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kvv77 [185]
3 years ago
13

The quantity of charge q (in coulombs) that has passed through a surface of area 2.00 cm2 varies with time

Physics
1 answer:
Makovka662 [10]3 years ago
7 0

Explanation:

We have,

Surface area, A=2\ cm^2=0.0002\ m^2

The current varies wrt time t as :

q(t) = 4t^3 + 5t + 6

(a) At t = 2 seconds, electrical charge is given by :

q(t) = 4t^3 + 5t + 6\\\\q(2) = 4(2)^3 + 5(2) + 6\\\\q=48\ C

(b) Current is given by :

I=\dfrac{dq}{dt}\\\\I=\dfrac{d(4t^3 + 5t + 6)}{dt}\\\\I=12t^2+5

Instantaneous current at t = 1 s is,

I=12(1)^2+5=17\ A

(c) Current is, I=12t^2+5

Current density is given by electric current per unit area.

J=\dfrac{I}{A}\\\\J=\dfrac{(12t^2+5)}{0.0002}\\\\J=5000(12t^2+5)\ A/m^2

Therefore, it is the required explanation.

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A generator with �# ' = 300 V and Zg = 50 Ω is connected to a load ZL = 75 Ω through a 50-Ω lossless line of length l = 0.15λ. (
ki77a [65]

Answer:

a. Zin = 41.25 - j 16.35 Ω

b. V₁ = 143. 6 e⁻ ¹¹ ⁴⁶

c.  Pin = 216 w

d. PL = Pin = 216 w

e. Pg = 478.4 w , Pzg = 262.4 w

Explanation:

a.

Zin = Zo * [ ZL + j Zo Tan (βl) ] / [ Zo + j ZL Tan (βl) ]  

βl = 2π / λ * 0.15 λ = 54 °

Zin = 50 * [ 75 + j 50 Tan (54) ] / [ 50 + j 75 Tan (54) ]

Zin = 41.25 - j 16.35 Ω

b.

I₁ = Vg / Zg + Zin ⇒ I₁ = 300 / 41.25 - j 16.35 = 3.24 e ¹⁰ ¹⁶

V₁ = I₁ * Zin = 3.24 e ¹⁰ ¹⁶ * ( 41.25 - j 16.35)

V₁ = 143. 6 e⁻ ¹¹ ⁴⁶

c.

Pin = ¹/₂ * Re * [V₁ * I₁]

Pin = ¹/₂ * 143.6 ⁻¹¹ ⁴⁶ * 3.24 e ⁻ ¹⁰ ¹⁶ = 143.6 * 3.24 / 2 * cos (21.62)

Pin = 216 w

d.

The power PL and Pin are the same as the line is lossless input to the line ends up in the load so

PL = Pin

PL = 216 w

e.

Pg Generator

Pg = ¹/₂ * Re * [ V₁ * I₁ ] = 486 * cos (10.16)

Pg = 478.4 w

Pzg dissipated

Pzg = ¹/₂ * I² * Zg = ¹/₂ * 3.24² * 50

Pzg = 262.4 w

4 0
4 years ago
What must be the pressure difference between the two ends of a 2.6 km section of pipe, 36 cm in diameter, if it is to transport
ad-work [718]

Answer: 1.13(10)^{3} Pa

Explanation:

This problem can be solved by the following equation:

\Delta P=\frac{8 \eta L Q}{\pi r^{4}}

Where:

\Delta P is the pressure difference between the two ends of the pipe

\eta=0.20 Pa.s is the viscosity of oil

L=2.6 km=2600 m is the length of the pipe

Q=900 \frac{cm^{3}}{s} \frac{1 m^{3}}{(100 cm)^{3}}=0.0009 \frac{m^{3}}{s} is the Rate of flow of the fluid

d=36 cm=0.36 m is the diameter of the pipe

r=\frac{d}{2}=0.18 m is the radius of the pipe

Soving for \Delta P:

\Delta P=\frac{8 (0.20 Pa.s)(2600 m)(0.0009 \frac{m^{3}}{s})}{\pi (0.18 m)^{4}}

Finally:

\Delta P=1135.26 Pa \approx 1.13(10)^{3} Pa

7 0
3 years ago
In a charging process, 1 × 10^13 electrons are removed from one small metal sphere and placed on a second identical sphere. Init
liraira [26]

Answer:

r = 0.303m

= 30.3cm

Explanation:

Given that,

The number of electrons transferred from one sphere to the other,  

n  = 1 ×10 ¹³e le c t r o n s

The electrostatic potential energy between the spheres,  

U = − 0.061 J

The charge on an electron,  

q = − 1.6 × 10 ⁻¹⁹C

The coulomb constant,  

K = 8.98755 × 10 ⁹ N ⋅ m ² / C 2²

Due to the transfer of electrons, both spheres become equally and oppositely.

The charge gained by the sphere due to the excess of the electron is:  

q ₁ = n q

   = 1 ×10 ¹³ *  − 1.6 × 10 ⁻¹⁹

   = -1.6  × 10⁻⁶C

The charge left on the first sphere is =

q ₂ = -q₁ = 1.6  × 10⁻⁶C

The electric potential energy between two point charges is given by the following equation:

U = K q ₁q ₂/r

q ₁ and  q ₂ are the two charges.

r  is the distance between the charge and the point.

K  =  8.98755  ×  10 ⁹ N ⋅ m ² / C ²

we have:

-0.061 =  (8.98755  ×  10 ⁹ * (-1.6  × 10⁻⁶)²) / r

r = (18.41 ×  10 ⁻³) / 0.061

r = 0.303m

= 30.3cm

4 0
3 years ago
Which factors are used to calculate the kinetic energy of an object?
kipiarov [429]

Answer:The factors which are used to calculate the kinetic energy of an object are mass and velocity. Explanation: The kinetic energy is the energy of the motion of an object. Here, m is the mass of the object and v is the velocity of the object.

Explanation:

hope it helps

3 0
3 years ago
Which type of lever is exemplified by the flexing of the forearm by the biceps brachii muscle?
ira [324]

Answer:

a third class lever

Explanation:

The third class or interpower lever is a lever that enables fast and dynamic movements. It places the power between the resistance and the support, so the resistance arm is longer than the power.

It is the most frequent type of lever in the human body and as an example we can put the action of the brachial biceps in the flexion of the elbow, where the biceps is inserted in the forearm between the elbow that is behind and the resistance that would be displaced towards the hand by the weight of the load attached to the weight of the forearm.

A good range of movements is achieved although with less force and is the most frequent type of lever in human movement, although the same joint can form different types of lever depending on the type of movement performed .

8 0
4 years ago
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