The electron beam inside a television picture tube is 0.40 {\rm mm} in diameter and carries a current of 50 {\rm \mu A}. This el
ectron beam impinges on the inside of the picture tube screen.
How many electrons strike the screen each second?
The electrons move with a velocity of 4.0\times10^7\;{\rm m/s}. What electric field strength is needed to accelerate electrons from rest to this velocity in a distance of 5.0 {\rm mm}?
Each electron transfers its kinetic energy to the picture tube screen upon impact. What is the power delivered to the screen by the electron beam? (Hint: What potential difference produced the field that accelerated electrons? This is an emf.)
How many electrons strike the screen each second?
The electrons move with a velocity of 4.0\times10^7\;{\rm m/s}. What electric field strength is needed to accelerate electrons from rest to this velocity in a distance of 5.0 {\rm mm}?
Each electron transfers its kinetic energy to the picture tube screen upon impact. What is the power delivered to the screen by the electron beam? (Hint: What potential difference produced the field that accelerated electrons? This is an emf.)
1 answer:
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Answer:
Explanation:
a) La fuerza neta que actúa sobre la caja en la dirección vertical es:
Fnet=Fg−f−Fp *sin45 °
aquí Fg representa la fuerza gravitacional, f es la fuerza de fricción, y Fp es la fuerza de empuje.
Fnet=ma
ma=Fg−f−Fp *sin45 °
a=
=0.24 m/s²
Vf =Vi +at
=0.48+0.24*2
Vf=2.98 m/s
b)
Fnet=Fg−f−Fp *sin45 °
=Fg−0.516Fp−Fp *sin45 °
=30-1.273Fp
Fnet=0 (Ya que la velocidad es constante)
Fp=30/1.273
=23.56 N
Answer:
The time required is 20 μs
Explanation:
Here is the data provided:
temperature = 20°C = 293 K
radius = 1 cm
atomic mass of air = 29 u
To determine
the duration for air to refill the vacuum space
solution:
We calculate the root mean square velocity of air particles. This can be expressed as:
where m indicates mass, t is temperature, v is speed, and R is the ideal gas constant, which is approximately 8.3145 (kg·m²/s²)/K·mol.
v = ............................1
v =
Resulting in v = 501.99 m/s.
<pNow, to cover the distance of 1 cm,<pThe duration needed for air is calculated as:time taken =
which gives us:
time taken =
so, time taken = 19.92 × seconds = 20μs.
Thus, the required time is 20 μs.
Answer:
Explanation:
We start with the fact that
Initially, the spacecraft was at rest, u = 0The final velocity of the rocket is given as v = 11 m/sThe distance that the rocket covers during acceleration is given as d = 213 mWe seek to determine the acceleration that the occupants of the spacecraft experience during launch, which is derived from the principles of kinematics. By applying thethird motion equation we can find the acceleration.
Thus, the acceleration felt by those inside the spacecraft is .