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Fiesta28 [93]
3 years ago
13

An aluminum wire with a diameter of 0.115 mm has a uniform electric field of 0.235 V/m imposed along its entire length. The temp

erature of the wire is 55.0°C. Assume one free electron per atom. Given that at 20 degrees, rhoo = 2.82x10-8 Ωm and α = 3.9x10-3 /C. Determine:
a) the resistivity of the wire.
b) the current density in the wire.
c) the total current in the wire.
d) the potential different that must exist between the ends of a 2m length of wire if the given electric field is to be produced.

Physics
2 answers:
RideAnS [48]3 years ago
7 0

Explanation:

Below are attachments containing the solution.

Aleonysh [2.5K]3 years ago
6 0

Answer:

Explanation:

a) To get the resistivity ρ at 50 Celsius, given the resitstivity at 20 Celsisus, use:

ρ = ρo(1 + α(T - To))

where To = 20 Celsius

b) Knowing the resistivity at 50 Celsius, and the (uniform) electric field E, you can determine the current density J using:

E = ρJ

(which is actually a density-averaged version of V = IR)

c) Assuming the current is uniform (which is should be in a uniform electric field and constant-diameter wire), the current i can be calculated using:

J = i/A --> i = JA

where A is the cross-sectional area of the wire (given by πr2); make sure to convert the given diameter to a radius, and the radius to base units

d) Since the electric field is given in volts per meter, and you have two meters of length in the wire, you can determine directly from that how many volts difference you need at the ends of the wire to get 0.2 volts per meter.

0.2 = V/d

with d = 2 m. This corresponds to a uniform electric field being related to voltage by V = Ed, where d is distance along the field line.

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Which forces tend to slow down an object
laila [671]
The answer is "friction and air resistance" gravity does some of the work by keeping the object from floating away, but friction and air resistance does the biggest part. Friction is how rough the ground it meaning on tile, dirt, grass, etc... that would slow down the object and air resistance is the gravity pushing on the object also making it stop. 

Hope this helps!
8 0
3 years ago
The outer layer of cable on a cable reel is 16.2 cm from the center of the reel. The reel is initially stationary and can rotate
ahrayia [7]

Answer:

B. w=12.68rad/s

C. α=3.52rad/s^2

Explanation:

B)

We can solve this problem by taking into account that (as in the uniformly accelerated motion)

\theta=\omega_{0}t+\frac{1}{2}\alpha t^{2}\\\theta = \frac{s}{r}      ( 1 )

where w0 is the initial angular speed, α is the angular acceleration, s is the arc length and r is the radius.

In this case s=3.7m, r=16.2cm=0.162m, t=3.6s and w0=0. Hence, by using the equations (1) we have

\theta=\frac{3.7m}{0.162m}=22.83rad

22.83rad=\frac{1}{2}\alpha (3.6s)^2\\\\\alpha=2\frac{(22.83rad)}{3.6^2s}=3.52\frac{rad}{s^2}

to calculate the angular speed w we can use\alpha=\frac{\omega _{f}-\omega _{i}}{t _{f}-t _{i}}\\\\\omega_{f}=\alpha t_{f}=(3.52\frac{rad}{s^2})(3.6)=12.68\frac{rad}{s}

Thus, wf=12.68rad/s

C) We can use our result in B)

\alpha=3.52\frac{rad}{s^2}

I hope this is useful for you

regards

3 0
3 years ago
Read 2 more answers
For the long life cells we have to connect them in ____ combination​
NISA [10]

Answer:

Parallel combination.

5 0
3 years ago
Read 2 more answers
You throw a ball upward with an initial speed of 4.3 m/s. When it returns to your hand 0.88 s later, it has the same speed in th
djverab [1.8K]

Answer:

The acceleration is -9.8 m/s²

Explanation:

Hi there!!

When you throw a ball upward, there is a downward acceleration that makes the ball return to your hand. This acceleration is produced by gravity.

The average acceleration is calculated as the variation of the speed over time. In this case, we know the time and the initial and final speed. Then:

acceleration = final speed - initial speed/ elapsed time

acceleration = -4.3 m/s - 4.3 m/s / 0.88 s

acceleration = -9.8 m/s²  

4 0
2 years ago
In a certain city, electricity costs $0.20 per kw·h. what is the annual cost for electricity to power a lamp-post for 8.00 hours
UNO [17]

Part a)

per day electricity power consumed when 100 W bulb is used for 8 hours

P = 8 * 100 = 800 Wh

for one year consumption

E = 365 * 800 = 292 kWH

now the cost will be given

cost = 0.20 * 292 = $ 58.4

now when other energy efficient light is used

P = 8 * 25 = 200 Wh

for one year consumption

E = 365 * 200 = 73 kWH

now the cost will be given

cost = 0.20 * 73 = $ 14.6

8 0
3 years ago
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