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stiv31 [10]
3 years ago
8

How many x-ray photons per second are created by an x-ray tube that produces a flux of x rays having a power of 1.00 W? Assume t

he average energy per photon in 78.0 keV.
Physics
1 answer:
gregori [183]3 years ago
4 0

Answer:

The correct solution is "8.012\times 10^{13} \ per/sec".

Explanation:

Given:

Power,

P = 1.00 W

Time,

t = 1 sec

Average energy,

E₀ = 78.0 KeV

Or,

    = 78.0\times 10^3\times 1.6\times 10^{-19}

    = 124.8\times 10^{-16} \ J

As we know,

⇒ Power =\frac{Total \ energy}{Time}

or,

⇒         P=\frac{nE_0}{t}

⇒      1.00=\frac{n(124.8\times 10^{-16})}{1}

⇒          n=8.012\times 10^{13} \ per/sec  

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A French submarine and a U.S. submarine move toward each other during maneuvers in motionless water in the North Atlantic. The F
Mandarinka [93]

Answer:

f_{U}= 1019.72hz

Explanation:

From the equation we are told that:

Velocity of French sub V_{U}= 43.00km/h

Velocity of U.S. sub at V_{F}|=64.00 km//h

French Wave Frequency  F_{F}=1000Hz

Velocity of wave  V_{s}=5470 km/h

Generally the equation for Signal's frequency as detected by the U.S. is mathematically given by

Doppler effect

 f_{U} = \frac{ f_F (vs + v_{U})}{(v_s - v_F) }

 f_{U}= \frac{ 1000 x ( 5470+64)}{(5470-43)}

 f_{U}= 1019.72hz

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3 years ago
A car is moving towards a student at constant speed. The student notices that the sound the car makes gets louder and louder as
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Answer:

C

Explanation:

On USA

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3 years ago
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A frog jumps vertically upward from a 20m tall building with an initial velocity of 8.1m/s. How high above the ground will the f
topjm [15]

Answer:

Explanation:

Consider the initial position of the frog (20 m above ground) as the reference position. All measurements are positive measured upward.

Therefore,

u = 10 m/s, initial upward velocity.

H = - 20 m, position of the ground.

g = 9.8 m/s², acceleration due to gravity.

Part (a)

When the frog reaches a maximum height of h from the reference position, its velocity is zero. Therefore

u² - 2gh = 0

h = u²/(2g) = 10²/(2*9.8) = 5.102 m

At maximum height, the frog will be 20 + 5.102 = 25.102 m above ground.

Answer: 25.1 m above ground

Part (b)

Let v = the velocity when the frog hits the ground. Then

v² = u² - 2gH

v² = 10² - 2*9.8*(-20) = 492

v = 22.18 m/s

Answer: The frog hits the ground with a velocity of 22.2 m/s

8 0
4 years ago
A. What frequency is received by a person watching an oncoming ambulance moving at 115 km/h and emitting a steady 753 Hz sound f
nasty-shy [4]

Answer:

A)828.8Hz

B)869.2Hz

Explanation:

Here is a complete question;

What frequency is received by a person watching an oncoming ambulance moving at 115 km/h and emitting a steady 753 Hz sound from its siren? Speed of sound is 345m/s

b. What frequency does she receive after the ambulance has passed?

Vs= speed of the ambulance

, We convert to m/s for unit consistency

= 115 km/h= 115km× 1000m/1m × 1hr/3600s= 31.94m/s

Dopler effect is when observed frequency of wave changes with respect to the source or when observed moves relative to transmitting medium can be expressed as

f'=[ (v + vo)/(v- vs)]*f

=[ (v )/(v- vs)]*f

The sign vo and vs depends on vthe direction of the velocity

f= frequency of ambulance siren= 753Hz

v= speed of sound in air= 345m/s

Vo= speed of observer= 0

A) we are to determine the f' of ambulance as heard by person as ambulance approaching.

To find the frequency f' observed by the person we use the expresion below

Then substitute the values

f'=[ (v )/(v- vs)]*f

=[ (345)/(345-31.94)]×753

= 828.8Hz

B)What frequency does she receive after the ambulance has passed?

To find the frequency f' observed by the person we use the expresion below

Then substitute the values

f'=[ (v )/(v + vs)]*f

=[ (345)/(345 + 31.94)]×753

= 869.2Hz

=

4 0
3 years ago
A 1-kilogram ball has a kinetic energy of 50 joules the speed of the ball is
sveta [45]

Answer:

10 m/s

Explanation:

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