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patriot [66]
3 years ago
15

A supercapacitor is an electrical energy storage device. A supercapacitor, initially charged to 2.1 thousand millivolts, supplie

s power to an emergency beacon. The power draw from the load causes the capacitor to lose 9.9% of its charge every minute. When the charge drops below 8 hundred millivolts the beacon will cease functioning. How long will t
Physics
1 answer:
svet-max [94.6K]3 years ago
6 0

Answer:

the time taken t is 9.25 minutes

Explanation:

Given the data in the question;

The initial charge on the supercapacitor = 2.1 × 10³ mV = 2.1 V

now, every minute, the charge lost is 9.9 %  

so we need to find the time for which the charge drops below 800 mV or 0.8 V

to get the time, we can use the formula for compound interest in basic mathematics;

A = P × ( (1 - r/100 )ⁿ

where A IS 0.8, P is 2.1, r is 9.9

so we substitute

0.8 = 2.1 × ( 1 - 0.099 )ⁿ

0.8/2.1 = 0.901ⁿ

0.901ⁿ = 0.381

n = 9.25 minutes

Therefore, the time taken t is 9.25 minutes

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If the volume is held constant, what happens to the pressure of a gas as temperature is decreased? Explain.
Lady bird [3.3K]

Answer:Decreases

Explanation:

Given

Volume is held constant that is it is a isochoric process.

We know that

PV=nRT

as n,V& R are constant therefore only variables are

P & T

so \frac{P_1}{T_1}=\frac{P_2}{T_2}

\frac{P_1}{P_2}=\frac{T_1}{T_2}

As T_1 is decreasing therefore Pressure must also decrease so that ratio remains constant.

6 0
3 years ago
an athlete in a hammer-throw event swings a 7.0-kilogram hammer in a horizontal circle at a constant speed of 12 meters per seco
Semenov [28]

Answer:

ac = 72 m/s²

Fc = 504 N

Explanation:

We can find the centripetal acceleration of the hammer by using the following formula:

a_c = \frac{v^2}{r}

where,

ac = centripetal acceleration = ?

v = constant speed = 12 m/s

r = radius = 2 m

Therefore,

a_c = \frac{(12\ m/s)^2}{2\ m}

<u>ac = 72 m/s²</u>

<u></u>

Now, the centripetal force applied by the athlete on the hammer will be:

F_c = ma_c\\F_c = (7\ kg)(72\ m/s^2)

<u>Fc = 504 N</u>

6 0
2 years ago
A lad, waiting for his friend walks in the sidewalk, in front of her house, from the front door, first, he moves towards the Pos
Andreas93 [3]

His total displacement from his original position is -1 m

We know that total displacement of an object from a position x to a position x', d = final position - initial position.

d = x' - x

If we assume the lad's initial position in front of her house is x = 0 m. The lad then moves towards the positive x-axis, 5 m. He then ends up at x' = 5 m. He then finally goes back 6 m.

Since displacement = final position - initial position, and his displacement is d' = -6 m (since he moves in the negative x - direction or moves back) from his initial position of x' = 5 m.

His final position, x" after moving back 6 m is gotten from

x" - x' = -6 m

x" = -6 + x'

x" = -6 + 5

x" = -1 m

Thus, his total displacement from his original position is

d = final position - initial position

d = x" - x

d = -1 m - 0 m

d = -1 m

So, his total displacement from his original position is -1 m

Learn more about displacement here:

brainly.com/question/17587058

3 0
3 years ago
TRUE OR FALSE? earth revolves around the sun tilted on its axis
Alex777 [14]

The axis of the Earth's rotation is tilted relative to the plain of the Earth's revolution around the Sun.

The question is worded very poorly, but you'd have to say it's TRUE.  

6 0
3 years ago
Read 2 more answers
To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V
aleksley [76]

Answer:

The maximum energy stored in the combination is 0.0466Joules

Explanation:

The question is incomplete. Here is the complete question.

Three capacitors C1-11.7 μF, C2 21.0 μF, and C3 = 28.8 μF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.

Energy stored in a capacitor is expressed as E = 1/2CtV² where

Ct is the total effective capacitance

V is the supply voltage

Since the capacitors are connected in series.

1/Ct = 1/C1+1/C2+1/C3

Given C1 = 11.7 μF, C2 = 21.0 μF, and C3 = 28.8 μF

1/Ct = 1/11.7 + 1/21.0 + 1/28.8

1/Ct = 0.0855+0.0476+0.0347

1/Ct = 0.1678

Ct = 1/0.1678

Ct = 5.96μF

Ct = 5.96×10^-6F

Since V = 125V

E = 1/2(5.96×10^-6)(125)²

E = 0.0466Joules

8 0
2 years ago
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