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2 years ago

Please help me!!!

2 answers:

7 0

<span>It is only possible if the fences has same voltage but different type of electric voltages(i.e. one, Ac voltage & another, Dc voltage).

As having the same magnitude they varies in their properties. among them Ac has sinosuidal variation of voltage. that's why, Ac has large amplitude then the mean voltage(rms).

+ Dc has constant magnitude. e.g. at home we know the recomended ac voltage is 220v(rms) but it raises upto 311

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!</span>

Leto [7]2 years ago

5 0

The voltage is the powerful force that will cause a superior or smaller electrical current dependent on the electrical resistance of the hurdles to the flow of electricity.

If one electric barrier has a greater confrontation to the stream of electricity, then with the formula we can construe that the current for the fence with the greater resistance will be smaller than for the fence with smaller resistance which will be greater.

The electrical shockwave rest on on the electric current which depends on the electrical resistance

hope it helps

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2 years ago

A horizontal pipe of diameter 1.03m has a smooth constriction to a section of diameter 0.618 m. The density of oil flowing in th

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Velocity of the oil in the pipe: 0.76 m/s, in the constricted section: 5.87 m/s

Explanation:

We can solve this problem by using Bernoulli's equation:

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$ (1)

where

$p_1 = 7340 N/m^2$ is the pressure in the pipe

$p_2 = 5505 N/m^2$ is the pressure in the constricted section

$\rho = 821 kg/m^3$ is the density of the oil

$v_1$ is the velocity of the oil in the pipe

$v_2$ is the velocity of the oil in the constricted section

Also, according to the continuity equation,

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the pipe, with

$r_1 = \frac{1.03}{2}=0.515 m$ is the radius

$A_2 = \pi r_2^2$ is the cross-sectional area of the constricted section, with

$r_2=\frac{0.618}{2}=0.309 m$ is the radius

So the equation becomes

$r_1^2 v_1 = r_2^2 v_2$

So we can write

$v_2=\frac{r_1^2}{r_2^2}v_1$

Substituting into eq.(1),

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho (\frac{r_1^2}{r_2^2}v_1)^2$

And solving the equation for $v_1$ :

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho \frac{r_1^4}{r_2^4}v_1^2\\v_1=\sqrt{\frac{p_2-p_1}{\frac{1}{2}\rho-\frac{1}{2}\rho \frac{r_1^4}{r_2^4}}}=0.76 m/s$

And the velocity in the constricted section is

$v_2=\frac{r_1^2}{r_2^2}v_1=5.87 m/s$

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2 years ago

According to Lenz's law, the induced current in a circuit always flows to oppose the external magnetic field through the circuit

Makovka662 [10]

Answer:

True.

Explanation:

According to Lenz's law, the induced current in a circuit always flows to oppose the external magnetic field through the circuit. This statement is true.

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$\epsilon=-\dfrac{d\phi}{dt}$