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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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Find its moment of inertia about an axis perpendicular to its plane and passing through the midpoint of the line connecting its
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A) Moment of inertia about an axis passing through the point where the two segments meet : $I_A=\frac{1}{12} M L^2$

B) Moment of inertia passing through the point where the midpoint of the line connects to its two ends: $I x=\frac{1}{3} M L^2$

What is Moment of inertia?

The term "moment of inertia" refers to a physical quantity that quantifies a body's resistance to having its speed of rotation along an axis changed by the application of a torque (turning force). The axis might be internal or exterior, fixed or not.

A) The moment of inertia about an axis passing through the point where the two segments meet is $I_A=\frac{1}{12} M L^2$given that the rod is bent at the center and distance from all the points to the axis remains the same, the moment of inertia about the center will remain the same.

B) Determine the moment of inertia about an axis passing through the point midpoint of the line which connects the two ends

First step: determine the distance between the ends ( d )

After applying Pythagoras theorem$\mathrm{d}=\frac{\sqrt{2}}{2} L$

Next step : determine distance between the two axis $(\mathrm{x})$

After applying Pythagoras theorem

\mathrm{x}=\frac{\sqrt{2}}{4} L$$

Final step : Calculate the value of $\mathrm{I}_{\mathrm{x}}$

applying Parallel Axis Theorem

$$I_x=I_8+M x^2$$

$$\begin{aligned}& =\frac{1}{12} M L^2+\frac{1}{4} M L^2 \\& \therefore \quad I x=\frac{1}{3} M L^2 \\&\end{aligned}$$

Hence we can conclude that Moment of inertia about an axis passing through the point where the two segments meet: $I_A=\frac{1}{12} M L^2$, Moment of inertia passing through the point where the midpoint of the line connects its two ends: $I x=\frac{1}{3} M L^2$

To learn more about moment of inertia visit:brainly.com/question/15246709

#SPJ4

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