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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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Answer:

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Explanation:

The complete question is:

" The potential energy between two atoms in a particular molecule has the form U(x) =(2.6)/x^8 −(5.1)/x^4 where the units of x are length and the num- bers 2.6 and 5.1 have appropriate units so that U(x) has units of energy. What is the equilibrium separation of the atoms (that is the distance at which the force between the atoms is zero)?  "

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- The correlation between force F and energy U is given as:

                                  F = - dU / dx

                                  F = - d[(2.6)/x^8 −(5.1)/x^4] / dx  

                                  F = 20.8 / x^9 - 20.4 / x^5

- The equilibrium separation distance between atoms is given when Force F is zero:

                                  0 = 20.8 / x^9 - 20.4 / x^5

                                  0 = 20.8 - 20.4*x^4

                                  x^4 = 20.8/20.4

                                  x = ( 20.8/20.4 )^0.25

                                  x = 1.00486 m

5 0
3 years ago
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