluminium = 24^oC

T_2 = temperature of coffee = 83^oC

c_1 = specific heat of aluminium = 0.904J/g^oC

c_2 = specific heat of coffee= 4.1801J/g^oC

Putting all the values in equation 1, we get:

39\times 0.904\times (T_{final}-24)=-[166\times 4.1801\times (T_{final}-83)]

T_{final}=80.14^oC

Hence, the final temperature of the solution is 80.14^oC

4 0
3 years ago
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