Analyze at the image below and answer the question that follows.

For a metal that has an electrical conductivity of 7.1 x 107 (Ω-m)-1, do the following: (a) Calculate the resistance (in Ω) of a

jonny [76]

Answer:

(a) 0.0178 Ω

(b) 3.4 A

(c) 6.4 x 10⁵ A/m²

(d) 9.01 x 10⁻³ V/m

Explanation:

(a)

σ = Electrical conductivity = 7.1 x 10⁷ Ω-m⁻¹

d = diameter of the wire = 2.6 mm = 2.6 x 10⁻³ m

Area of cross-section of the wire is given as

A = (0.25) π d²

A = (0.25) (3.14) (2.6 x 10⁻³)²

A = 5.3 x 10⁻⁶ m²

L = length of the wire = 6.7 m

Resistance of the wire is given as

$R=\frac{L}{A\sigma }$

$R=\frac{6.7}{(5.3\times10^{-6})(7.1\times10^{7}) }$

R = 0.0178 Ω

(b)

V = potential drop across the ends of wire = 0.060 volts

i = current flowing in the wire

Using ohm's law, current flowing is given as

$i = \frac{V}{R}$

$i = \frac{0.060}{0.0178}$

i = 3.4 A

(c)

Current density is given as

$J = \frac{i}{A}$

$J = \frac{3.4}{5.3\times10^{-6}}$

J = 6.4 x 10⁵ A/m²

(d)

Magnitude of electric field is given as

$E = \frac{J}{\sigma }$

$E = \frac{6.4 \times 10^{5}}{ 7.1 \times 10^{7}}$

E = 9.01 x 10⁻³ V/m

5 0

2 years ago

A car travels 30 miles in 1 hour on a winding mountain road. Which of the following is a true statement?

siniylev [52]

Answer:

The true statement is:

"(C) The magnitude of the average velocity is equal to 30 m.p.h."

Explanation:

Given that a car travels 30 miles in 1 hour on a winding mountain road.

Let' check all the statements one by one:

(A) The magnitude of the total displacement is larger than the distance traveled.

Since the entire motion of the car is not exactly given in the question, so it is not possible to tell whether the magnitude of the total displacement is larger than the distance traveled or not.

Thus, this statement is not true.

(B) The magnitude of the average velocity is greater than 30 m.p.h.

The average velocity of an object is defined as the total displacement covered by the particle divided by the total time taken in covering that displacement.

Total distance covered by the car = 30 miles.

Total time taken by the car to cover this distance = 1 hour.

Therefore, the average velocity of the car for this time interval = $\rm \dfrac{30\ miles}{1\ hour }= 30\ m.p.h.$

Thus, this statement is also not true.

(C) The magnitude of the average velocity is equal to 30 m.p.h.

As is cleared in part (B) section above, the average velocity of the car in the given time interval is 30 m.p.h.

Thus, this statement is true.

(D)The magnitude of the average velocity is less than to 30 m.p.h.

Since. the average velocity of the car is 30 m.p.h.

Thus, this statement is not true.

(E)The car traveled with a constant speed of 30 m.p.h.

The motion of the car on the mountain road is not thoroughly given in the question, so again it is not possible to tell whether the car traveled with a constant speed of 30 m.p.h. or not.

Thus, this statement is also not true.

4 0

2 years ago

Read 2 more answers

When jumping, a flea reaches a takeoff speed of 1.0 m/s over a distance of 0.47 mm .What is the flea's acceleration during the j

garri49 [273]

We can use kinematics here if we assume a constant acceleration (not realistic, but they want a single value answer, so it's implied). We know final velocity, vf, is 1.0 m/s, and we cover a distance, d, of 0.47mm or 0.00047 m (1m = 1000mm for conversion). We also can assume that the flea's initial velocity, vi, is 0 at the beginning of its jump. Using the equation vf^2 = vi^2 + 2ad, we can solve for our acceleration, a. Like so: a = (vf^2 - vi^2)/2d = (1.0^2 - 0^2)/(2*0.00047) = 1,064 m/s^2, not bad for a flea!

8 0

2 years ago

Read 2 more answers

Compressional stress on rock can cause strong and deep earthquakes, usually at _____.

valentinak56 [21]

The answer is reverse faults.

7 0

2 years ago

A charge of 4.9 x 10-11 C is to be stored on each plate of a parallel-plate capacitor having an area of 150 mm2 and a plate sepa

Triss [41]

Answer:2.53*10^-10F

Explanation:

C=£o£r*A/d

Where £ is the permitivity of a constant

£o= 8.85*10^-12f/m

£r=6.3

A=150mm^2=0.015m^2

d=3.3mm= 0.0033m

C=8.85*10^-12*6.3*0.015/0.0033

C=8.85*6.3*10^-12*0.015/0.0033

C=55.755*0.015^-12/0.003

C=8.36/3.3*10^-13+3

C=2.53*10^-10F

7 0

2 years ago