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:
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Answer:
The average magnetic flux through each turn of the inner solenoid is
Explanation:
Given that,
Number of turns = 22 turns
Number of turns another coil = 330 turns
Length of solenoid = 21.0 cm
Diameter = 2.30 cm
Current in inner solenoid = 0.140 A
Rate = 1800 A/s
Suppose For this time, calculate the average magnetic flux through each turn of the inner solenoid
We need to calculate the magnetic flux
Using formula of magnetic flux
Put the value into the formula
Hence, The average magnetic flux through each turn of the inner solenoid is
Answer:
a)W=8.333lbf.ft
b)W=0.0107 Btu.
Explanation:
<u>Complete question</u>
The force F required to compress a spring a distance x is given by F– F0 = kx where k is the spring constant and F0 is the preload. Determine the work required to compress a spring whose spring constant is k= 200 lbf/in a distance of one inch starting from its free length where F0 = 0 lbf. Express your answer in both lbf-ft and Btu.
Solution
Preload = F₀=0 lbf
Spring constant k= 200 lbf/in
Initial length of spring x₁=0
Final length of spring x₂= 1 in
At any point, the force during deflection of a spring is given by;
F= F₀× kx where F₀ initial force, k is spring constant and x is the deflection from original point of the spring.
Change to lbf.ft by dividing the value by 12 because 1ft=12 in
100/12 = 8.333 lbf.ft
work required to compress the spring, W=8.333lbf.ft
The work required to compress the spring in Btu will be;
1 Btu= 778 lbf.ft
?= 8.333 lbf.ft----------------cross multiply
(8.333*1)/ 778 =0.0107 Btu.