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2 years ago

14

I am confused. Magnetic flux linkage and magnetic flux don't have the same definition, but they have same units. But there formu

lae are different, what are they?

1 answer:

ss7ja [257]2 years ago

6 0

Answer:

magnetic flux linkage includes number of turns, N while magnetic flux has no N

Since number of turns are not significant to units, they become similar.

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What is an isotope ?

Step2247 [10]

<span>An isotope is a form of a chemical element whose atomic nucleus contains a specific number of neutrons in addition to the number of protons that distinctively defines the element. The nuclei of most atoms have neutrons as well as protons.</span>

8 0

3 years ago

What happens to the atomic number of an atom when the number of neutrons in the nucleus of that atom increases? a It decreases b

kupik [55]

Answer:

It remains the same

Explanation:

It remains the same. This is because the number of protons doesn't change and the number of protons determines the atomic number.

8 0

2 years ago

Most comets have short periods and orbit close to the ecliptic plane. (T/F)

Lapatulllka [165]

Answer:

False

Explanation:

Most comets are located outside the solar system, in part of the cloud that originated from dust and gas that has remained virtually untouchable for billions of years. The orbit of these comets can reach the order of a light year. Thus, they are called long-period comets.

3 0

3 years ago

An ideal spring with spring constant k is hung from the ceiling. The initial length of the spring, with nothing attached to the

hram777 [196]

The mass m of the object = 5.25 kg

<h3>Further explanation</h3>

Given

k = spring constant = 3.5 N/cm

Δx= 30 cm - 15 cm = 15 cm

Required

the mass m

Solution

F=m.g

Hooke's Law

F = k.Δx

$\tt m.g=k.\Delta x\\\\m.10=3.5\times 15\\\\m=5.25~kg$

7 0

2 years ago

You wish to produce an emf of 41.0 mV using an inductor whose inductance is 13.0 H. You start with a current of 1.50 mA through

Molodets [167]

Answer:

The current through the inductor at the end of 2.60s is 9.7 mA.

Explanation:

Given;

emf of the inductor, V = 41.0 mV

inductance of the inductor, L = 13 H

initial current in the inductor, I₀ = 1.5 mA

change in time, Δt = 2.6 s

The emf of the inductor is given by;

$V = L\frac{di}{dt} \\\\V = \frac{L(I_1-I_o)}{dt} \\\\L(I_1-I_o) = V*dt\\\\I_1-I_o = \frac{V*dt}{L}\\\\I_1 = \frac{V*dt}{L} + I_o\\\\I_1 = \frac{41*10^{-3}*2.6}{13} +1.5*10^{-3}\\\\I_1 = 8.2*10^{-3} + 1.5*10^{-3}\\\\I_1 = 9.7 *10^{-3} \ A\\\\ I_1 = 9.7 \ mA$

Therefore, the current through the inductor at the end of 2.60 s is 9.7 mA.

6 0

3 years ago