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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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irina1246 [14]

Answer with Step-by -step explanation:

We are given that

b.\mid A\mid=46 m

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x-component of vector A=A_x=\mid A\mid cosx=46cos 340=46\times 0.94=43.24

y-Component of vector A=A_y=\mid A\mid sinx=46sin340=46(-0.34)=-15.64

Magnitude of vector B=86 m

The vector B makes angle with positive x- axis=x'=42^{\circ}

x-component of vector B=B_x=86cos42=63.64

y-Component of vector B=B_y=86sin42=57.62

Vector A=A_xi+A_yj=43.24i-15.64j

Vector B=B_xi+B_yj=63.64i+57.62j

Vector C=A+B

Substitute the values

C=43.24i-15.64j+63.64i+57.62j

C=106.88i+41.98j

c.Direction=\theta=tan^{-1}(\frac{y}{x})=tan^{-1}(\frac{41.98}{106.88})=21.5^{\circ}

The direction of the vector C=21.5 degree

6 0
3 years ago
A girl and a boy are riding on a merry go round that is turning at a constant rate. The girl is near the outer edge, and the boy
galina1969 [7]

Answer:

The girl has greater tangential acceleration

Explanation:

The angular acceleration (\alpha) of the merry go round is equal to the rate of the change of the angular velocity, \omega:

\alpha = \frac{d\omega}{dt}

Since all the points of the merry go round complete 1 circle in the same time, the angular velocity of each point of the merry go round is the same, and so all the points also have the same angular acceleration.

The tangential acceleration instead is given by

a_t = \alpha r

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\alpha is the angular acceleration

r is the distance from the centre of the merry go round

Since the girl is near the outer edge and the boy is closer to the centre, the value of r for the girl is larger than for the boy, so the girl has greater tangential acceleration.

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A hollow cylinder with an inner radius of 5 mm and an outer radius of 26 mm conducts a 4-A current flowing parallel to the axis
bearhunter [10]

Answer:

B = 38.2μT

Explanation:

By the Ampere's law you have that the magnetic field generated by a current, in a wire, is given by:

B=\frac{\mu_o I_r}{2\pi r}     (1)

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r: distance from the center of the cylinder, in which B is calculated

Ir: current for the distance r

In this case, you first calculate the current Ir, by using the following relation:

I_r=JA_r

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Ar: cross sectional area for r in the hollow cylinder

Ar is given by  A_r=\pi(r^2-R_1^2)

The current density is given by the total area and the total current:

J=\frac{I_T}{A_T}=\frac{I_T}{\pi(R_2^2-R_1^2)}

R2: outer radius = 26mm = 26*10^-3 m

R1: inner radius = 5 mm = 5*10^-3 m

IT: total current  = 4 A

Then, the current in the wire for a distance r is:

I_r=JA_r=\frac{I_T}{\pi(R_2^2-R_1^2)}\pi(r^2-R_1^2)\\\\I_r=I_T\frac{r^2-R_1^2}{R_2^2-R_1^2}  (2)

You replace the last result of equation (2) into the equation (1):

B=\frac{\mu_oI_T}{2\pi r}(\frac{r^2-R_1^2}{R_2^2-R_1^2})

Finally. you replace the values of all parameters:

B=\frac{(4\pi*10^{-7}T/A)(4A)}{2\PI (12*10^{-3}m)}(\frac{(12*10^{-3})^2-(5*10^{-3}m)^2}{(26*10^{-3}m)^2-(5*10^{-3}m)^2})\\\\B=3.82*10^{-5}T=38.2\mu T

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8 0
3 years ago
How does the electrostatic force compare with the strong nuclear force in the
diamong [38]

Answer:

Strong nuclear force is 1-2 order of magnitude larger than the electrostatic force

Explanation:

There are mainly two forces acting between protons and neutrons in the nucleus:

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F_E=\frac{kq_1 q_2}{r^2}

where k is the Coulomb's constant, q1 and q2 are the  charges of the two particles, r is the separation between the particles.

The force is attractive for two opposite charges and repulsive for two same charges: therefore, the electrostatic force between two protons is repulsive.

- The strong nuclear force, which is the force exerted between nucleons. At short distance (such as in the nucleus), it is attractive, therefore neutrons and protons attract each other and this contributes in keeping the whole nucleus together.

At the scale involved in the nucleus, the strong nuclear force (attractive) is 1-2 order of magnitude larger than the electrostatic force (repulsive), therefore the nucleus stays together and does not break apart.

3 0
3 years ago
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Robert has just bought a new model rocket, and is trying to measure its flight characteristics. The rocket engine package claims
777dan777 [17]

Answer:

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Put value of KE

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Therefore, the work done by the drag force is given by 29.96 J

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