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Vesnalui [34]
3 years ago
10

A bird lands on a bare copper wire carrying a current of 51

Physics
1 answer:
nirvana33 [79]3 years ago
7 0
The resistance of the piece of wire is
R= \frac{\rho L}{A}
where
\rho = 1.68 \cdot 10^{-8}\Omega m is the resistivity of the copper
L=5.1 cm=0.051 m is the length of the piece of wire
A=0.13 cm^2 = 0.13 \cdot 10^{-4} m^2 is the cross sectional area of the wire
By substituting these values, we find the value of R:
R= \frac{\rho L}{A}=6.6 \cdot 10^{-5} \Omega

Then, by using Ohm's law, we find the potential difference between the two points of the wire:
V=IR=(51 A)(6.6 \cdot 10^{-5} \Omega )=3.4 \cdot 10^{-3} V
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Understanding the benefits of an activity can __________.
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Answer:A.

Increase your motivation to continue doing it

Explanation:

benefits help you

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3 years ago
A mass weight of 120N is hung from two strings. what is the tension?
kramer
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3 years ago
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1 poir
djverab [1.8K]

Answer:

D) 735 J(oules)

Explanation:

Work is defined as force * distance

Force is defined as mass * acceleration

Given a mass of 15 kg and a gravitational acceleration of 9.8 m/s² since the box is being lifted up, the force being applied to the box is 15 kg * 9.8 m/s² = 147 N

Since the distance is 5 meters, the work done is 147 N * 5 m = 735 N/m = 735 J, making D the correct answer.

4 0
3 years ago
A flywheel with a diameter of 1.63 m is rotating at an angular speed of 79.9 rev/min. (a) What is the angular speed of the flywh
Studentka2010 [4]

Answer:

(a) 8.362 rad/sec

(b) 6.815 m/sec

(c) 9.446 rad/sec^2

(d) 396.22 revolution

Explanation:

We have given that diameter d = 1.63 m

So radius r=\frac{d}{2}=\frac{1.63}{2}=0.815m

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(a) We know that angular speed in radian per sec

\omega =\frac{2\pi N}{60}=\frac{2\times 3.14\times 79.9}{60}=8.362rad/sec

(b) We know that linear speed is given by

v=r\omega =0.815\times 8.362=6.815m/sec

(c) We have given final angular velocity \omega _f=675rev/min

And \omega _i=79.9rev/min

Time t = 63 sec

Angular acceleration is given by \alpha =\frac{\omega _f-\omega _i}{t}=\frac{675-79.9}{63}=9.446rad/sec^2

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7 0
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