Answer:
245 divided by 5.14=47.6653696 or 47.66
Explanation:
Answer:
The time is 0.5 sec.
Explanation:
Given that,
Voltage V= 12.00 V
Inductance L= 1.20 H
Current = 3.00 A
Increases rate = 8.00 A
We need to calculate change in current

We need to calculate the time interval
Using formula of inductor


Where,
= change in current
V = voltage
L = inductance
Put the value into the formula


Hence, The time is 0.5 sec.
Answer:
W = 8.01 × 10^(-17) [J]
Explanation:
To solve this problem we need to know the electron is a subatomic particle with a negative elementary electrical charge (-1,602 × 10-19 C), The expression to calculate the work is given by:
W = q*V
where:
q = charge = 1,602 × 10^(-19) [C]
V = voltage = 500 [V]
W = work [J]
W = 1,602 × 10^(-19) * 500
W = 8.01 × 10^(-17) [J]
Answer:
q₁ = + 1.25 nC
Explanation:
Theory of electrical forces
Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.
Known data
q₃=5 nC
q₂=- 3 nC
d₁₃= 2 cm
d₂₃ = 4 cm
Graphic attached
The directions of the individual forces exerted by q1 and q₂ on q₃ are shown in the attached figure.
For the net force on q3 to be zero F₁₃ and F₂₃ must have the same magnitude and opposite direction, So, the charge q₁ must be positive(q₁+).
The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs ,then. F₁₃ is directed to the left (-x).
The force (F₂₃) of q₂ on q₃ is attractive because the charges have opposite signs. F₂₃ is directed to the right (+x)
Calculation of q1
F₁₃ = F₂₃

We divide by (k * q3) on both sides of the equation



q₁ = + 1.25 nC