The electron beam inside an old television picture tube is 0.40 mm in diameter and carries a current of 50 μA. This electron bea
m impinges on the inside of the picture tube screen.
Part A. What is the current density in the electron beam?
Express your answer with the appropriate units.
Part B. The electrons move with a velocity of 3.8 ×107 m/s. What electric field strength is needed to accelerate electrons from rest to this velocity in a distance of 5.0 mm?
Express your answer with the appropriate units.
Part C. Each electron transfers its kinetic energy to the picture tube screen upon impact. What is the powerdelivered to the screen by the electron beam?
Express your answer with the appropriate units.
Part A. What is the current density in the electron beam?
Express your answer with the appropriate units.
Part B. The electrons move with a velocity of 3.8 ×107 m/s. What electric field strength is needed to accelerate electrons from rest to this velocity in a distance of 5.0 mm?
Express your answer with the appropriate units.
Part C. Each electron transfers its kinetic energy to the picture tube screen upon impact. What is the powerdelivered to the screen by the electron beam?
Express your answer with the appropriate units.
1 answer:
Answer:
A) J= 398 A/m²
B) E= 1.6×10⁶ N/C
C) P= ×10⁴ W
Explanation:
My work is in the attachment. Comment with questions or if something seems wrong with my work. (Honestly, they seem little high but it could just be the given numbers being unrealistic.) Below I have explanations of each part to match up with the image as well.
Part A:
Current density (J) is defined as the amount of current in a particular cross-sectional area. To get this, we simply need to divide the current (I) by the cross-sectional area of the electron beam tube (A).
Part B:
This one took the most work for me. I used a kinematic equation (yes they apply to electrons) to find the electric field (E). I used a modified form of the familiar: ∆d=V₀τ+aτ²/2
We can use the fact that τ= V/a, a=(qE/m), and V₀=0 here to rewrite the equation in terms of values we know and/or can look up. From there we solve for E and plug in the values.
Part C:
Power (P) is simply work (W) over time (τ). We know what τ is from before and can take W= mV²/2. Plugging these in and reducing some values gives us an equation for power as well.
You might be interested in
The defined friction force allows you to find which is the correct result for what type of wax you should use:
B. One where the coefficient of static friction is larger.
The friction force is a macroscopic force that opposes the relative movement of bodies, it is the sum of many interactions at the microscope level.
In general for solid bodies the expression of the out of friction is
Where fr is the friction force, μ the friction coefficient and N the normal.
The friction coefficient can be of two types:
- Static for when the relative velocity between the two bodies is zero.
- Dynamic when the relative velocity between the two bodies is different from zero.
In general, the dynamic coefficient is less than the static one.
In this case, as the temperature increases, the friction coefficient decreases, so the skier begins to slide downwards. To prevent this sliding, he must use a wax with a higher friction coefficient.
If you only want to stand on the hill, a wax with a static coefficient must be used, while standing if you want to advance on the hill you must use a wax with a dynamic coefficient since there is a relative movement between the skier and the hill.
Let's examine the different alternatives:
A. False. The friction coefficient must be higher, so that there is more friction force.
B. True. The friction increases and can be stopped on the hill and move up the hill.
C. False In this case, you can go up the hill, but if you stop, you can go back.
D. False. The coefficient must be higher, to increase friction.
Consequently we see that the only type of wax that will allow you to stand or advance on the hill is
B. A wax with a higher statutory coefficient.
In conclusion, using the defined friction force we can find the correct result for what type of wax to use is:
B. One where the coefficient of static friction is larger.
Learn more here: brainly.com/question/4230804
Answer:
m= 1.47 kg
Explanation:
Given:
spring constant, K = 23.15 N/m
Displacement, x= 19.79 cm = 0.1979 m
at, x₁= 7.417 cm, v₁= 0.7286 m/s
Now,
x = 19.79 cos ( ωt)
on substituting the values, we get
7.417 x 10⁻² = 19.79 x 10⁻² cos (ωt)
or
cos(ωt) = 0.374
or
ωt = 67.98°
also
v = -0.1979×ωsin (ωt)
also
on substituting the values in the above equation, we get
0.7286 = -0.1979 sin ( 67.98°)
3.68 =-(0.927)
or
T = 1.582 sec
also,
where, m is the mass
on substituting the values, we have
on squaring both sides and solving, we have
m= 1.47 kg