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3 years ago

What quantities are related by Ohm's law? Check all that apply. voltage conductivity current resistance insulation

Physics

2 answers:

IrinaK [193]3 years ago

8 0

Answer:

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Explanation:

Free_Kalibri [48]3 years ago

7 0

Answer: Current, resistance and voltage are the quantities which are related by Ohm's law.

Explanation:

A law which states that electric current is directly proportional to voltage and inversely proportional to resistance is called Ohm's law.

Mathematically, it is represented as follows.

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

where,

I = current

V = voltage

R = resistance

This means that the quantities related by Ohm's law include current, voltage and resistance.

Thus, we can conclude that current, resistance and voltage are the quantities which are related by Ohm's law.

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A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end.

Veronika [31]

Answer:

A) ω = 6v/19L

B) K2/K1 = 3/19

Explanation:

Mr = Mass of rod

Mb = Mass of bullet = Mr/4

Ir = (1/3)(Mr)L²

Ib = MbRb²

Radius of rotation of bullet Rb = L/2

A) From conservation of angular momentum,

L1 = L2

(Mb)v(L/2) = (Ir+ Ib)ω2

Where Ir is moment of inertia of rod while Ib is moment of inertia of bullet.

(Mr/4)(vL/2) = [(1/3)(Mr)L² + (Mr/4)(L/2)²]ω2

(MrvL/8) = [((Mr)L²/3) + (MrL²/16)]ω2

Divide each term by Mr;

vL/8 = (L²/3 + L²/16)ω2

vL/8 = (19L²/48)ω2

Divide both sides by L to obtain;

v/8 = (19L/48)ω2

Thus;

ω2 = 48v/(19x8L) = 6v/19L

B) K1 = K1b + K1r

K1 = (1/2)(Mb)v² + Ir(w1²)

= (1/2)(Mr/4)v² + (1/3)(Mr)L²(0²)

= (1/8)(Mr)v²

K2 = (1/2)(Isys)(ω2²)

I(sys) is (Ir+ Ib). This gives us;

Isys = (19L²Mr/48)

K2 =(1/2)(19L²Mr/48)(6v/19L)²

= (1/2)(36v²Mr/(48x19)) = 3v²Mr/152

Thus, the ratio, K2/K1 =

[3v²Mr/152] / (1/8)(Mr)v² = 24/152 = 3/19

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3 years ago

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ohaa [14]

Answer:

Answer is Option 1. decreases

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3 years ago

Read 2 more answers

Determine the minimum angle at which a roadbedshould be banked

poizon [28]

To solve this problem, apply the concepts related to the relationship given between the centripetal Force and the Weight.

The horizontal force component is equivalent to the weight of the car, while the vertical component is linked to the centripetal force exerted on the car, therefore,

$T cos\theta = mg \rightarrow T = \frac{mg}{cos\theta}$

$Tsin\theta = \frac{mv^2}{r} \rightarrow T = \frac{mv^2/r}{sin\theta}$

Equating both equation we have that,

$\frac{mv^2}{r} = mgtan\theta$

$tan\theta = \frac{v^2}{rg}$

Rearranging to find the angle we have that,

$\theta = tan^{-1} (\frac{v^2}{rg})$

Our values are given as,

$r = 2.00*10^2m$

$v = 20m/s$

$\theta = tan^{-1}(\frac{20^2}{(2*10^{2})(9.8)})$

$\theta = 11.53 \°$

Therefore the minimum angle will be 11.53°

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3 years ago

Your friend, Tim, is playing with his sled. He ties a rope to his sled and attaches it to Lar's snowmobile. Tim has a mass of 71

kykrilka [37]

The maximum speed of Tim is 16.95 m/s.

The given parameters:

Mass of the rope, m = 71 kg
Tension on the rope, T = 220 N
Coefficient of kinetic friction, = 0.1
Time of motion, t = 8 s

<h3>What is Newton's second law of motion?</h3>

Newton's second law of motion states that, the force applied to an object is directly proportional to the product of mass and acceleration of the object.

The net force on Tim is calculated by applying Newton's second law of motion as follows;

$T - \mu _k F_n = ma\\\\T - \mu _k F_n = m\frac{v}{t} \\\\T - \mu_k mg = m\frac{v}{t} \\\\t(\frac{T - \mu_k mg}{m} )= v\\\\8 (\frac{220 \ -\ 0.1 \times 71 \times 9.8}{71} ) =v \\\\ 16.95 \ m/s = v$