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3 years ago

13

Household wiring often uses 2.0 mm diameter copper wires. The wires can get rather long as they snake through the walls from the

fuse box to the farthest corners of your house.
What is the potential difference across a 16 m long, 2.0 mm diameter copper wire carrying a 7.3 A current?

1 answer:

MAXImum [283]3 years ago

3 0

Answer:

0.637 V

Explanation:

From Ohm's law,

V = IR ................ Equation 1

Where V = Voltage, I = current, R = resistance.

Also

R = Lρ/A........... Equation 2

Where L = Length of the copper wire, ρ = resistivity of the copper wire, A = cross sectional area of the copper wire.

But,

A = πd²/4.................. Equation 2

Where d = diameter.

Substitute equation 2 into equation 1

R = 4Lρ/πd²............... Equation 3

Substitute equation 3 into equation 1

V = I(4Lρ/πd²)........... Equation 4

Given: I = 7.3 A, L = 16 m, d = 2 mm = 0.002 m, π = 3.14,

constant: ρ = 1.72 x 10⁻⁸ Ωm

Substitute into equation 4

V = 7.3(4×16×1.72x10⁻⁸)/(3.14×0.002²)

V = 8.04×10⁻⁶/(1.256×10⁻⁵)

V = 0.637 V

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The rotational inertia of a thin rod about one end is 1/3 ML2. What is the rotational inertia of the same rod about a point loca

zlopas [31]

Answer:

The value is $I = 0.0932 ML ^2$

Explanation:

From the question we are told that

The rotational inertia about one end is $I_R = \frac{1}{3} ML^2$

The location of the axis of rotation considered is $d = 0.4 L$

Generally the mass of the portion of the rod from the axis of rotation considered to the end of the rod is $0.4 M$

Generally the length of the rod from the its beginning to the axis of rotation consider is

$k = 1 - 0.4 L = 0.6L$

Generally the mass of the portion of the rod from the its beginning to the axis of rotation consider is

$m = 1- 0.4 M = 0.6 M$

Generally the rotational inertia about the axis of rotation consider for the first portion of the rod is

$I_{R1} = \frac{1}{3} (0.6 M )(0.6L)^2$

$I_{R1} = \frac{1}{3} (0.6 M )L^2 0.6^2$

Generally the rotational inertia about the axis of rotation consider for the second portion of the rod is

$I_{R2} = \frac{1}{3} (0.6 M )(0.6L)^2$

=> $I_{R2} = \frac{1}{3} (0.4 M )(0.4L)^2$

=> $I_{R2} = \frac{1}{3} (0.4 M )L^2 0.4^2$

Generally by the principle of superposition that rotational inertia of the rod at the considered axis of rotation is

$I = \frac{1}{3} (0.6 M )L^2 0.6^2 + \frac{1}{3} (0.4 M )L^2 0.4^2$

=> $I = \frac{1}{3} ML ^2 [0.6 * (0.6)^2 + 0.4 * (0.4)^2 ]$

=> $I = 0.0932 ML ^2$

8 0

3 years ago

Youare on a train travelling northat 80.0m/s relative to the ground. The air is still relativetothegroundwhen you hear the whist

AURORKA [14]

To solve this problem we will apply the concepts related to the Doppler effect. This is understood as the change in apparent frequency of a wave produced by the relative movement of the source with respect to its observer. Mathematically this is given as,

$f = \frac{v \pm v_r}{v \pm v_s}(f_0)$

Here,

v = Speed of the waves in the middle

$v_r$ = Speed of the receiver in relation to the medium (Positive if the receiver is moving towards the transmitter or vice versa)

$v_s$ = Speed of the source with respect to the medium (Positive if the source moves away from the receiver or vice versa)

Our values are given as,

$v = 342m/s$

$f_0 = 262Hz$

$v_r = 80m/s$

$f = 350Hz$

Replacing,

$350 = \frac{342+80}{342-v} (262)$

Solving for the velocity of the source,

$v = 26.1m/s$

Therefore the speed of the other train is 26.1m/s

3 0

3 years ago

A scientist wants to study the effects of morning temperatures on flower width. Which is most likely a source of error in this s

ratelena [41]

<span>The flower width data were collected at about the same time every day. </span>

According to the research, the responder made with the same query as this question it is: <span><span>

1. </span>The flower width data were collected at about the same time every day. </span>
Why this choice because, the study is delving into the effects of morning temperatures on flower width hence, if the scientists collect flower at the same time –hour or minutes everyday during the period of the study the variable of time is not allotted from sunrise period to high noon since it defines time as morning, the variant of time is only inapt.

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4 years ago

ladessa [460]

Answer:

maibi.... D

Explanation:

I think is D

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4 years ago

In which part of the ear are the vibrations occurring in a gas, liquid, and solid?

UkoKoshka [18]

Answer:

eardrum is the correct answer

hope this helps ❤️

Explanation:

Sound waves enter the ear canal and cause the eardrum to vibrate. Three small bones transmit these vibrations to the cochlea. This produces electrical signals which pass through the auditory nerve to the brain, where they are interpreted as sound.

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3 years ago