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Alina [70]
2 years ago
6

When operated on a household 110.0 V line, typical hair dryers draw about 1650 W of power. The current can be modeled as a long,

straight wire in the handle. During use, the current is about 1.95 cm from the user's hand.
a.What is the current in the dryer?
b.What is the resitance of the dryer?
c.What magnetic field does the dryer produce at the users hand?
Physics
1 answer:
Andre45 [30]2 years ago
4 0

Explanation:

Given that,

Voltage of household line, V = 110 V

Power of the hairdryer, P = 1650 W

During use, the current is about 1.95 cm from the user's hand.

(a) Power is given by :

P=V\times I\\\\I=\dfrac{P}{V}\\\\I=\dfrac{1650\ W}{110\ V}\\\\I=15\ A

(b) Again the power is given by :

P=\dfrac{V^2}{R}

R is resistance of the dryer

R=\dfrac{V^2}{P}\\\\R=\dfrac{(110)^2}{1650}\\\\R=7.34\ \Omega

(c) The magnetic field produced by the dryer at the user's hand is given by :

B=\dfrac{\mu_o I}{2\pi r}\\\\B=\dfrac{4\pi \times 10^{-7}\times 15}{2\pi \times 1.95\times 10^{-2}}\\\\B=1.53\times 10^{-4}\ T

Hence, this is the required solution.

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Explanation:

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A satellite that orbits Earth with a speed of v0 must be in an orbit of radius 8RE to maintain a circular orbit, where RE is the
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1.024 × 10⁸ m

Explanation:

The velocity v₀ of the orbit 8RE is v₀ = 8REω where ω = angular speed.

So, ω =  v₀/8RE

For the orbit with radius R for it to maintain a circular orbit and velocity 2v₀, we have

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substituting ω =  v₀/8RE into the equation, we have

2v₀ = v₀R/8RE

dividing both sides by v₀, we have

2v₀/v₀ = R/8RE

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So, R = 2 × 8RE

R = 16RE

substituting RE = 6.4 × 10⁶ m

R = 16RE

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The average kinetic energy of the molecules of an ideal gas at 10∘C has the value K10. At what temperature T1 (in degrees Celsiu
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Answer:

A) T1 = 566 k = 293°C

B) T2 = 1132 k = 859°C

Explanation:

A)

The average kinetic energy of the molecules of an ideal gas is givwn by the formula:

K.E = (3/2)KT

where,

K.E = Average Kinetic Energy

K = Boltzman Constant

T = Absolute Temperature

At 10°C:

K.E = K10

T = 10°C + 273 = 283 K

Therefore,

K10 = (3/2)(K)(283)

FOR TWICE VALUE OF K10:

T = T1

Therefore,

2 K10 = (3/2)(K)(T1)

using the value of K10:

2(3/2)(K)(283) = (3/2)(K)(T1)

<u>T1 = 566 k = 293°C</u>

<u></u>

B)

The average kinetic energy of the molecules of an ideal gas is given by the formula:

K.E = (3/2)KT

but K.E is also given by:

K.E = (1/2)(m)(vrms)²

Therefore,

(3/2)KT = (1/2)(m)(vrms)²

vrms = √(3KT/m)

where,

vrms = Root Mean Square Velocity of Molecule

K = Boltzman Constant

T = Absolute Temperature

m = mass

At

T = 10°C + 273 = 283 K

vrms = √[3K(283)/m]

FOR TWICE VALUE OF vrms:

T = T2

Therefore,

2 vrms = √(3KT2/m)

using the value of vrms:

2√[3K(283)/m] = √(3KT2/m)

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Squaring on both sides:

(4)(283) = T2

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2 years ago
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Answer:

Explained in explanation

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4 0
2 years ago
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