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3 years ago

200 Coulombs of charge passes through a point in a circuit for 0.6 minutes. what is the magnitude of the current flowing

Physics

1 answer:

Tasya [4]3 years ago

5 0

Answer:

5.56 A

Explanation:

From the question,

Q = it.............. Equation 1

Where Q = charges, i = current, t = time.

Make i the subject of the equation

i = Q/t.............. Equation 2

Given: Q = 200 coulombs, t = 0.6 minutes = (0.6×60) seconds

Substitite these values into equation 2

i = 200/(0.6×60)

i = 5.56 A

Hence the magnitude of the current flowing through the circuit is 5.56 A

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Explanation:

From the information given:

The equation of the motion can be represented as:

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$m_2$ = mass of the body 2

$\hat u$ = acceleration

k = spring constant

u = displacement

g = acceleration due to gravity

Recall that the formula for natural frequency $\omega _n = \sqrt{\dfrac{k}{m_1+m_2}}$

And the equation for the general solution can be represented as:

$u(t) = A cos \omega_nt + B sin \omega _n + \dfrac{m_2g}{k} --- (2)$

To determine the initial velocity, we have:

$\hat u_2^2 = 2gh$

$\hat u_2 = \sqrt{2gh}$

where h = height

Suppose we differentiate equation (2) with respect to time t; we have the following illustration:

$\hat u (t) = - \omega_n A sin \omega_n t+ \omega_n B cos \omega _n t + 0$

now if t = 0

Then

$\hat u (0) = - \omega_n A sin \omega_n (0)+ \omega_n B cos \omega _n (0) + 0$

$= \omega _n B$

Using the law of conservation of momentum on the impact;

$m_2 \hat u_2=(m_1+m_2) \hat u (0)$

By replacing the value of $\hat u_2$ with $\sqrt{2gh$

Then the above equation becomes:

$m_2 \times \sqrt{2gh}=(m_1+m_2) \ u(0)$

Making u(0) the subject of the formula, we have:

$u(0)= \dfrac{ m_2 \times \sqrt{2gh}}{(m_1+m_2)}$

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3 years ago

200 Coulombs of charge passes through a point in a circuit for 0.6 minutes. what is the magnitude of the current flowing​

200 Coulombs of charge passes through a point in a circuit for 0.6 minutes. what is the magnitude of the current flowing