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netineya
1 month ago
8

Four wires are made of the same highly resistive material, cut to the same length, and connected in series. Wire 1 has resistanc

e R1 and cross-sectional area A. Wire 2 has resistance R2 and cross-sectional area 2A. Wire 3 has resistance R3 and cross-sectional area 3A. Wire 4 has resistance R4 and cross-sectional area 4A. A voltage V0 is applied across the series, as shown in the figure.Find the voltage V2 across wire 2.Give your answer in terms of V0, the voltage of the battery.
Physics
1 answer:
ValentinkaMS [3.4K]1 month ago
8 0

Answer:

V_2 = \frac{R_2 V_0}{R_1+R_2+R_3+R_4}

Explanation:

The four wires are arranged in series: this setup indicates that the same current passes through them, while the total voltage from the battery, V0, equals the combined voltages across each resistor:

V_0=V_1+V_2+V_3+V_4

Additionally, the total resistance for this series configuration is

R_{eq}=R_1+R_2+R_3+R_4

Using Ohm's law, we can determine the voltage V2 across wire 2:

V_2 = R_2 I (1)

Here, I represents the entire current flowing through the circuit, which is calculated as:

I=\frac{V_0}{R_{eq}}=\frac{V_0}{R_1+R_2+R_3+R_4}

By substituting into equation (1), we derive V2:

V_2 = \frac{R_2 V_0}{R_1+R_2+R_3+R_4}

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A shuttle on Earth has a mass of 4.5 E 5 kg. Compare its weight on Earth to its weight while in orbit at a height of 6.3 E 5 met
ValentinkaMS [3465]

Response:

83%

Clarification:

At the surface, the weight can be expressed as:

W = GMm / R²

where G denotes the gravitational constant, M represents the Earth's mass, m signifies the shuttle's mass, and R is the Earth's radius.

When in orbit, the weight is given by:

w = GMm / (R+h)²

where h indicates the shuttle's altitude above Earth's surface.

The weight ratio is as follows:

w/W = R² / (R+h)²

w/W = (R / (R+h))²

For R = 6.4×10⁶ m and h = 6.3×10⁵ m:

w/W = (6.4×10⁶ / 7.03×10⁶)²

w/W = 0.83

Thus, the shuttle maintains 83% of its weight as it orbits.

4 0
2 months ago
What is the sources of error and suggestion on how to overcome it in the hooke's law experiment?
Yuliya22 [3333]
In the study of physics, Hooke's law can be expressed as:

F = kx

This law indicates that the spring force F is proportional to the extension x, with k being the spring constant.

In experiments, this is often examined using the setup illustrated in the included figure. The spring is tested, and a known weight is applied underneath it. This weight exerts a gravitational pull, essentially its weight, on the spring. While the spring elongates, the displacement can be measured using a ruler.

Several potential errors can arise during this experiment. Firstly, the person's measurement reading may be faulty. Digital scales offer greater accuracy as they reduce human error, while ruler readings can be subjective, especially if not viewed at eye level. Additionally, the object's weight may be inaccurately measured if the scale is untrustworthy. Lastly, the measuring equipment may not be correctly calibrated.

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Joule’s law is a linear relationship, that is, the more heat you provide, the greater the temperature change. However, in the pr
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the term is hydrogen bridge.
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