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Temka [501]
3 years ago
10

uniform electric field exists between two parallel plates separated by 2.0 cm. The intensity of the field is 15 kN/C. What is th

e potential difference between the plates?
Physics
1 answer:
adoni [48]3 years ago
7 0

Answer:

potential difference V= 300 volts

Explanation:

Given:

d= 2.0 cm = 0.02m

E = 15 kN/C = 15 × 10³ N/C

For a uniform field between two plates, the Electric Filed Intensity (E) is proportional to the potential difference (V) and inversely  proportional to distance between the plates.

E= V/d

⇒ V= E×d = 15 × 10³ N/C × 0.02 m = 300 volts  (∴1 Nm/C = 1 J/C= 1 volts)

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Answer:

Q=977216.256\ J=977.216\ kJ

Explanation:

Given:

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  • temperature of resultant water, T_{fw}=31^{\circ}C

We have,

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<em>When we cool the steam of 100°C then firstly it loses its latent heat to convert into water of 100°C and the further cools the water.</em>

<u>Now the heat removed from steam to achieve the final state of water:</u>

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Q=m(L+c.\Delta T)

Q=384(2256+4.186\times (100-31))

Q=977216.256\ J=977.216\ kJ

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Something stopped the force and was able to recreate the same amount of force to send it back to her. Example: A pole
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A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .
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To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,

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Where,

m = Mass of spacecraft

M = Mass of Earth

r = Radius (Orbit)

G = Gravitational Universal Music

v = Velocity

Re-arrange to find the velocity

\frac{GM}{r^2} = \frac{v^2}{r}

\frac{GM}{r} = v^2

v = \sqrt{\frac{GM}{r}}

PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}

v = 4564.42m/s

From the speed it is possible to use find the formula, so

T = \frac{2\pi r}{v}

T = \frac{2\pi (6.371*10^6)}{4564.42}

T = 8770.05s\approx 146min\approx 2.4hour

Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.

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KE = \frac{1}{2} mv^2

KE = \frac{1}{2} (100)(4564.42)^2

KE = 1.0416*10^9J

Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.

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Answer:

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