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shusha [124]
3 years ago
9

A car battery with a 12 V emf and an internal resistance of 0.037 Ω is being charged with a current of 45 A. Note that in this p

rocess, the battery is being charged.
(a) What is the potential difference (in V) across its terminals?
(b) At what rate is thermal energy being dissipated in the battery (in W)?
(c) At what rate is electric energy being converted to chemical energy (in W)?
(d) What is Pr in this case?
Physics
1 answer:
matrenka [14]3 years ago
6 0

Answer:

Part a)

V = 13.665 V

Part b)

P = 74.9 Watt

Part c)

P = 540 watt

Part d)

P_r = 0.878

Explanation:

Part a)

As we know that the during the charging process of the battery the terminal voltage of the cell is given as

V = E + iR

V = 12 + (0.037)(45)

V = 13.665 V

Part b)

Thermal energy dissipated in the battery is due to its internal resistance

so it is given as

P = i^2 R

here we have

P = 45^2 (0.037)

P = 74.9 Watt

Part c)

rate of energy conversion in the in the battery is given as

P = E. i

P = 12(45)

P = 540 watt

Part d)

percentage of the power conversion is given as

Pr = \frac{P_{out}}{P_{total}}

Pr = \frac{540}{540 + 74.9}

P_r = 0.878

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A water wheel rotates with a period of 2.26 s. If the water wheel has a radius of l.94 m,
Dafna11 [192]

Answer:

The correct answer is B)

Explanation:

When a wheel rotates without sliding, the straight-line distance covered by the wheel's center-of-mass is exactly equal to the rotational distance covered by a point on the edge of the wheel.  So given that the distances and times are same, the translational speed of the center of the wheel amounts to or becomes the same as the rotational speed of a point on the edge of the wheel.

The formula for calculating the velocity of a point on the edge of the wheel is given as

V_{r} = 2π r / T

Where

π is Pi which mathematically is approximately 3.14159

T is period of time

Vr is Velocity of the point on the edge of the wheel

The answer is left in Meters/Seconds so we will work with our information as is given in the question.

Vr = (2 x 3.14159 x 1.94m)/2.26

Vr = 12.1893692/2.26

Vr = 5.39352619469

Which is approximately 5.39

Cheers!

7 0
2 years ago
A 3.00 kg mass is traveling at an initial speed of 25.0 m/s. What is the
Nitella [24]

Answer:

The magnitude of the force required to bring the mass to rest is 15 N.

Explanation:

Given;

mass, m = 3 .00 kg

initial speed of the mass, u = 25 m/s

distance traveled by the mass, d = 62.5 m

The acceleration of the mass is given as;

v² = u² + 2ad

at the maximum distance of 62.5 m, the final velocity of the mass = 0

0 = u² + 2ad

-2ad = u²

-a = u²/2d

-a = (25)² / (2 x 62.5)

-a = 5

a = -5 m/s²

the magnitude of the acceleration = 5 m/s²

Apply Newton's second law of motion;

F = ma

F = 3 x 5

F = 15 N

Therefore, the magnitude of the force required to bring the mass to rest is 15 N.

4 0
3 years ago
An increase in the magnitude of velocity is??
bazaltina [42]
It means the speed of the object is increasing
and
there is a positive acceleration in the direction of the velocity
hence
there is a force acting on the object, in the direction of the velocity
8 0
3 years ago
A point charge with charge q1 = 3.40 μC is held stationary at the origin. A second point charge with charge q2 = -4.90 μC moves
expeople1 [14]

Answer:

-0.79 J

Explanation:

We are given that

q_1=3.4\mu C=3.4\times 10^{-6} C

1\mu C=10^{-6} C

q_2=-4.9\mu C=-4.9\times 10^{-6} C

x_1=0.125,y_1=0

x_2=0.280,y_2=0.235

We have to find the work done by the electric force on the moving point charge.

r_1=\sqrt{x^2_1+y^2_1}=\sqrt{(0.125)^2+0}=0.125

r_2=\sqrt{(0.280)^2+(0.235)^2}=0.366

Work done,W=kq_1q_2(\frac{1}{r_1}-\frac{1}{r_2})

Where k=9\times 10^9

Using the formula

W=9\times 10^9\times 3.4\times 10^{-6}\times(-4.9\times 10^{-6})(\frac{1}{0.125}-\frac{1}{0.366})

W=-0.79 J

5 0
3 years ago
A listener increases his distance from a sound source by a factor of 4.49.
noname [10]

Answer: Δβ (dB) = -13.1dB

Explanation:

The intensity of sound is inversely proportional to the square of the distance between them.

I ∝ 1/r²

I₁/I₂= r₂²/r₁² .....1

When the listener increases his distance from the source by a factor of 4.49.

Then,

r₂/r₁= 4.49

From equation 1

I₁/I₂ = (4.49)²

I₁/I₂ = 20.16

I₂/I₁ = 1/20.16

The change in sound intensity in dB can be given as

Δβ (dB) = 10 log(I₂/l₁) = 10log(1/20.6) = -13.1dB

6 0
3 years ago
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