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topjm [15]
3 years ago
6

Your toaster has a power cord with a resistance of 2.2x10^−2 Ω connected in series with a 9.7 Ω nichrome heating element. The po

tential difference between the terminals of the toaster is 120 V.
1. How much power is dissipated in the power cord?
Pcord =_______ W
2. How much power is dissipated in the heating element?
Phe= ________ kW
Physics
1 answer:
kati45 [8]3 years ago
5 0

Answer:

(A) 654.545 Kw

(B) 2.885\times 10^5kw

Explanation:

We have given resistance of the toaster R_1=2.2\times 10^{-2}ohm=0.022ohm

Resistance of nichrome heating element R_2=9.7ohm

Both the resistances are connected in series so same current will flow through the circuit

Potential difference across the toaster V = 120 volt

So current i=\frac{120}{0.022}=5454.5454A

(a) Power dissipated in toaster P=i^2R=5454.5454^2\times 0.022=654545.441W=654.545kw

(B) Power dissipated in heating element  P=i^2R=5454.5454^2\times 9.7=2.885\times 10^8W=2.885\times 10^5kw

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Explanation:

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E = 4.8×10^-17/1.6×10^-19

E = 300 N/C

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What is the car's average velocity
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Two tiny particles having charges of +5.00 μC and +7.00 μC are placed along the x-axis. The +5.00-µC particle is at x = 0.00 cm,
Liula [17]

Answer:

The third charged particle must be placed at x = 0.458 m = 45.8 cm

Explanation:

To solve this problem we apply Coulomb's law:  

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:  

F = \frac{k*q_1*q_2}{d^2} Formula (1)  

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁, q₂: Charges in Coulombs (C)  

d: distance between the charges in meters (m)

Equivalence  

1μC= 10⁻⁶C

1m = 100 cm

Data

K = 8.99 * 10⁹ N*m²/C²

q₁ = +5.00 μC = +5.00 * 10⁻⁶ C

q₂= +7.00 μC = +7.00 * 10⁻⁶ C

d₁ = x (m)

d₂ = 1-x (m)

Problem development

Look at the attached graphic.

We assume a positive charge q₃ so F₁₃ and F₂₃ are repulsive forces and must be equal so that the net force is zero:

We use formula (1) to calculate the forces F₁₃ and F₂₃

F_{13} = \frac{k*q_1*q_3}{d_1^2}

F_{23} = \frac{k*q_2*q_3}{d_2^2}

F₁₃ = F₂₃

\frac{k*q_1*q_3}{d_1^2} = \frac{k*q_2*q_3}{d_2^2} We eliminate k and q₃ on both sides

\frac{q_1}{d_1^2}= \frac{q_2}{d_2^2}

\frac{q_1}{x^2}=\frac{q_2}{(1-x)^2}

\frac{5*10^{-6}}{x^2}=\frac{7*10^{-6}}{(1-x)^2} We eliminate 10⁻⁶ on both sides

(1-x)^2 = \frac{7}{5} x^2

1-2x+x^2=\frac{7}{5} x^2

5-10x+5x^2=7 x^2

2x^2+10x-5=0

We solve the quadratic equation:

x_1 = \frac{-b+\sqrt{b^2-4ac} }{2a} = \frac{-10+\sqrt{10^2-4*2*(-5)} }{2*2} = 0.458m

x_2 = \frac{-b-\sqrt{b^2-4ac} }{2a} = \frac{-10-\sqrt{10^2-4*2*(-5)} }{2*2} = -5.458m

In the option x₂, F₁₃ and F₂₃ will go in the same direction and will not be canceled, therefore we take x₁ as the correct option since at that point the forces are in  opposite way .

x = 0.458m = 45.8cm

8 0
3 years ago
I NEED HELP ASAP!!!
Savatey [412]
I think option C is correct..hope it helps
3 0
2 years ago
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