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3 years ago

What is the energy (in electron volts) of a 50 ghz (1 gigahertz = 109 hz) microwave photon?

1 answer:

8 0

The frequency of the microwave photon is:
$f=50 GHz=50 \cdot 10^9 Hz$
and its energy is given by:
$E=hf$
where
$h=6.6 \cdot 10^{-34}Js$ is the Planck constant
f is the photon frequency

By substituting the numbers into the formula, we find the photon's energy:
$E=hf=(6.6 \cdot 10^{-34}Js)(50 \cdot 10^9 Hz)=3.3 \cdot 10^{-23}J$

Now we want to convert it into electronvolts (eV). We know that
$1 eV= 1.6 \cdot 10^{-19}J$
Therefore, we should simply divide our energy by this conversion factor, and we get:
$E= \frac{3.3 \cdot 10^{-23} J}{1.6 \cdot 10^{-19} J/eV}=2.1 \cdot 10^{-4}eV$

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In an oscillating lc circuit, when 75.0% of the total energy is stored in the inductor's magnetic field, (a) what multiple of th

olga2289 [7]

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3 years ago

An ice chest at a beach party contains 12 cans of soda at 4.05 °C. Each can of soda has a mass of 0.35 kg and a specific heat ca

BARSIC [14]

Below is the solution:

Heat soda=heat melon
m1*cp1*(t-t1)=m2*cp2*(t2-t); cp2=cpwater 
12*0.35*3800*(t-5)=6.5*4200*(27-t) 
15960(t-5)=27300(27-t) 
15960t-136500=737100-27300t 
43260t=873600 
t=873600/43260 
t=20.19 deg celcius

4 0

3 years ago

Which of the following air conditions would be least likely to have precipitation?

Lynna [10]

The best and most correct answer among the choices provided by the question is the first choice "warm, dry air"

In meteorology, precipitation is any product of the condensation of atmospheric water vapor that falls under gravity. The main forms of precipitation include drizzle, rain, sleet, snow, graupel and hail.

I hope my answer has come to your help. God bless and have a nice day ahead!

7 0

3 years ago

Read 2 more answers

Find the angle formed by two forces of 7N and 15N respectively if its result is worth 20N

nadezda [96]

First, you need to make certain assumptions before solving this question. Why? Because there are no information given about the direction of these forces. In such questions as above, ALWAYS make the following assumptions:

1) Take first force, say $F_{1}$ , and assume that it is pointing towards the x-direction.

Let us take the 7N force! By keeping the above assumption in our minds, the force vector would be like:
$F_{1}$ = $7i$ , where $i$ = Unit vector in the x-direction.

2) Take the second force, say $F_{2}$ , and assume that it is making an angle $\alpha$ with the first force $F_{1}$ .

Let us take the 15N force! By keeping the above assumption in our minds, the forces vector would be like:

$F_{2} = (15*cos \alpha)i + (15*sin \alpha )j$

Now from simple vector addition, we know that,
$F_{R} = F_{1} + F_{2}$ --- (A)

Where $F_{R}$ = Resultant vector.
NOTE: In equation (A), all forces are in vector notation. Assume that there is an arrow head on top of them.

Let us find $F_{1}+F_{2}$ first!
$F_{1}+F_{2}$ =   $7i+(15*cos \alpha)i + (15*sin \alpha )j$

=> $F_{1}+F_{2}$ =   $(7+15*cos \alpha)i + (15*sin \alpha )j$

Now the magnitude of $F_{1}+F_{2}$ is,
| $F_{1}+F_{2}$ | = $\sqrt{ (7+ 15*cos \alpha)^{2} + (15*sin \alpha )^{2}}$

=> | $F_{1}+F_{2}$ | = $\sqrt{ 49 + 225*(cos \alpha)^{2} + 210*(cos \alpha)+ 255*(sin \alpha )^{2}}$

Since $(sin \alpha)^{2} + (cos \alpha)^{2} = 1$ , therefore,

=> | $F_{1}+F_{2}$ | = $\sqrt{ 49 + 225 + 210*(cos \alpha)}$

Since  | $F_{1}+F_{2}$ | = | $F_{R}$ |, and the magnitude of the resultant force is 20N, therefore,

| $F_{R}$ | = | $F_{1}+F_{2}$ |
20 = $\sqrt{ 49 + 225 + 210*(cos \alpha)}$

Take square on both sides,
400 = $49 + 225 + 210*(cos \alpha)$
$(cos \alpha) = \frac{3}{5}$

$\alpha = 53.13^{o}$

Ans: Angle formed by the two forces, 7N and 15N, is: 53.13°

-israr

4 0

2 years ago

A student estimated a mass to be 325 g, but the actual mass is 342 g. What is the percent error?

Dovator [93]

Answer:

Answer:

5.2307 %

Explanation:

(acutal mass- estimated mass) / ( estimated mass)

3 0

3 years ago