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alisha [4.7K]
4 years ago
13

If the current through a 20-Ω resistor is 8.0 A , how much energy is dissipated by the resistor in 1.0 h ? Express your answer w

ith the appropriate units. ΔEΔ E = nothing nothing Request Answer Provide Feedback
Physics
1 answer:
marshall27 [118]4 years ago
5 0

Answer:

P(3600)=593.247W

Explanation:

First, let's find the voltage through the resistor using ohm's law:

V=IR=20*8=160V

AC power as function of time can be calculated as:

P(t)= V*I*cos(\phi)-V*I*cos(2 \omega t-\phi)  (1)

Where:

\phi=Phase\hspace{3}angle\\\omega= Angular\hspace{3}frequency

Because of the problem doesn't give us additional information, let's assume:

\phi=0\\\omega=2 \pi f=2*\pi *(60)=120\pi

Evaluating the equation (1) in t=3600 (Because 1h equal to 3600s):

P(3600)=160*8*cos(0)-160*8*cos(2*120\pi*3600-0)\\P(3600)=1280-1280*cos(2714336.053)\\P(3600)=1280-1280*0.5365255751\\P(3600)=1280-686.7527361=593.2472639\approx=593.247W

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A pilot heads his jet due east. The jet has a speed of 425 mi/h relative to the air. The wind is blowing due north with a speed
Alja [10]

Answer:

(a)Velocity of wind = 35 j

(b)Velocity of jet relative to air = 425 i

(c)True velocity of jet = 425 i + 35 j

(d)True speed of jet = \sqrt{425^{2}+35^{2}} = 426.88 mi/h

    Direction of jet is, ∅ = tan^{-1} \frac{40}{425} = 5.38°

Explanation:

We can represent east direction by i and north direction by j.

The jet has a relative speed of 425 mi/h relative to the air.

The wind is blowing due north with a speed of 35 mi/h = 35 j

425 mi/h is the relative speed with respect to wind that is

Velocity of jet wrt wind= V_{jw} = V_{j}-V_{w}

                               425 i = V_{j} - 35 j

V_{j} = 425 i + 35 j

(a)Velocity of wind = 35 j

(b)Velocity of jet relative to air = 425 i

(c)True velocity of jet = 425 i + 35 j

(d)True speed of jet = \sqrt{425^{2}+35^{2}} = 426.88 mi/h

    Direction of jet is,

     ∅ = tan^{-1} \frac{40}{425} = 5.38°

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4 years ago
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What are the units for the spring constant, k? A. newton meters B. newton seconds C. newtons/meter D. newtons/second E. newtons/
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Answer:

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Please let me know if you want me to explain further!

Thanks!

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3 years ago
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