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NNADVOKAT [17]
3 years ago
5

Calculate the current in a circuit if 500 C of charge passes through it in 10 minutes. *

Physics
1 answer:
Sloan [31]3 years ago
7 0

Answer:

Choice a. Approximately 0.83\; \rm A on average.

Explanation:

The electric current through a wire is the rate at which electric charge flows through a cross-section of this wire.

Assume that electric charge of size Q flowed through a wire cross-section over a period of time t. The average current in that wire would be:

\displaystyle I = \frac{Q}{t}.

For this question:

  • Q = 500\; \rm C, whereas
  • t = 10\; \rm \text{minutes}.

Therefore, the average current in this circuit would be:

\displaystyle I = \frac{Q}{t} = \frac{500\; \rm C}{10\; \text{minutes}} = 50\; \rm C /\text{minute}.

However, the units in the choices are all in \rm A (for Amperes.) One Ampere is equal to one \rm C / \text{second}. It will take some unit conversations to change the unit of I = 50\; \rm C/ \text{minute} (coulombs-per-minute) to coulombs-per-second.

\begin{aligned}I &= 50\; \rm C/ \text{minute} \\ &= \frac{50\; \rm C}{1\; \rm \text{minute}} \times \frac{1 \; \text{minute}}{60\; \rm \text{seconds}} \approx 0.83\; \rm C/ \text{second} = 0.83 \; \rm A\end{aligned}.

Hence, the most accurate choice here would be choice a.

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if two point charges are separated by 1.5 cm and have charge values of 2.0 and -4.0, respectively, what is the value of the mutu
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Complete question:

if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.

Answer:

The mutual force between the two point charges is 319.64 N

Explanation:

Given;

distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m

value of the charges, q₁ and q₂ = 2 μC and - μ4 C

Apply Coulomb's law;

F = \frac{k|q_1||q_2|}{r^2}

where;

F is the force of attraction between the two charges

|q₁| and |q₂| are the magnitude of the two charges

r is the distance between the two charges

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N

Therefore, the mutual force between the two point charges is 319.64 N

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How much kinetic energy does a 50 kg object have if its moving at a velocity of 2 m/s?
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Answer:

C) 100 joules

Explanation:

The kinetic energy of an object is given by:

K=\frac{1}{2}mv^2

where m is the mass of the object and v its speed.

In this problem, we have an object of mass m = 50 kg and v = 2 m/s, so by using the formula we can find its kinetic energy:

K=\frac{1}{2}(50 kg)(2 m/s)^2=100 J

3 0
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