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mart [117]
3 years ago
6

An electric heating element is connected to a 110 V circuit and a current of 3.2 A is flowing through the element. How much ener

gy is used up during a period of 5 hours by the element
Physics
2 answers:
UkoKoshka [18]3 years ago
6 0

Power (rate of using energy) = (voltage) x (current)

Power of this heating element = (120 V) x (3.2 A) = 384 watts


The easy way:

     384 watts = 0.384 kilowatt

     (0.384 kilowatt) x (5 hours) = 1.92 kilowatt-hour


Another way:
     
                            384 watts = 384 joules per second .

        (384 joule/sec) x (3600 sec/hour) x (5 hours)  

=      (384 x 3600 x 5) (joules)  =  6,912,000 joules.

dangina [55]3 years ago
6 0
Wouldn't it be 110V * 3.2A * 5 hours? 

The problem shows 110V not 120V

So it would be 110V * 3.2A = 352

And 352 * 5 = 1760
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Answer: High pass filters, and in particular LC high pass filters are used in many RF applications where they block the lower frequencies and allow through higher frequency signals. Typically LC filters are used for the higher radio frequencies where active filters are not so manageable, and inductors more appropriate.

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3 years ago
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an electron and a 0.0320-kg bullet each have a velocity of magnitude 510 m/s, accurate to within 0.0100%. within what lower limi
MArishka [77]

for this we apply, Heisenberg's uncertainty principle.

it states that physical variables like position and momentum, can never simultaneously know both variables at the same moment.

the formula is,

Δp * Δx = h/4π

m(e).Δv * Δx = h/4π

by rearranging,

Δx = h / 4π * m(e).Δv

Δx = (6.63*10^-34) / 4 * 3.142 * 9.11*10^-31 * 5.10*10^-2

Δx = 6.63*10^-34 / 583.9 X 10 ⁻³¹

Δx = 0.011 X 10⁻³

for the bullet

Δx = (6.63*10^-34) / 4 * 3.142 * 0.032*10^-31 * 5.10*10^-2

Δx = 6.63*10^-34 /2.05

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6 0
1 year ago
2. A rock is dropped off a bridge. How fast is the rock
zhenek [66]

Answer:

a) The velocity of rock at 1 second, v = 9.8 m/s

b) The velocity of rock at 3 second,  v = 29.4 m/s

c) The velocity of rock at 5.5 second,  v = 53.9 m/s

Explanation:

Given data,

The rock is dropped from a bridge.

The initial velocity of the rock, u = 0

a) The velocity of rock at 1 second,

   Using the first equation of motion

                         v = u + gt

                         v = 0 + 9.8 x 1

                          v = 9.8 m/s

b) The velocity of rock at 3 second,

                         v = u + gt

                         v = 0 + 9.8 x 3

                          v = 29.4 m/s

c) The velocity of rock at 5.5 second,

                         v = u + gt

                         v = 0 + 9.8 x 5.5

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3 years ago
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Substituting numbers into the equation, we find

W=(0.1 kg)(9.81 m/s^2)(0.3 m)=0.3 J


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(3) 0.3 J

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