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.
B. force, distance, and time
Take a look at the definition of a Joule (SI unit of work) and the definition of a Watt (SI unit of power). They're (kg*m^2)/s^2 for work and (kg*m^2)/s^3 for power. Another definition for work is Newton Meter which is force times distance, and since you can define work as force times distance, then power is work per second. So it looks like you need force and distance to calculate work, and then time since power is work over time. So of the 4 choices, we've been given, let's see if any of them allow us to calculate both work and power.
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a. energy, force, and time
* OK. Force will get us Newtons. But how much work do you have, don't know. Since work is force times distance. So can't get work. And without getting work, can't get power. Wrong answer.
b. force, distance, and time
* Force over distance nicely defines work. And time is essential since power is work over time. So this looks to be very good choice.
c. force, mass, and distance
* Have a problem here. Time is pretty essential since all of the SI units for work and power have seconds hiding somewhere in their definition. So this is the wrong answer.
d. mass, force, and energy
* Same issue, no time element here. So wrong answer.
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Answer: hello your question is incomplete below is the complete question
answer : Angular acceleration = 7.88 rad/s^2 ( clockwise )
minimum coefficient of friction = -0.2225
Explanation:
First step : we calculate acceleration at point O
a = rα
acceleration at point A
aA = ( 2r ) α = 2α
∑Fy ( force in y axis ) = mass of block * aA
∴ T = mBlock ( g - 2α )
= 8.7 ( 9.81 - 2α )
attached below is the remaining part of the solution
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[] - Hello ! - []</h2>
Answer:
The control subsystem of a vehicle can have several parts. Two examples are the braking and steering wheel. If the subsystem were to malfunction, driving could potentially be dangerous due to the fact that the car could go out of control and make impact with another car. For example, if the brakes aren’t working you cannot stop the vehicle in time and if the steering wheels gone you can't move the vehicle.
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- I hope this helped
- Brainliest appreciated!
*If this answer is wrong or displeases you please tell me in the comment below*
<h3>Thank you have a wonderful and blessed day! </h3>
I think the answer is C. The speed increases. I hope this helps. I'm sorry if I'm wrong :)
