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Karolina [17]
2 years ago
13

An x-ray tube is an evacuated glass tube that produces electrons at one end and then accelerates them to very high speeds by the

time they reach the other end. The acceleration is accomplished using an electric field. The high-speed electrons hit a metal target at the other end, and the violence of the collision converts their kinetic energy into high-energy light rays, commonly known as x-rays.
Required:
a. Through what potential difference should electrons be accelerated so that their speed is 2.7 % of the speed of light when they hit the target?
b. What potential difference would be needed to give protons the same kinetic energy as the electrons?
c. What speed would this potential difference give to protons?
Physics
1 answer:
laila [671]2 years ago
5 0

Answer:

a)  V = 1.866 10² V ,  b)   V = 3.424 10⁵ V , c)   v = 8.1 10⁶ m / s

Explanation:

a) the potential difference is requested to accelerate the electrons up to 2.7% of the speed of light

           v = 0.027 c

           v = 0.027 3 10⁸

           v = 8.1 10⁶ m / s

for this part we can use the conservation of mechanical energy

starting point. When electrons are at rest

           Em₀ = U = q V

final point. Electrons with maximum speed

          Em_f = K = ½ m v2

          Em₀ = Em_{f}

          e V = ½ m v²

          V = ½ m v² / e

let's calculate

          V = ½  9.1 10⁻³¹ (8.1 10⁶)² / 1.6 10⁻¹⁹

          V = 1.866 10² V

           V = 1866 V

         

b) if this acceleration protons is the mass of the proton is m_{p} = 1.67 10-27

          V = ½ 1.67 10⁻²⁷ (8.1 10⁶)² / 1.6 10⁻¹⁹

           V = 3.424 10⁵ V

           V = 342402 V

c)   this potential difference should give the protons the same speed as the electrons

             v = 8.1 10⁶ m / s

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Find the time t2 that it would take the charge of the capacitor to reach 99.99% of its maximum value given that r=12.0ω and c=50
defon

Answer:

Explanation:

Given that, .

R = 12 ohms

C = 500μf.

Time t =? When the charge reaches 99.99% of maximum

The charge on a RC circuit is given as

A discharging circuit

Q = Qo•exp(-t/RC)

Where RC is the time constant

τ = RC = 12 × 500 ×10^-6

τ = 0.006 sec

The maximum charge is Qo,

Therefore Q = 99.99% of Qo

Then, Q = 99.99/100 × Qo

Q = 0.9999Qo

So, substituting this into the equation above

Q = Qo•exp(-t/RC)

0.9999Qo = Qo•exp(-t / 0.006)

Divide both side by Qo

0.9999 = exp(-t / 0.006)

Take In of both sodes

In(0.9999) = In(exp(-t / 0.006))

-1 × 10^-4 = -t / 0.006

t = -1 × 10^-4 × - 0.006

t = 6 × 10^-7 second

So it will take 6 × 10^-7 a for charge to reached 99.99% of it's maximum charge

8 0
2 years ago
A research group at Dartmouth College has developed a Head Impact Telemetry (HIT) System that can be used to collect data about
Olin [163]

Answer:

6.05 cm

Explanation:

The given equation is

2 aₓ(x-x₀)=( Vₓ²-V₀ₓ²)

The initial head velocity V₀ₓ =11 m/s

The final head velocity  Vₓ is 0

The accelerationis given by =1000 m/s²

the stopping distance = x-x₀=?

So we can wind the stopping distance by following formula

2 (-1000)(x-x₀)=[0^{2} -11^{2}]

x-x₀=6.05*10^{-2} m

       =6.05 cm

3 0
2 years ago
An ice rescue team pulls a stranded hiker off a frozen lake by throwing him a rope and pulling him horizontally across the essen
Anuta_ua [19.1K]

Answer:T=116.84 N

Explanation:

Given

Weight of hiker =1040 N

acceleration a=1.1 m/s^2

Force exerted by Rope is equal to Tension in the rope

F_{net}=T=ma_{net}

T=\frac{1040}{g}\times 1.1

T=116.84 N

8 0
2 years ago
The human heart is a powerful and extremely reliable pump. Each day it takes in and discharges about 7500 L of blood. Assume tha
gregori [183]

The answers are:

a) Work=125,923.61J

b) Power=1.46watt

Why?

It seems that you forgot to write the questions of the problem, however, in order to help you, I will try to complete it.

The questions are:

a) How much work does the heart do in a day?

b) What is its power output in watts?

So, solving we have:

We need to convert from liter to cubic meters in order to use the given information, so:

1L=0.001m^{3}\\\\7500L*\frac{0.001m^{3} }{1L}=7.5m^{3}

Also, we need to find the mass given the density of the blood.

1050}\frac{kg}{m^{3}}*7.5m^{3}=7875kg

Now, calculating how much work does the heart do in a day, we have:

Work=Fd=mgh\\\\Work=7875kg*9.81\frac{m}{s^{2}}*1.63m=125,923.61J

Then, calculating what is the power output and its horsepower, we have:

Power=\frac{Work}{time}\\\\Power=\frac{125,923.61J}{86,400s}=1.46watt

Have a nice day!

7 0
2 years ago
A power station burns 75 kilograms of coal per second. Each kg of coal contains 27 million joules of energy.
Kaylis [27]

Answer:

Explanation:

a )

one kg of coal gives energy of 27 x 10⁶ J

75 kg of coal gives energy of 27 x 10⁶ x 75 J

So rate which energy is coming out of coal per second

= 27 x 10⁶ x 75 J

= 2025 x 10⁶ J /s

2025 million watts .

b ) energy output = 800 million watts

efficiency = (800 / 2025) x 100

= 39.5 % .

3 0
2 years ago
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