Um elétron é lançado entre duas placas eletrizadas como mostra a figura. Sejam v= 6x10^6 m/s, ângulo 45°, E= 2x10^3 N/C, d= 3 cm
, L=12cm, m=10^-30 kg, e=1,6x10-19 C. Despreze a ação gravitacional.
a) o eletron atingirá a placa negativa?
b) determinar a posicao em que o eletron atinge uma das placas.
a) o eletron atingirá a placa negativa?
b) determinar a posicao em que o eletron atinge uma das placas.
1 answer:
You might be interested in
Answer:
4
Explanation:
Given parameters:
Force input = 30N
Force output = 120N
Unknown:
Mechanical advantage of the lever = ?
Solution:
The mechanical advantage is the quantity that describes how much a simple machine is able to multiply force.
This is expressed as:
Mechanical advantage = Force output / Force input
Mechanical advantage = = 4
The radius of the given circular path moved by the proton which is perpendicular to a magnetic field of 0.87 T determined as 4 cm.
<h3>Magnetic force of the proton</h3>The magnitude of the magnetic force on the proton is determined using the following formula;
F = qvBsin(θ)
<h3>Centripetal force on the proton</h3>The centripetal force of the circular path is given as;
F = mv²/r
Solve the two equations together,
Where;
- v is the speed of the proton
The speed of the proton is determined from its kinetic energy;
The radius of the circular path is calculated as follows;
Learn more about magnetic force here: brainly.com/question/13277365