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Klio2033 [76]
3 years ago
12

You have a 7.04 W, 236 Ω resistor.

Physics
1 answer:
Over [174]3 years ago
3 0

Answer

the current is 0.17amps

Explanation:

This problem bothers power of an electrical circuit.

Electric power is defined as the electric heat transferred per second. This can be expressed as

P=IV

P= V²/R

P= I²R

Given data

Power p= 7.04w

Resistance R=236 ohms

Using the third equation and Substituting our data we have

7.04= I²* 236

I²= 7.04/236

I= √0.029

I= 0.17amps

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How do lone pairs of electrons affect the bond angle differently than electrons shared in a bond?
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How do Newton's laws of motion explain why it is important to keep the ice smooth on a hockey rink so that players can
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I'm not sure..but please refer to your teacher later.

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Answer:

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Explanation:

By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

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m\cdot g \cdot h_{1} = \frac{1}{2}\cdot m\cdot v_{1}^{2} +W_{1, loss} (1)

From the bottom to the top of the second hill

\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss} (2)

Where:

m - Mass of the roller coaster car, in kilograms.

v_{1} - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

g - Gravitational acceleration, in meters per square second.

h_{1} - Height of the first top of the hill with respect to the bottom, in meters.

W_{1, loss} - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

v_{2} - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

h_{2} - Height of the second top of the hill with respect to the bottom, in meters.

W_{2, loss} - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

By using (1) and (2), we reduce the system of equation into a sole expression:

m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss} (3)

Where W_{loss} is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

If we know that m = 175\,kg, g = 9.807\,\frac{m}{s^{2}}, h_{1} = 18\,m, h_{2} = 8\,m and v_{2} = 11\,\frac{m}{s}, then the work done by non-conservative force is:

W_{loss} = m\cdot\left[ g\cdot \left(h_{1}-h_{2}\right)-\frac{1}{2}\cdot v_{2}^{2} \right]

W_{loss} = 6574.75\,J

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

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