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olga nikolaevna [1]
3 years ago
14

The resistivity of gold is 2.44 × 10-8 Ω · m at room temperature. A gold wire that is 1.8 mm in diameter and 11 cm long carries

a current of 170 mA. How much power is dissipated in the wire?
Physics
1 answer:
tester [92]3 years ago
6 0
Power in a wire where current is flowing can be calculated from the product of the square of the current and the resistance. Resistance is equal to the product of resistivity and length divided by the area of the wire. We do as follows:

Resistance  =  2.44 × 10-8 ( 0.11) / (π)(0.0009)^2 = 1.055x10^-3 <span>Ω

P = I^2R = .170^2 (</span>1.055x10^-3 ) = 3.048x10^-5 W
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Maximum power transfer theorem iz related to which chapter of physics<br> Class 12
Tpy6a [65]
DaddyFed is right, it would be all of them.
8 0
3 years ago
A passenger on a stopped bus notices that rain is falling vertically just outside the window. When the bus moves with constant v
nekit [7.7K]

Answer:

1)0.325

2)6.17\ \rm m/s

Explanation:

<u>Given:</u>

The angle that falling raindrops make with the vertical=18^\circ

Let V_R be the velocity of the raindrops and V_B be the velocity of the bus.

1)

\dfrac{V_R}{V_B}=\tan 18^\circ\\\dfrac{V_R}{V_B}=0.315\\

2)Speed of the raindrops=0.315\times 19

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5 0
3 years ago
An ideal monatomic gas initially has a temperature of 300 K and a pressure of 5.79 atm. It is to expand from volume 420 cm3 to v
maxonik [38]

Answer:

a) The final pressure is 1.68 atm.

b) The work done by the gas is 305.3 J.

Explanation:

a) The final pressure of an isothermal expansion is given by:

T = \frac{PV}{nR}

T_{i} = T_{f}

\frac{P_{i}V_{i}}{nR} = \frac{P_{f}V_{f}}{nR}

Where:

P_{i}: is the initial pressure = 5.79 atm

P_{f}: is the final pressure =?

V_{i}: is the initial volume = 420 cm³

V_{f}: is the final volume = 1450 cm³

n: is the number of moles of the gas

R: is the gas constant

P_{f} = \frac{P_{i}V_{i}}{V_{f}} = \frac{5.79 atm*420 cm^{3}}{1450 cm^{3}} = 1.68 atm

Hence, the final pressure is 1.68 atm.

b) The work done by the isothermal expansion is:

W = P_{i}V_{i}ln(\frac{V_{f}}{V_{i}}) = 5.79 atm*\frac{101325 Pa}{1 atm}*420 cm^{3}*\frac{1 m^{3}}{(100 cm)^{3}}ln(\frac{1450 cm^{3}}{420 cm^{3}}) = 305.3 J

Therefore, the work done by the gas is 305.3 J.

I hope it helps you!        

3 0
3 years ago
Two particles, each of charge Q, are fixed at opposite corners of a square that lies in the plane of the page. A positive test c
amid [387]

Answer:

The magnitude of the net force is √2F.

Explanation:

Since the two particles have the same charge Q, they exert the same force on the test charge; both attractive or repulsive. So, the angle between the two forces is 90° in any case. Now, as we know the magnitude of these forces and that they form a 90° angle, we can use the Pythagorean Theorem to calculate the magnitude of the resultant net force:

F_N=\sqrt{F^{2}+F^{2}}\\\\F_N=\sqrt{2F^{2}}\\\\F_N=\sqrt{2}F

Then, it means that the net force acting on the test charge has a magnitude of √2F.

7 0
3 years ago
Why can an object still be seen when it is at absolute zero?
zavuch27 [327]
<span>As the temperature goes down, the chaotic motion (velocity) of atoms start decreasing. If the temperature hits the absolute zero (which, in reality, is impossible to achieve), the atoms of the body would freeze, making the body still and stiff. One thing to note here is that the atoms do not get destroyed when the temperature reaches the absolute zero. That is the reason why the object can still be seen when it is at absolute zero.</span>
6 0
3 years ago
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