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sergiy2304 [10]
3 years ago
8

Determine the maximum voltage that can be applied to a Teflon-filled parallel-plate capacitor having a plate area of 180 cm2 and

insulation thickness of 0.0770 mm.

Physics
2 answers:
SOVA2 [1]3 years ago
8 0

Answer:

Recall that

V=Ed

Were E is the dielectric contast for Teflon which is 60MV/m

V=60*10^6*0.077*10^-4=462V

Explanation:

Liula [17]3 years ago
8 0

Answer: 4620 volts

Explanation:

in the attachment

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Answer:

Earth's atmosphere traps energy from the sun. Which is a direct result of the trapping of energy by earth's atmosphere? Earth has moderate temperatures.

Explanation:

7 0
3 years ago
A layer of oil (n = 1.38) floats on an unknown liquid. A ray of light originates in the oil and passes into the unknown liquid.
Vinil7 [7]

Answer:

Refractive index of unknown liquid = 1.56

Explanation:

Using Snell's law as:

n_i\times {sin\theta_i}={n_r}\times{sin\theta_r}

Where,  

{\theta_i}  is the angle of incidence  ( 65.0° )

{\theta_r} is the angle of refraction  ( 53.0° )

{n_r} is the refractive index of the refraction medium  (unknown liquid, n=?)

{n_i} is the refractive index of the incidence medium (oil, n=1.38)

Hence,  

1.38\times {sin65.0^0}={n_r}\times{sin53.0^0}

Solving for {n_r},

Refractive index of unknown liquid = 1.56

4 0
4 years ago
You kick a soccer ball straight up into the air with a speed of 21.2 m/s. How long does it take the soccer ball to reachbits hig
Dimas [21]
Gravity increased the downward speed (or decreases the upward speed) by 9.8 m/s every second.

21.2/9.8 = 2.2 seconds
7 0
3 years ago
Read 2 more answers
Water enters a typical garden hose of diameter 0.016 m with a velocity of 3 m/s. Calculate the exit velocity of water from the g
Vladimir79 [104]

Answer:

v₂ = 306.12 m/s

Explanation:

We know that the volume flow rate of the water or any in-compressible liquid remains constant throughout motion. Therefore, from continuity equation, we know that:

A₁v₁ = A₂v₂

where,

A₁ = Area of entrance pipe = πd₁²/4 = π(0.016 m)²/4 = 0.0002 m²

v₁ = entrance velocity = 3 m/s

A₂ = Area of nozzle = πd₂²/4 = π(0.005 m)²/4 = 0.0000196 m²

v₂ = exit velocity = ?

Therefore,

(0.0002 m²)(3 m/s) = (0.0000196 m²)v₂

v₂ = (0.006 m³/s)/(0.0000196 m²)

<u>v₂ = 306.12 m/s</u>

5 0
3 years ago
An accepted value for the acceleration due to gravity is 9.801 m/s2. In an experiment with pendulums, you calculate that the val
Fed [463]

g Generally the accepted value of acceleration due to gravity is 9.801 m/s^2

as per the question the acceleration due to gravity is found to be 9.42m/s^2 in an experiment performed.

the difference between the ideal and observed value is 0.381.

hence the error is -\frac{0.381}{9.801} *100

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the error is not so high,so it can be  accepted.

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As the experiment is done under air resistance,so it will affect to the time period.hence the time period will be more which in turn decreases the value of g.

if this experiment is done in a environment of zero air resistance,we will get the value of g which must be approximately equal to 9.801  m/s^2

5 0
4 years ago
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