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ki77a [65]
3 years ago
12

. You have completed your design of a new surgical headlamp. You have decided to power the headlamp using four AA rechargeable b

atteries. Each battery is rated at 1.2V, 2550 mAh. Your design uses an LED light that consumes a constant 3W. The LED is driven by a current source that takes power from the batteries and provides a constant current to the LED. The current source is 95% efficient. How long will the LED run before the batteries are depleted?
Physics
1 answer:
SSSSS [86.1K]3 years ago
5 0

Answer:

LED run before the batteries are depleted is 3.87 hours

Explanation:

given data

battery = 4 AA

battery is rated = 1.2V, 2550 mAh

power = 3 W

efficient = 95%

to find out

How long will the LED run before the batteries are depleted

solution

we consider here power delivered by battery is = x

so power = x × efficient

3 = x 95%

x = 3.15 W

and

voltage by 4 battery is = 4 × 1.2 = 4.8 V

so current will be = \frac{x}{volatge}

current = \frac{3.15}{4.8}

current = 0.6577 A

so total discharge hours = \frac{2550}{current}

total discharge hours = \frac{2550}{657.7}

time = 3.87 hours

so LED run before the batteries are depleted is 3.87 hours

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A) 2.4\cdot 10^{-16}kg

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B) 1.5\cdot 10^{-18}C

The potential difference across the electrodes is

V=17.8 V

and the distance between the plates is

d=11 mm=0.011 m

So the electric field between the electrodes is

E=\frac{V}{d}=\frac{17.8 V}{0.011 m}=1618.2 V/m

The droplet hangs motionless between the electrodes if the electric force on it is equal to the weight of the droplet:

qE=mg

So, from this equation, we can find the charge of the droplet:

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The droplet is hanging near the upper electrode, which is positive: since unlike charges attract each other, the droplet must be negatively charged. So the real charge on the droplet is

q=-1.5\cdot 10^{-18}C

we can think this charge has made of N excess electrons, so the net charge is given by

q=Ne

where

e=-1.6\cdot 10^{-19}C is the charge of each electron

Re-arranging the equation for N, we find:

N=\frac{q}{e}=\frac{-1.5\cdot 10^{-18}C}{-1.6\cdot 10^{-19}C}=9.4 \sim 9

so, a surplus of 9 electrons.

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How do I find the x and y components and the resultant force?​
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Explanation:

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2) the diagonal force times the cosine of the angle it makes gives the horizontal component or x component. do this to get the x component of all the diagonal forces.

add all the x components as well as the horizontal forces together to get the final x component

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