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nikdorinn [45]
4 years ago
14

A capacitor is connected to an ac generator that has a frequency of 2.3 kHz and produces a rms voltage of 1.5 V. The rms current

in the capacitor is 31 mA. When the same capacitor is connected to a second ac generator that has a frequency of 4.8 kHz, the rms current in the capacitor is 85 mA. What rms voltage does the second generator produce
Physics
1 answer:
miv72 [106K]4 years ago
7 0

Answer:

Explanation:

Given that,

AC frequency of 2.3KHz

f=2.3×10³Hz

Vrms produce is

Vrms=1.5V

Current rms

Irms= 31mA

The capacitor is reconnected to a generator of frequency

f=4.8KHz =4800Hz

The current rms becomes

Irms= 85mA

Vrms=?

Solution

First genrator

The capacitive reactance is given as

Xc=Vrms/Irms

Xc=1.5/31×10^-3

Xc=48.39 ohms

Now, to know the capacitance of the capacitor

Xc=1/2πfC

Then,

C=1/2πfXc

So,

C=1/2×π×2300×48.39

C=1.43×10^-6C

C=1.43μF

Note: the capacitance of the capacitor did not change,

Now for generator two.

The reactance are given as

Xc=1/2πfC

Xc=1/2×π×4800×1.43×10^-6

Xc=23.19ohms

Then,

Vrms2=Irms2 ×Xc

Vrms2=85×10^-3×23.19ohms

Vrms2=1.97V

Vrms2=1.97Volts

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One of the most efficient engines built so far has the following characteristics: combustion chamber temperature = 1900°C, exhau
suter [353]

Answer:

actual efficiency is  47.78 %

Carnot efficiency  is 67.65 %

power output is 5.20 × 10^3 hp

Explanation:

given data

temperature = 1900°C = 1900+ 273 K = 2173 K

exhaust temperature = 430°C = 430 + 273 K = 703 K

fuel = 7.0 × 10^9 cal

work = 1.4 × 10^10 J

to find out

actual efficiency  and Carnot efficiency and power output of engine

solution

first we find actual efficiency that is = work / heat input

put the value and

input energy  = 7.0 × 10^9 cal  (4.184 J/1 cal)  = 29.29 × 10^9 J

actual efficiency  =  1.4 × 10^10 / ( 29.29 × 10^9 )

actual efficiency  =  0.4778

actual efficiency is  47.78 %

and

Carnot efficiency  is = 1 - ( 703 / 2173 )

so Carnot efficiency  is  = 0.67648

Carnot efficiency  is 67.65 %

and

power output  = work / time

power output  =  1.4 × 10^10 / 3600 sec

power output = 3.88 × 10^6 W

power output = 3.88 × 10^6 W / 746 hp

so power output is 5.20 × 10^3 hp

5 0
3 years ago
A small but measurable current of 5.8 × 10-10 A exists in a copper wire whose diameter is 3.0 mm. The number of charge carriers
swat32

Answer:

a) The current density ,J = 2.05×10^-5

b) The drift velocity Vd= 1.51×10^-15

Explanation:

The equation for the current density and drift velocity is given by:

J = i/A = (ne)×Vd

Where i= current

A = Are

Vd = drift velocity

e = charge ,q= 1.602 ×10^-19C

n = volume

Given: i = 5.8×10^-10A

Raduis,r = 3mm= 3.0×10^-3m

n = 8.49×10^28m^3

a) Current density, J =( 5.8×10^-10)/[3.142(3.0×10^-3)^2]

J = (5.8×10^-10) /(2.83×10^-5)

J = 2.05 ×10^-5

b) Drift velocity, Vd = J/ (ne)

Vd = (2.05×10^-5)/ (8.49×10^28)(1.602×10^-19)

Vd = (2.05×10^-5)/(1.36 ×10^10)

Vd = 1.51× 10^-5

8 0
3 years ago
Read 2 more answers
When a cube is inscribed in a sphere of radius r, the length Lof a side of the cube is . If a positive point charge Qis placed a
Nana76 [90]

Answer:

  Ф_cube /Ф_sphere = 3 /π

Explanation:

The electrical flow is

      Ф = E A

where E is the electric field and A is the surface area

Let's shut down the electric field with Gauss's law

       Фi = ∫ E .dA = q_{int} / ε₀

the Gaussian surface is a sphere so its area is

        A = 4 π r²

the charge inside is

        q_{int} = Q

we substitute

       E 4π r² = Q /ε₀

       E = 1 / 4πε₀   Q / r²

To calculate the flow on the two surfaces

* Sphere

       Ф = E A

        Ф = 1 / 4πε₀  Q / r² (4π r²)

        Ф_sphere = Q /ε₀

* Cube

Let's find the side value of the cube inscribed inside the sphere.

In this case the radius of the sphere is half the diagonal of the cube

          r = d / 2

We look for the diagonal with the Pythagorean theorem

         d² = L² + L² = 2 L²

         d = √2 L

         

we substitute

          r = √2 / 2 L

          r = L / √2

          L = √2  r

now we can calculate the area of ​​the cube that has 6 faces

          A = 6 L²

          A = 6 (√2  r)²

          A = 12 r²

the flow is

          Ф = E A

          Ф = 1 / 4πε₀  Q/r²  (12r²)

          Ф_cubo = 3 /πε₀  Q

the relationship of these two flows is

         Ф_cube /Ф_sphere = 3 /π

8 0
4 years ago
In order to be _______________ forces, their effects must cancel each other out and not cause a change in the object's motion
Tju [1.3M]
Balancing.

When the same force is applied from both sides,the forces cancel out each other.
4 0
3 years ago
The lanthanides are all radioactive true or false
Oksi-84 [34.3K]

Answer:

False

Explanation:

All the lanthanides are not radioactive in nature. Just one of the lanthanides are radioactive.

  • The lanthanides belong to the f-block on the periodic table
  • Only Promethium in this group are radioactive in nature.
  • Other elements in the lanthanide series might have radioactive isotopes.
3 0
3 years ago
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