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3 years ago

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The potential difference between the plates of a parallel plate capacitor is 4.00 V. If the plate separation for this capacitor

is 6.00 cm, what is the intensity of the electric field between the plates of this capacitor?

1 answer:

shtirl [24]3 years ago

4 0

<h2>Answer:</h2>

66.67 V / m

<h2>Explanation:</h2>

The electric field intensity (E) between the plates of a capacitor is the ratio of the potential difference (V) between the plates to the distance (d) of separation between the plates. Electric field intensity can be measured in V/m. It can be written as follows;

E = V / d ----------------------------(i)

<em>From the question;</em>

V = 4.00V

d = 6.00cm = 0.06m

<em>Substitute these values into the equation above. i.e equation (i) as follows;</em>

E = 4.00 / 0.06

E = 66.67 V / m

Therefore, the electric field intensity between the plates of the capacitor is 66.67 V / m

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luda_lava [24]

The downwards component of weight of the object would be 35sin25 degrees = 14.8

F=14.8 - 8 = 6.8 N                                         m= 35/9.8= 3.57 kg
F=mA
Therefore,
3.57A= 6.8 N
=> A= 6.8/3.57
=> A= 1.902 ms^-2

F(max)= (U)R                       where (U)= coefficient of friction
                                                   and R= Normal reaction force

R= 35cos25
   = 31.72 N
Since, F(max)= 8
          8= (U)* 31.72
      =>(U)= 8/31.72
      =>(U)= 0.25

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3 years ago

Suppose a parachutist is falling toward the ground, and the downward force of gravity is exactly equal to the upward force of ai

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Answer:

Option b. is correct.

Explanation:

In the given question a parachutist is falling toward the ground .

Also, the downward force of gravity is exactly equal to the upward force of air resistance.

So, net force applied to the parachutist is equal to zero ( because both force acts in opposite direction ).

Now by first law of motion :

An object will be in rest or in constant speed unless and until no external force is applied on it .

So, in the question the velocity of the parachutist is not changing with time.

Therefore, option b. is correct.

Hence, this is the required solution.

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The energy of microwaves is less than the energy of ultraviolet light. Which comparison of the energies of

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Answer:

A. Gamma rays have higher energy than microwaves because gamma rays have shorter wavelengths.

Explanation:

Electromagnetic waves are waves produced by the interaction between both magnetic and electric fields. These waves have some properties that make them to be arranged in a definite form producing an electromagnetic spectrum.

The spectrum has a general property of which as the wavelength of the waves increases, the frequency decreases. And vice versa.

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2 years ago

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efrigerant-134a is expanded isentropically from 600 kPa and 70°C at the inlet of a steady-flow turbine to 100 kPa at the outlet.

PolarNik [594]

Answer:

Inlet : $v_i=0.0646\frac{m}{s}$

Outlet: $v_o=0.171\frac{m}{s}$

Explanation:

1) Notation and important concepts

Flow of mass represent "the mass of a substance which passes per unit of time".

Flow rate represent "a measure of the volume of liquid that moves in a certain amount of time"

Specific volume is "the ratio of the substance's volume to its mass. It is the reciprocal of density."

Isentropic process is a "thermodynamic process, in which the entropy of the fluid or gas remains constant".

We know that the flow of mass is given by the following expression

$\dot{m}=\frac{\dot{V}}{\upsilon}$ , where $\dot{V}$ represent the flow rate and $\upsilon$ the specific volume at the pressure and temperature given.

$A_i=0.5m^2$ is the inlet area

$P_i=600Kpa$ pressure at the inlet area

$T_i=70C$ temperature at the inlet area

$A_o=1m^2$ is the outlet area

$P_o=100Kpa$ pressure at the outlet area

$T_o=C$ temperature at the outlet area

$\dot{m}=0.75\frac{kg}{s}$ represent the flow of mass

If we look at the first figure attached Table A-13 we see that the specific volume for the inlet condition is

$\upsilon_i =0.04304\frac{kg}{m^3}$ and the entropy is $h_i=1.0645\frac{KJ}{KgK}=h_o$

With the value of entropy and the outlet pressure of 100 Kpa we can find we specific volume at the outlet condition since w ehave the entropy $h_o=1.0645\frac{KJ}{KgK}$

Since on the table we don't have the exact value we need to interpolate between these two values (see the second figure attached)

$h_1=1.0531\frac{KJ}{KgK} , \upsilon_1=0.22473\frac{kg}{m^3}$

$h_2=1.0829\frac{KJ}{KgK} , \upsilon_2=0.23349\frac{kg}{m^3}$

Our interest value would be given using interpolation like this:

$\upsilon=0.22473+\frac{(0.23349-0.22473)}{(1.0829-1.0531)}(1.0645-1.0531)=0.228\frac{kg}{m^3}$

2) Solution to the problem

Now since we have all the info required to solve the problem we can find the velocities on this way.

We know from the definition of flow of mass that $\dot{m}=\frac{\dot{V}}{\upsilon}$ , but since $\dot{V}=Av$ we have this:

$\dot{m}=\frac{Av}{\upsilon}$

If we solve from the velocity v we have this:

$v=\frac{\upsilon \dot{m}}{A}$ (*)

And now we just need to replace the values into equation (*)

For the inlet case:

$v_i=\frac{\upsilon_i \dot{m}}{A_i}=\frac{0.043069\frac{kg}{m^3}(0.75\frac{kg}{s})}{0.5m^2}=0.0646\frac{m}{s}$

For the oulet case:

$v_o=\frac{\upsilon_o \dot{m}}{A_o}=\frac{0.228\frac{kg}{m^3}(0.75\frac{kg}{s})}{1m^2}=0.171\frac{m}{s}$

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3 years ago

Standards for all managers ethical responsibilities are covered in a company's?

katen-ka-za [31]

Answer:

C. Code of ethics

Explanation:

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