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LiRa [457]
3 years ago
9

A certain type of D-cell battery that costs $0.50 is capable of producing 1.3 V and a current of 0.3 A for a period of 75 hours.

Determine the cost of the energy delivered by this battery per kilowatt hour. (For comparison, the approximate cost of energy purchased from electric utilities in the United States is $0.12 per kilowatt hour.)
Physics
1 answer:
Arte-miy333 [17]3 years ago
6 0

Answer:

$3.51

Explanation:

The energy (E) delivered by a system is the product of the power (P) dissipated by the system and the time (t) taken for the dissipation. i.e

E = P x t -----------------------(i)

Where;

The power (P) is the product of the current (I) flowing through the system and the voltage (V) across the system. i.e

P = I x V

Substituting P = I x V into equation (i) gives;

E = (I x V) x t

=> E = IVt    --------------------(ii)

<em>From the question;</em>

Current (I) = 0.3A

Voltage (V) = 1.3V

time (t) = 75 hours = 75h

<em>Substituting these values into equation (ii) gives;</em>

=> E = 0.3 x 1.3 x 75

=> E = 29.25 Kwh

The energy delivered is 29.25Kwh

But in the U.S;

$0.12 = 1Kwh

=> 29.25Kwh = 29.25 x $0.12

=> $3.51

Therefore, the cost of the energy delivered by this battery per kilowatt hour is $3.51

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andrew-mc [135]

Answer:

a) True. The number of photoelectrons is proportional to the amount (intensity) of the incident beam. From the expression above we see that threshold frequency cannot emit electrons.

b)   λ = c / f

Therefore, as the wavelength increases, the frequency decreases and therefore the energy of the photoelectrons emitted,

c)  threshold energy

        h f =Ф

Explanation:

It's photoelectric effect was fully explained by Einstein by the expression

       Knox = h f - fi

Where K is the kinetic energy of the photoelectrons, f the frequency of the incident radiation and fi the work function of the metal

a) True. The number of photoelectrons is proportional to the amount (intensity) of the incident beam. From the expression above we see that threshold frequency cannot emit electrons.

b) wavelength is related to frequency

         λ = c / f

Therefore, as the wavelength increases, the frequency decreases and therefore the energy of the photoelectrons emitted, so there is a wavelength from which electrons cannot be removed from the metal.

c) As the work increases, more frequency radiation is needed to remove the electrons, because there is a threshold energy

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3 years ago
A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,
viktelen [127]
Part A)
First of all, let's convert the radii of the inner and the outer sphere:
r_A = 11.0 cm = 0.110 m
r_B = 16.5 cm=0.165 m
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C=4 \pi \epsilon _0  \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=
=3.67\cdot 10^{-11}F

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
E\cdot (4 \pi r^2) =  \frac{Q}{\epsilon _0}
from which
E(r) =  \frac{Q}{4 \pi \epsilon_0 r^2}
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
E(0.111m)=2680 N/C

And so, the energy density at r=0.111 m is
U= \frac{1}{2} \epsilon _0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: Q=3.67 \cdot 10^{-9}C. We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C

And therefore, the energy density at this distance from the center is
U= \frac{1}{2}\epsilon_0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3
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Answer:

Explanation:

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We can see from the figure that to solve for the value of the horizontal component, we have to make use of the sin function. That is:

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3 0
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