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babymother [125]
3 years ago
6

Quan sát sóng dừng trên sợi dây dài 60cm cố định hai đầu, thấy có 3 nút (bao gồm cả hai

Physics
1 answer:
stiks02 [169]3 years ago
7 0

Answer:

Quan sát sóng dừng trên sợi dây dài 60cm cố định hai đầu, thấy có 3 nút (bao gồm cả hai

nút hai đầu) và 2 bụng. Bước sóng của sóng lan truyền trên sợi dây là:

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Please help in finding the Total Resistence, Total Current, Req Current Center of Parallel and V40 olms
Blizzard [7]
Total Current = 2 Amps
Req total = 60 ohms
Current on EF mesh = 1/3 Amp
Current on 24 ohms resistor = 1/6 Amp

Voltage on the 40ohms resistor is lefting.

3 0
3 years ago
If a series circuit contains a 12-V battery, a 6-ohm resistor, and a 4-ohm resistor, what is the current in the circuit?
Shalnov [3]

In a series circuit the total current is the same throughout resistors and so:

I_{total}=I_1=I_2

The voltage is distributed throughout the resistors and so:

V_{total}=V_1+V_2

and the total resistance can be calculated by adding up the resistors resistance:

R_{total}=R_1+R_2

First thing is to calculate the total resistance and so:

R_{total}=6\Omega + 4\Omega = 10\Omega

And by Omh's law V=IR we have:

V_{total}=I_{total}R_{total}\\\\I_{total}=\frac{V_{total}}{R_{total}}= \frac{12V}{10\Omega} =1.2A

And so the total current of the circuit is 1.2 amps i.e. 1.2 A.


6 0
3 years ago
Read 2 more answers
Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500◦C, and 80 m/s, and the exit
Cerrena [4.2K]

Answer:

a) ΔEC=-23.4kW

b)W=12106.2kW

c)A=0.01297m^2

Explanation:

A)

The kinetic energy is defined as:

\frac{m*vel^2}{2} (vel is the velocity, to differentiate with v, specific volume).

The kinetic energy change will be: Δ (\frac{mvel^2}{2})=\frac{m*vel_2^2}{2}-\frac{m*vel_1^2}{2}

Δ (\frac{mvel^2}{2})=\frac{m}{2}*(vel_2^2-vel_1^2)

Where 1 and 2 subscripts mean initial and final state respectively.

Δ(\frac{mvel^2}{2})=\frac{12\frac{kg}{s}}{2}*(50^2-80^2)\frac{m^2}{s^2}=-23400W=-23.4kW

This amount is negative because the steam is losing that energy.

B)

Consider the energy balance, with a neglective height difference: The energy that enters to the turbine (which is in the steam) is the same that goes out (which is in the steam and in the work done).

H_1+\frac{m*vel_1^2}{2}=H_2+\frac{m*vel_2^2}{2}+W\\W=m*(h_1-h_2)+\frac{m}{2} *(vel_1^2-vel_2^2)

We already know the last quantity: \frac{m}{2} *(vel_1^2-vel_2^2)=-Δ (\frac{mvel^2}{2})=23400W

For the steam enthalpies, review the steam tables (I attach the ones that I used); according to that, h_1=h(T=500C,P=4MPa)=3445.3\frac{kJ}{kg}

The exit state is a liquid-vapor mixture, so its enthalpy is:

h_2=h_f+xh_{fg}=289.23+0.92*2366.1=2483.4\frac{kJ}{kg}

Finally, the work can be obtained:

W=12\frac{kg}{s}*(3445.3-2438.4)\frac{kJ}{kg} +23.400kW)=12106.2kW

C) For the area, consider the equation of mass flow:

m=p*vel*A where p is the density, and A the area. The density is the inverse of the specific volume, so m=\frac{vel*A}{v}

The specific volume of the inlet steam can be read also from the steam tables, and its value is: 0.08643\frac{m^3}{kg}, so:

A=\frac{m*v}{vel}=\frac{12\frac{kg}{s}*0.08643\frac{m^3}{kg}}{80\frac{m}{s}}=0.01297m^2

Download pdf
7 0
3 years ago
When a laser shines on a screen after passing through two closely spaced slits, you see ___
Aleonysh [2.5K]

Answer:

diffraction

Explanation:

diffraction occurs when light passes sharp edges or goes through narrow slits the rays are deflected and produce fringes of light and dark bands

6 0
3 years ago
A uniformly charged solid disk of radius R = 0.45 m carries a uniform charge density of σ = 175 μC/m². A point P is located a di
siniylev [52]

Answer:

1408.685 KN/C

Explanation:

Given:

R = 0.45 m

σ = 175 μC/m²

P is located a distance a = 0.75 m

k = 8.99*10^9

  • The Electric Field Strength E of a uniformly solid disk of charge at distance a perpendicular to disk is given by:

                                  E = 2*pi*k*o * (1 - \frac{a}{\sqrt{a^2 + R^2} })\\

part a)

Electric Field strength at point P: a = 0.75 m

E = 2*pi*8.99*10^9*175*10^-6 * (1 - \frac{0.75}{\sqrt{0.75^2 + 0.45^2} })\\\\E = 9885021.285*(0.1425070743)\\\\E = 1408.685 KN/C

part b)

Since, R >> a, we can approximate a / R = 0 ,

Hence, E simplified relation becomes:

E = 2*pi*k*o * (1 - \frac{a/R}{\sqrt{a^2/R^2 + 1} })\\\\E = 2*pi*k*o * (1 - \frac{0}{\sqrt{0 + 1} })\\\\E = 2*pi*k*o

E = σ / 2*e_o

part c)

Since, a >> R, we can approximate. that the uniform disc of charge becomes a single point charge:

Electric Field strength due to point charge is:

E = k*δ*pi*R^2 / a^2  

Since, R << a, Surface area = δ*pi

Hence,

E = (k*δ*pi/a^2)

 

6 0
3 years ago
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