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Mrrafil [7]
3 years ago
12

A wire carrying an electric current is often likened to a pipe carrying water. What part of this analogy is incorrect?

Physics
1 answer:
tamaranim1 [39]3 years ago
6 0

In a direct current (DC) electrical circuit, the voltage (V in volts) is an expression of the available energy per unit charge which drives the electric current (I in amperes) around a closed circuit. Increasing the resistance (R in ohms) will proportionately decrease the current which may be driven through the circuit by the voltage.

Each quantity and each operational relationship in a battery-operated DC circuit has a direct analog in the water circuit. The nature of the analogies can help develop an understanding of the quantities in basic electric ciruits. In the water circuit, the pressure P drives the water around the closed loop of pipe at a certain volume flow rate F. If the resistance to flow R is increased, then the volume flow rate decreases proportionately. You may click any component or any relationship to explore the the details of the analogy with a DC electric circuit.

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Does an object travel farther on a smooth or slippery surface or on a rough surface? Why?
Vitek1552 [10]

Answer:

There is much more friction on the rough surface than there is on the smooth surface.

Explanation:

8 0
3 years ago
Autotrophic plants do not require which of the following? solar energy carbon dioxide water oxygen NextReset
alukav5142 [94]
<span>Autotrophic plants do not require "Oxygen" as it is a waste product of the process of photosynthesis which they do.

In short, Your Final Answer would be Option D

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4 0
3 years ago
A 60kg student traveling in a 1000kg car with a constant velocity has a kinetic energy of 1.2 x 10^4 J. What is the speedometer
777dan777 [17]

Answer:

17.64 km/h

Explanation:

mass of car, m = 1000 kg

Kinetic energy of car, K = 1.2 x 10^4 J

Let the speed of car is v.

Use the formula for kinetic energy.

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

By substituting the values

1.2\times 10^{4} = \frac{1}{2}\times 1000\times v^{2}

v = 4.9 m/s

Now convert metre per second into km / h

We know that

1 km = 1000 m

1 h = 3600 second

So, v = \left (\frac{4.9}{1000}   \right )\times \left ( \frac{3600}{1} \right )

v = 17.64 km/h

Thus, the reading of speedometer is 17.64 km/h.

4 0
3 years ago
Can someone please help with this. "A ball is thrown vertically upward with a speed of 26.6 m/s. How high does it rise? The acce
lubasha [3.4K]

v^2=u^2+2as

0=26.6^2-2x9.8xs

-26.6^2/-2x9.8=s

calculator

8 0
3 years ago
A simple pendulum has a period of 3.45 second, when the length of the pendulum is shortened by 1.0m, the period is 2.81 second c
den301095 [7]

Answer:

Original length = 2.97 m

Explanation:

Let the original length of the pendulum be 'L' m

Given:

Acceleration due to gravity (g) = 9.8 m/s²

Original time period of the pendulum (T) = 3.45 s

Now, the length is shortened by 1.0 m. So, the new length is 1 m less than the original length.

New length of the pendulum is, L_1=L-1

New time period of the pendulum is, T_1=2.81\ s

We know that, the time period of a simple pendulum of length 'L' is given as:

T=2\pi\sqrt{\frac{L}{g}}-------------- (1)

So, for the new length, the time period is given as:

T_1=2\pi\sqrt{\frac{L_1}{g}}------------ (2)

Squaring both the equations and then dividing them, we get:

\dfrac{T^2}{T_1^2}=\dfrac{(2\pi)^2\frac{L}{g}}{(2\pi)^2\frac{L_1}{g}}\\\\\\\dfrac{T^2}{T_1^2}=\dfrac{L}{L_1}\\\\\\L=\dfrac{T^2}{T_1^2}\times L_1

Now, plug in the given values and calculate 'L'. This gives,

L=\frac{3.45^2}{2.81^2}\times (L-1)\\\\L=1.507L-1.507\\\\L-1.507L=-1.507\\\\-0.507L=-1.507\\\\L=\frac{-1.507}{-0.507}=2.97\ m

Therefore, the original length of the simple pendulum is 2.97 m

4 0
3 years ago
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