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kiruha [24]
2 years ago
8

What does vf stand for a.fringe velocity b.first velocity c.final velocity

Physics
1 answer:
Elza [17]2 years ago
5 0
The correct answer is C. Final Velocity

Hope this helped!
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How to calculate the average value of the indirect measurements of KE
Tom [10]

Answer:

1/2 m*(2(h)/t)*g

Explanation:

8 0
3 years ago
What is the period of 60.0 hz electrical power?
Anarel [89]

Answer:

0.017 s

Explanation:

The period of a periodic signal is defined as the reciprocal of the frequency:

T=\frac{1}{f}

where

T is the period

f is the frequency

For the electrical power, the frequency is

f = 60.0 Hz

Substituting into the previous equation, we find the period:

T=\frac{1}{60.0 Hz}=0.017 s

8 0
3 years ago
12.51 A parallel RLC circuit, which is driven by a variable frequency 2-A current source, has the following values: R = 1 kΩ, L
Anastaziya [24]

Answer:

BW = 100 rad/s

wlow = 452.49 rad/s

whigh = 552.49 rad/s

V(jwlow) =1414.21 < 45°V

V(jwhigh) =1414.21 <-45°V

Explanation:

To calculate bandwidth we have formula

BW = 1/RC

BW = 1/ 1000x10x10^¯6

BW = 100 rad/s

We will first calculate resonant frequency and quality factor for half power frequencies.

For resonant frequency

wo = 1/(SQRT LC)

wo = 1/SQRT 400×10¯³ × 10×10^¯6

wo = 500 rad/s

For Quality

Q = wo / BW

Q = 500/100

Q = 5

wlow = wo [-1/2Q+ SQRT (1/2Q)² + 1]

wlow = 500 [-1/2×5 + SQRT (1/2×5)² + 1]

wlow = 452.49 rad/s

whigh = wo [1/2Q+ SQRT (1/2Q)² + 1]

whigh = 500 [1/2×5 + SQRT (1/2×5)² + 1]

whigh = 552.49 rad/s

We will start with admittance at lower half power frequency

Y(jwlow) = (1/R) + (1/jwlow L) + (jwlow C)

Y(jwlow) = (1/1000) + (1/j×452.49×400×10¯³) + (j×452.49×10×10^¯6)

Y(jwlow) = 0.001 - j5.525×10¯³ + j4.525×10¯³

Y(jwlow) = (1-j).10¯³ S

Voltage across the network is calculated by ohm's law

V(jwlow) = I/Y(jwlow)

V(jwlow) = 2/(1-j).10¯³

V(jwlow) = 1414.2 < 45°V

Now we will calculate the admittance at higher half power frequency

Y(jwhigh) = (1/R) + (1/jwhigh L) + (jwhigh C)

Y(jwhigh) = (1/1000) + (1/j×552.49×400×10¯³) + (j×552.49×10×10^¯6)

Y(jwhigh) = 0.001 - j4.525×10¯³ + j5.525×10¯³

Y(jwhigh) = (1+j).10¯³ S

Voltage across network will be calculated by ohm's law

V(jwhigh) = I/Y(jwhigh)

V(jwhigh) = 2/(1+j).10¯³

V(jwhigh) = 1414.2 < - 45°V

6 0
3 years ago
What are one of the ways the atmosphere recycle gases through interactions with the biosphere?
tino4ka555 [31]

Answer:

Though the process photosynthesis

Explanation:

One of the important ways by which the atmosphere recycle gases through interactions with the the biosphere is by photosynthesis.

The biosphere is the living component on the earth. The atmosphere is the gaseous envelope round the earth.

  • Photosynthesis is the process whereby green plants manufacture their food using carbon dioxide and water in the presence of sunlight to produce glucose and oxygen gas.
  • This way, atmospheric carbon dioxide is exchanged for oxygen gas.

This way, there is a sustained interaction between the atmosphere and biosphere.

3 0
3 years ago
Read 2 more answers
Business are necessary in order to preform which of the following functions
NeTakaya

Answer:

Hire, organize, and supply workers.

Explanation:

7 0
2 years ago
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