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arsen [322]
3 years ago
14

What is the emf of a battery that increases the electric potential energy of 0.060 CC of charge by 0.70 JJ as it moves it from t

he negative to the positive terminal?
Physics
2 answers:
Evgen [1.6K]3 years ago
5 0

Answer:

emf = 11.667 V

Explanation:

Given: charge q = 0.060 C, electric potential energy E =0.70 J,

Solution :

by definition 1 volt = 1 joule per coulomb

so Voltage = emf = E/C

emf =  0.70 J / 0.060 C

emf = 11.667 V

Anastasy [175]3 years ago
5 0

Answer:

11.67 V.

Explanation:

Workdone by a charge, W = q × v

Where,

W = 0.7 J

q = 0.06 C

v = W/q

= 0.7/0.06

= 11.67 V

Emf = 11.67 V

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White blood cell,also called leukocyte or white corpuscle,a cellular component of the blood that lacks hemoglobin,has a nucleus,is capable of motility,and defends the body against infection and disease by ingesting foreign materials and cellular debris,by destroying infectious agents and cancer cells.
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3 years ago
The ability to navigate, hunt, and communicate by electric current in water most likely helps make up for the effect of dark, mu
fgiga [73]
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6 0
4 years ago
The sound level at a distance of 1.48 m from a source is 120 dB. At what distance will the sound level be 70.7 dB?
Tju [1.3M]

Answer:

The second distance of the sound from the source is 431.78 m..

Explanation:

Given;

first distance of the sound from the source, r₁ = 1.48 m

first sound intensity level, I₁ = 120 dB

second sound intensity level, I₂ = 70.7 dB

second distance of the sound from the source, r₂ = ?

The intensity of sound in W/m² is given as;

dB = 10 Log[\frac{I}{I_o} ]\\\\For \ 120 dB\\\\120 = 10Log[\frac{I}{1*10^{-12}}]\\\\12 =  Log[\frac{I}{1*10^{-12}}]\\\\10^{12} = \frac{I}{1*10^{-12}}\\\\I = 10^{12} \ \times \ 10^{-12}\\\\I = 1 \ W/m^2

For \ 70.7 dB\\\\70.7 = 10Log[\frac{I}{1*10^{-12}}]\\\\7.07 =  Log[\frac{I}{1*10^{-12}}]\\\\10^{7.07} = \frac{I}{1*10^{-12}}\\\\I = 10^{7.07} \ \times \ 10^{-12}\\\\I = 1 \times \ 10^{-4.93} \ W/m^2

The second distance, r₂, can be determined from sound intensity formula given as;

I = \frac{P}{A}\\\\I = \frac{P}{\pi r^2}\\\\Ir^2 =  \frac{P}{\pi }\\\\I_1r_1^2 = I_2r_2^2\\\\r_2^2 = \frac{I_1r_1^2}{I_2} \\\\r_2 = \sqrt{\frac{I_1r_1^2}{I_2}} \\\\r_2 =   \sqrt{\frac{(1)(1.48^2)}{(1 \times \ 10^{-4.93})}}\\\\r_2 = 431.78 \ m

Therefore, the second distance of the sound from the source is 431.78 m.

7 0
3 years ago
In the graph, which two regions show the particle undergoing zero acceleration and negative acceleration respectively?
GuDViN [60]
 when velocity and time both are constant and when velocity will decrease the acceleration will be negative
7 0
4 years ago
Suppose the earth suddenly came to halt and ceased revolving around the sun. The gravitational force would then pull it directly
AleksAgata [21]

Answer:

613373.65233 m/s

Explanation:

M = Mass of Sun = 1.989\times 10^{30}\ kg

m = Mass of Earth

v = Velocity of Earth

r = Distance between Earth and Sun = 147.12\times 10^{9}\ m

r_e = Radius of Earth = 6.371\times 10^6\ m

r_s = Radius of Sun = 695.51\times 10^6\ m

In this system it is assumed that the potential and kinetic energies are conserved

\dfrac{1}{2}Mv_2-\dfrac{GMm}{r_e+r_s}=0-\dfrac{GMm}{r}\\\Rightarrow v=\sqrt{2GM(\dfrac{1}{r_e+r_s}-\dfrac{1}{r})}\\\Rightarrow v=\sqrt{2\times 6.67\times 10^{-11}\times 1.989\times 10^{30}(\dfrac{1}{6.371\times 10^6+695.51\times 10^6}-\dfrac{1}{147.12\times 10^{9}})}\\\Rightarrow v=613373.65233\ m/s

The velocity of Earth would be 613373.65233 m/s

3 0
3 years ago
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