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ELEN [110]
3 years ago
10

A maser is a laser-type device that produces electromagnetic waves with frequencies in the microwave and radio-wave bands of the

electromagnetic spectrum. You can use the radio waves generated by a hydrogen maser as a standard of frequency. The frequency of these waves is 1,420,405,751.786 hertz. (A hertz is another name for one cycle per second.) A clock controlled by a hydrogen maser is off by only 1 s in 100,000 years. For the following questions, use only three significant figures. (The large number of significant figures given for the frequency simply illustrates the remarkable accuracy to which it has been measured.) (a) What is the time for one cycle of the radio wave? (b) How many cycles occur in 1 h? (c) How many cycles would have occurred during the age of the earth, which is estimated to be 4.6×109 years? (d) By how many seconds would a hydrogen maser clock be off after a time interval equal to the age of the earth?By how many seconds would a hydrogen maser clock be off after a time interval equal to the age of the earth
Physics
1 answer:
rusak2 [61]3 years ago
7 0

Answers:

a) T=7.04(10)^{-10} s

b) 5.11(10)^{12} cycles

c) 2.06(10)^{26} cycles

d) 46000 s

Explanation:

<h2>a) Time for one cycle of the radio wave</h2>

We know the maser radiowave has a frequency f of 1,420,405,751.786 cycles/s

In addition we know there is an inverse relation between frequency and time T:

f=\frac{1}{T} (1)

Isolating  T: T=\frac{1}{f} (2)

T=\frac{1}{1,420,405,751.786 cycles/s} (3)

T=7.04(10)^{-10} s (4) This is the time for 1 cycle

<h2>b) Cycles that occur in 1 h</h2>

If 1h=3600s and we already know the amount of cycles per second 1,420,405,751.786 cycles/s, then:

1,420,405,751.786 \frac{cycles}{s}(3600s)=5.11(10)^{12} cycles This is the number of cycles in an hour

<h2>c) How many cycles would have occurred during the age of the earth, which is estimated to be 4.6(10)^{9} years?</h2>

Firstly, we have to convert this from years to seconds:

4.6(10)^{9} years \frac{365 days}{1 year} \frac{24 h}{1 day} \frac{3600 s}{1 h}=1.45(10)^{17} s

Now we have to multiply this value for the frequency of the maser radiowave:

1,420,405,751.786 cycles/s (1.45(10)^{17} s)=2.06(10)^{26} cycles This is the number of cycles in the age of the Earth

<h2>d) By how many seconds would a hydrogen maser clock be off after a time interval equal to the age of the earth?</h2>

If we have 1 second out for every 100,000 years, then:

4.6(10)^{9} years \frac{1 s}{100,000 years}=46000 s

This means the maser would be 46000 s off after a time interval equal to the age of the earth

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A) atmospheric pollutants. Nuclear power plants do not produce air pollution as carbon dioxide, sulfur dioxide.

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3 0
3 years ago
Invader Zim’s spaceship is sitting at rest in outer space. The ship then accelerates at a uniform rate of 12 m/s2 for 10 seconds
melisa1 [442]

Answer:

120 m/s

Explanation:

Given:

v₀ = 0 m/s

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t = 10 s

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v = at + v₀

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v = 120 m/s

6 0
3 years ago
You observe a plane approaching overhead and assume that its speed is 600 miles per hour. The angle of elevation of the plane is
scZoUnD [109]

Answer:4.34 miles

Explanation:

first  Elevation =19^{\circ}

After 1 minute Elevation changes to 59^{\circ}

Ditsance travelled in 1 minute =600\times \frac{1}{60}=10 mile

Now

tan59=\frac{H}{x}

H=xtan59

tan19=\frac{H}{10+x}

H=\left ( 10+x\right )tan19

Equating H

we get

1.319x=10tan19

x=2.61 miles

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4 0
3 years ago
Good evening! Can someone please answer this, ill give you brainliest and your earning 50 points. Would be very appreciated.
BARSIC [14]

Answer:

Gases that are very good at absorbing long wave photons of infrared light

Explanation:

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7 0
2 years ago
A mosquito can fly with a speed of 1.10 m/s with respect to the air. Suppose a mosquito flies east at this speed across a swamp.
sineoko [7]

Answer:

The velocity of Mosquito with respect to earth will be 0.302m/s

Explanation:

V(ma) = 1.10 m/s, east  Velocity of mosquito with respect to air

V(ae) = 1.4 m/s at 35°  Velocity of air with respect to Earth in west of south direction.

Velocity of Mosquito with respect to earth will be  

V(me) = V(ma) + V(ae)

We need to find the mosquito’s speed with respect to Earth in the x direction.

V(x, me) = V(x, ma) + V(x, ae) = V(ma) + V(ae)(cos theta(ae) )

Angle (ae) = –90.0° − 35°=−125°

V(x, me) = 1.10 + (1.4)Cos(-125)

             = 1.10 + 1.4(-0.57)

             = 1.10 -0.798

              = 0.302

So the velocity of Mosquito with respect to earth will be 0.302m/s

7 0
3 years ago
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