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2 years ago

The peak current passing through an inductor is 2. 0 aa. part a what is the peak current if the peak emf e0e0 is doubled?

Mathematics

1 answer:

kumpel [21]2 years ago

3 0

Answer:

I remember this question in school

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NEED HELP!!!

AnnyKZ [126]

Answer:

see explanation

Step-by-step explanation:

(1)

$\frac{6}{2}$ = $\frac{4}{p}$ ( cross- multiply )

6p = 8 ( divide both sides by 6 )

p = $\frac{8}{6}$ = $\frac{4}{3}$

(2)

$\frac{n}{4}$ = $\frac{8}{7}$ ( cross- multiply )

7n = 32 ( divide both sides by 7 )

n = $\frac{32}{7}$

(3)

$\frac{5}{3}$ = $\frac{x}{4}$ ( cross- multiply )

3x = 20 ( divide both sides by 3 )

x = $\frac{20}{3}$

6 0

3 years ago

I ONLY HAVE 5 minutes HELP!!!

Serjik [45]

Answer:

685.75

Step-by-step explanation:

650*0.055+650

4 0

3 years ago

Jack's Plumbing charges customers $100 per hour for repair jobs. A-1 Plumbing charges according to

s2008m [1.1K]

Answer:

37/3

Step-by-step explanation:

4 0

3 years ago

Will give brainliest

tresset_1 [31]

4=76

2=38

TOTAL=266 cars

4 0

3 years ago

Given that f(x) = 19x2 + 152, solve the equation f(x) = 0

telo118 [61]

Option A

The solution is:

$x = \pm 2i \sqrt{2}$

Solution:

$f(x) = 19x^2+152$

We have to solve the equation f(x) = 0

Let f(x) = 0

$0=19x^2+152$

Solve the above equation

$19x^2 + 152 = 0$

$\mathrm{Subtract\:}152\mathrm{\:from\:both\:sides}\\\\19x^2+152-152=0-152\\\\Simplify\ the\ above\ equation\\\\19x^2 = -152\\\\\mathrm{Divide\:both\:sides\:by\:}19\\\\\frac{19x^2}{19} = \frac{-152}{19}\\\\x^2 = -8$

Take square root on both sides

$x = \pm \sqrt{-8}\\\\x = \pm \sqrt{-1}\sqrt{8}\\\\\mathrm{Apply\:imaginary\:number\:rule}:\quad \sqrt{-1}=i\\\\x = \pm i\sqrt{8}\\\\x = \pm i \sqrt{2 \times 2 \times 2}\\\\x = \pm 2i\sqrt{2}$

Thus the solution is found

5 0

3 years ago