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FromTheMoon [43]
2 years ago
12

IF THE CAPACITOR HAS PLATE AREA OF 0.2m² AND DISTANCE BETWEEN THEM IS 0.04 m. What is the capacitance if dielectric constant is

E=0.001 C/VM.
please i need it hurry if correct mark u brainliest.please hurry.​
Physics
2 answers:
ipn [44]2 years ago
8 0

Answer:

The capacitance is 1.75 nF

Explanation:

From the question we are given that

   The inner radius is

    The outer radius is 000.1

   Length of the capacitor is L = 1m

   The dielectric constant is Di = 2

  The dielectric constant is  Di2 = 4

Generally the capacitance of a capacitor can be mathematically represented as

1.75nF

               

                 

                 

                 

                 

               

                 

Step2247 [10]2 years ago
5 0

\\ \rm\Rrightarrow C=\dfrac{\epsilon_oA}{d}

\\ \rm\Rrightarrow C=\dfrac{0.2(0.001)}{0.04}

\\ \rm\Rrightarrow C=\dfrac{0.0002}{0.04}

\\ \rm\Rrightarrow C=0.005F

\\ \rm\Rrightarrow C=5mF

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ddd [48]

Answer:

28.57 Mpc

Explanation:

This question is going to be solved by applying Hubble's Law.

This Hubble's Law is actually an observation in physical cosmology. This observation makes it clear that galaxies are moving away from the Earth, and are doing so at speeds proportional to their distance. This essentially means that the farther they are from the Earth, the faster they are moving away from Earth.

It is represented by this formula

v = H(0)D, where

v = speed

H(0) = Constant of proportionality, or otherwise, Hubble's constant.

D = Distance to a galaxy

Applying the given parameters to the formula, we have

v = H(0).D

D = v / H(0)

D = 2000 / 70

D = 28.57 Mpc

3 0
3 years ago
A banked circular highway curve is designed for traffic moving at 58 km/h. The radius of the curve is 201 m. Traffic is moving a
skelet666 [1.2K]

Answer:0.077

Explanation:

Given

banked designed for traffic moving at 58 km/h\approx 16.11 m/s

Radius of the curve 201 m

Actual traffic velocity =37 km/h approx  10.27 m/s

For banking of road

tan\theta =\frac{v^2}{rg}

tan\theta =\frac{16.11^2}{201\times 9.81}

\theta =7.49^{\circ}

Centripetal acceleration is given by

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

Taking component of centripetal acceleration

along and perpendicular to surface

a_{parallel}=\frac{v^2cos\theta }{r}

a_{perpendicular}=\frac{v^2sin\theta }{r}

From FBD

mgsin\theta -f_s=ma_{parallel}

f_s=mgsin\theta -ma_{parallel}----1

where f_s is frictional force

N-mgcos\theta =ma_{perpedicular}

N=mgcos\theta +ma_{perpedicular}----2

and we know coefficient of friction is given by

\mu =\frac{f_s}{N}

\mu =\frac{mgsin\theta -ma_{parallel}}{mgcos\theta +ma_{perpedicular}}

\mu =\frac{gsin\theta -\frac{v^2cos\theta }{r}}{gcos\theta +\frac{v^2sin\theta }{r}}

\mu =\frac{1.2804-0.5202}{9.726+0.068}

\mu =0.077

7 0
3 years ago
Two drag cars race. They line up at the starting line at rest. The winning car accelerates at a constant rate a and reaches the
Mila [183]

Answer:d=\frac{v^2}{8a}

Explanation:

Given

winning car accelerates with a and its final velocity is v

considering they both start from rest

time taken by winning car is

v=u+at

where u=initial velocity

a=acceleration

t=time

v=at

t=\frac{v}{a}

Now loosing car is accelerating with \frac{a}{4}

Distance traveled by loosing car in time t

s_1=ut+\frac{at^2}{2\cdot 4}

s_1=0+\frac{a}{8}\times (\frac{v}{a})^2

s_1=\frac{v^2}{8a}

Thus distance d traveled by loosing car is given by d=\frac{v^2}{8a}

5 0
4 years ago
An nfl linebacker can go from 0 m/s to 2.5 m/s in 2.5s. What is his acceleration?
dybincka [34]

Answer:

a= 1 m/s^2

Explanation:

a= (Vf-Vi)/t

a=(2.5m/s - 0m/s)/2.5 s

a=2.5 m/s / 2.5 s

a= 1 m/s^2

5 0
3 years ago
the wattage ratting of a light bulb is the power it sonsumers when it is connected accross a 120 V potential difference what is
igor_vitrenko [27]

Answer:

240 Ω

Explanation:

Resistance: This can be defined as the opposition to the flow of current in an electric field. The S.I unit of resistance is ohms (Ω).

The expression for resistance power and voltage is give as,

P = V²/R.......................... Equation 1

Where P = Power, V = Voltage, R = Resistance

Making R the subject of the equation,

R = V²/P.................... Equation 2

Given: V = 120 V, P = 60 W.

Substitute into equation 2

R = 120²/60

R = 240 Ω

Hence the resistance of the bulb = 240 Ω

8 0
3 years ago
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