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Natasha_Volkova [10]
3 years ago
11

If an X-ray tube is operating at a current of 30.0 mA:a) How many electrons are striking the target per second? b) If the potent

ial difference between the anode and cathode of this tube is 100.0 kV, how much power is expended
Physics
1 answer:
Zina [86]3 years ago
3 0

Answer:

Explanation:

A. Q= It

30E-3A=q

Ne = q

30E-3/1.6*10-19= N

N= 1.8*10^16 electrons

B. Power= I x v

= 30*10^-3A x 100*10^3v

= 3000watts

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Monochromatic light falls on two very narrow slits 0.048 mm apart. successive fringes on a screen 5.00 m away are 6.5 cm apart n
atroni [7]
In a double-slit interference experiment, the distance y of the maximum of order m from the center of the observed interference pattern on the screen is
y= \frac{m \lambda D}{d}
where D=5.00 m is the distance of the screen from the slits, and 
d=0.048 mm=0.048 \cdot 10^{-3}m is the distance between the two slits.
The fringes on the screen are 6.5 cm=0.065 m apart from each other, this means that the first maximum (m=1) is located at y=0.065 m from the center of the pattern.
Therefore, from the previous formula we can find the wavelength of the light:
\lambda =  \frac{yd}{mD}= \frac{(0.065 m)(0.048 \cdot 10^{-3}m)}{(1)(5.00 m)}=  6.24 \cdot 10^{-7}m

And from the relationship between frequency and wavelength, c=\lambda f, we can find the frequency of the light:
f= \frac{c}{\lambda}= \frac{3 \cdot 10^8 m/s}{6.24 \cdot 10^{-7}m}=4.81 \cdot 10^{14}Hz
4 0
3 years ago
An object with total mass mtotal = 16.2 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.7 k
vichka [17]

Answer:

1.) 0 kgm/s

2) 6.3 kg

3) -0.0978 m/s

4)

5)

6)

An object with total mass mtotal = 16.2 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.7 kg moves up and to the left at an angle of θ1 = 23° above the –x axis with a speed of v1 = 25.4 m/s. A second piece with mass m2 = 5.2 kg moves down and to the right an angle of θ2 = 28° to the right of the -y axis at a speed of v2 = 23.8 m/s.

2) What is the mass of the third piece?

3) What is the x-component of the velocity of the third piece?

4) What is the y-component of the velocity of the third piece?

5) What is the magnitude of the velocity of the center of mass of the pieces after the collision?

6) Calculate the increase in kinetic energy of the pieces during the explosion

Explanation:

Since explosions and collisions follow the law of conservation of Momentum.

1) Magnitude of the final momentum of the system = Magnitude of the initial momentum of the system

Since the body was initially at rest,

Magnitude of the initial momentum of the system = 0 kgm/s

Hence, Magnitude of the final momentum of the system is also equal to 0 kgm/s.

2) mass of the third piece

Sum of all the masses = 16.2 kg

4.7 + 5.2 + c = 16.2

c = 6.3 kg

3) doing an x-component balance on momentum

(4.7)×(-25.4 cos 23) + (5.2)×(23.8 cos 28) + (6.3)(v) = 0

-0.6163 + 6.3v = 0

v = -0.0978 m/s

Hope this Helps!!!

4 0
3 years ago
Which elements are found in all four components of Earth?​
kifflom [539]

Answer:

oxygen silicon aluminun iron

3 0
2 years ago
​A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oC. The air undergoes a process to a state where th
vlada-n [284]

Answer:

Explanation:

The process is isothermic,  as P V = constant .

work done = 2.303 n RT log P₁ / P₂

= 2.303 x 5 / 29 x 8.3 x 303  log 2 / 1 kJ

= 300.5k J

This energy in work done by the gas will come fro heat supplied as internal energy is constant due to constant temperature.

heat supplied  = 300.5k J

specific volume is volume per unit mass

v / m

pv = n RT

pv  = m / M  RT

v / m = RT / p M

specific volume = RT / p M

option B is correct.

5 0
3 years ago
Rebecca heated 50mL of water from 0 degrees Celsius to 60 degrees Celsius. How much energy did she use to heat the water? Rememb
GuDViN [60]

Answer:

Q = 12540  J

Explanation:

It is given that,

Mass of water, m = 50 mL = 50 g

It is heated from 0 degrees Celsius to 60 degrees Celsius.

We need to find the energy required to heat the water. The formula use to find it as follows :

Q=mc\Delta T

Where c is the specific heat of water, c = 4.18 J/g°C

Put all the values,

Q=50\times 4.18\times (60-0)\\Q=12540\ J

So, 12540 J of energy is used to heat the water.

7 0
2 years ago
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