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Ksju [112]
3 years ago
9

two parallel plates are charged with 4.98*10^-7 C of charge. What must the area of the plates be to create an electric field of

8720 N/C
Physics
1 answer:
77julia77 [94]3 years ago
3 0

Answer:

6.45

Explanation:

Equation: E= Q/A(8.85*10^-12)

A= Q/E / (8.85*10^-12)

A= (4.98*10^-7)/8720 / (8.85*10^-12)

A=6.45

Got it right on acellus

Big truzz  

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A water main pipe of diameter 10 cm enters a house 2 m below ground. A smaller diameter pipe carries water to a faucet 5 m above
Lemur [1.5K]

Explanation:

Given that,

Diameter = 10 cm

Distance = 2 m

Speed v_{1}= 2\ m/s

Speed v_{2}=7\ m/s

Pressure in main pipe P_{1}=2\times10^{5}\ Pa

(I). We need to calculate the diameter

Using equation of continuity

Av_{1}=Av_{2}

\pi(\dfrac{d_{1}}{2})^2\times v_{1}=\pi(\dfrac{d_{2}}{2})^2\times v_{2}

(\dfrac{10}{2})^2\times2=(\dfrac{d_{2}}{2})^2\times7

d_{2}=\sqrt{\dfrac{25\times2\times4}{7}}

d_{2}=5.345\ cm

(II). We need to calculate the pressure the gauge pressure

Using Bernoulli equation

P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}

P_{2}=P_{1}+\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)-\rho g(h_{1}-h_{2})

P_{2}=2\times10^{5}+\dfrac{1}{2}\times1000(4-49)-1000\times 9.8\times(5)

P_{2}=1.28500\times10^{5}\ Pa

(III).  If it is possible to carry water to a faucet 17 m above ground,

Using Bernoulli equation

P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh=P_{3}+\dfrac{1}{2}\rho v_{3}^2+\rho g h_{3}

P_{3}=P_{1}+\dfrac{1}{2}\rho v_{1}^2-\rho g(h_{1}-h_{3})

Here, h_{3}=0

Put the value in the equation

P_{3}=2\times10^{5}+\dfrac{1}{2}\times1000\times4-1000\times 9.8\times17

P_{3}=3.5400\times10^{5}\ Pa

Hence, This is required solution.

7 0
3 years ago
A ball is launched with initial speed v from the ground level up a frictionless hill. The hill becomes steeper as the ball slide
Triss [41]

Answer:

H(max) = (v²/2g)

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The maximum height the ball will climb will be when there is no friction at all on the surface of the hill.

Normally, the conservation of kinetic energy (specifically, the work-energy theorem) states that, the change in kinetic energy of a body between two points is equal to the work done in moving the body between the two points.

With no frictional force to do work, all of the initial kinetic emergy is used to climb to the maximum height.

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W = - mgH

ΔK.E = W

0 - (1/2)(m)(v²) = - mgH

mgH = mv²/2

gH = v²/2

H = v²/2g.

7 0
3 years ago
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