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WITCHER [35]
3 years ago
13

The manufacturer of a 9V dry-cell flashlight battery says that the battery will deliver 20 mA for 80 continuous hours. During th

at time the voltage will drop from 9 V to 6 V. Assume the drop in voltage is linear with time. How much energy does the battery deliver in this 80 h interval
Physics
1 answer:
Maksim231197 [3]3 years ago
5 0

Answer:

17280 J or 17.28 kJ

Explanation:

Given that the voltage drop,

U = U2 - U1

U = 9 - 6

U = 3V

Also, we're told that the current, I is equal to 20 mA with the discharge time, t being 80 hrs.

Converting the time from h oi urs to seconds, we have

t = 80 * 3600

t = 288000

Now, to find the energy needed, we're going to use the formula

w = pt, where p = U * I

p = 3 * 20*10^-3

p = 60*10^-3

w = 60*10^-3 * 288000

w = 17280 J or 17.28 kJ

Therefore, the total energy the battery delivers in the 80 hrs is 17.28 kJ

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The power that the light is able to utilize out of the supply is only 0.089 of the given.
                           Power utilized = (0.089)(22 W)
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The energy required in this item is the product of the power utilized and the time. That is,
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4 years ago
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A ball is thrown upward with a speed of 40 m/s.  Gravity decreases its upward speed (increases its downward speed) by 9.8 m/s every second.

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Just like so many other aspects of life, the downward fall is an exact "mirror image" of the upward trip.  After another 4.08 seconds, the ball has returned to the height of the hand which flung it.  In total, the ball is in the air for <em>8.16 seconds</em> up and down.

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A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the c
puteri [66]

Correct question:

A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the center of the solenoid is 2.8 x 10⁻² T, what is the number of turns per meter for this solenoid?

Answer:

the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.

Explanation:

Given;

length of solenoid, L= 0.35 m

diameter of the solenoid, d = 0.04 m

current through the solenoid, I = 5.0 A

magnetic field in the center of the solenoid, 2.8 x 10⁻² T

The number of turns per meter for the solenoid is calculated as follows;

B =\mu_o  I(\frac{N}{L} )\\\\B =  \mu_o  I(n)\\\\n = \frac{B}{\mu_o I} \\\\n = \frac{2.8 \times 10^{-2}}{4 \pi \times 10^{-7} \times 5.0} \\\\n = 4.5 \times 10^3 \ turns/m

Therefore, the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.

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Explanation:

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