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enyata [817]
3 years ago
9

A loop circuit has a resistance of R1 and a current of 2.2 A. The current is reduced to 1.6 A when an additional 2.8 Ω resistor

is added in series with R1. What is the value of R1? Assume the internal resistance of the source of emf is zero. Answer in units of Ω
Physics
1 answer:
kvasek [131]3 years ago
7 0

Answer:

Value of R_1 is 7.466 ohm

Explanation:

Let emf of the battery is e

In first case when only R_1 is there current through battery is 2.2 A

So 2.2=\frac{e}{R_1}

e=2.2R_1......eqn 1

In second case current is reduced to 1.6 A when additional resistance of 2.8 ohm is added in series

So 1.6=\frac{e}{R_1+2.8}

e=1.6R_1+4.48....eqn 2

Fro eqn 1 and eqn 2

2.2R_1=1.6R_1+4.48

0.6R_1=4.48

R_1=7.466ohm

So value of R_1 is 7.466 ohm

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What is the density of 18.0-karat gold that is a mixture of 18 parts gold, 5 parts silver, and 1 part copper? (These values are
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Density of 18.0-karat gold mixture is 15.58 g/cm^3.

Explanation:

A mixture of 18 parts gold, 5 parts silver, and 1 part copper.

Let mass of gold be 18x

Let the mass of silver be 5x

Let the mass of copper be 1x

The density of gold = 19.32g/cm^3

The density of silver = 10.1g/cm^3

The density of copper =8.8g/cm^3

Volume=\frac{Mass}{Density}

Volume of the gold in the mixture = V_1=\frac{18x}{19.32 g/cm^3}

Volume of the silver in the mixture = V_2=\frac{5x}{10.1 g/cm^3}

Volume of the copper in the mixture = V_3=\frac{1x}{8.8 g/cm^3}

Mass of the mixture = M = 18x+5x+1x =24x

Volume of the mixture = V_1+V_2+V_3

Density of the mixture:

\frac{M}{V_1+V_2+V_3}=15.58 g/cm^3

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Three packing crates of masses, M1 = 6 kg, M2 = 2 kg and M3 = 8 kg are connected by a light string of negligible mass that passe
never [62]

Answer:

39.81 N

Explanation:

I attached an image of the free body diagrams I drew of crate #1 and #2.  

Using these diagram, we can set up a system of equations for the sum of forces in the x and y direction.

∑Fₓ = maₓ

∑Fᵧ = maᵧ

Let's start with the free body diagram for crate #2. Let's set the positive direction on top and the negative direction on the bottom. We can see that the forces acting on crate #2 are in the y-direction, so let's use Newton's 2nd Law to write this equation:

  • ∑Fᵧ = maᵧ  
  • T₁ - m₂g = m₂aᵧ

Note that the tension and acceleration are constant throughout the system since the string has a negligible mass. Therefore, we don't really need to write the subscripts under T and a, but I am doing so just so there is no confusion.

Let's solve for T in the equation...

  • T₁ = m₂aᵧ + m₂g
  • T₁ = m₂(a + g)

We'll come back to this equation later. Now let's go to the free body diagram for crate #1.

We want to solve for the forces in the x-direction now. Let's set the leftwards direction to be positive and the rightwards direction to be negative.

  • ∑Fₓ = maₓ
  • F_f - F_g sinΘ = maₓ

The normal force is equal to the x-component of the force of gravity.

  • (F_n · μ_k) - m₁g sinΘ = m₁aₓ
  • (F_g cosΘ · μ_k) - m₁g sinΘ = m₁aₓ
  • [m₁g cos(30) · 0.28] - [m₁g sin(30)] = m₁aₓ
  • [(6)(9.8)cos(30) · 0.28] - [(6)(9.8)sin(30)] = (6)aₓ
  • [2.539595871] - [-58.0962595] = 6aₓ
  • 60.63585537 = 6aₓ
  • aₓ = 10.1059759 m/s²

Now let's go back to this equation:

  • T₁ = m₂(a + g)  

We have 3 known variables and we can solve for the tension force.

  • T = 2(10.1059759 + 9.8)
  • T = 2(19.9059759)
  • T = 39.8119518 N

The tension force is the same throughout the string, therefore, the tension in the string connecting M2 and M3 is 39.81 N.

3 0
2 years ago
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