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kvv77 [185]
3 years ago
15

Look at your data. With three foods, Flock v

Physics
1 answer:
wolverine [178]3 years ago
6 0

Answer:the first one was x

the second one is y

Explanation:

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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
Concept map for kinetic energy, work and power
Helen [10]

Kinetic energy: the energy of motion

Work: the change in kinetic energy

Power: the rate of work done

Explanation:

The kinetic energy of an object is the energy possessed by the object due to its motion. Mathematically, it is given by:

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

The work done an object is the amount of energy transferred; according to the energy-work theorem, it is equal to the change in kinetic energy of an object:

W=K_f - K_i

where

K_f is the  final kinetic energy

K_i is the initial kinetic energy

Finally, the power is the rate of work done per unit time. Mathematically, ti can be expressed as

P=\frac{W}{t}

where

W is the work done

t is the time elapsed

Learn more about kinetic energy, work and power:

brainly.com/question/6536722

brainly.com/question/6763771

brainly.com/question/6443626

brainly.com/question/7956557

#LearnwithBrainly

8 0
3 years ago
In a river, sediment is deposited _____. on the inside bends on the outside bends by fast moving water by oxbows
N76 [4]
Hey There!

Your answer is on the inside bends!

When rocks are deposition on the bottom of the river pressure can cause the rock to break down in sentiments!

If you need anymore help with your work feel free to ask me!

Hope this Helps!
3 0
3 years ago
How much potential energy does a 40n block medicine ball gain when it is lifted 5.m
MAVERICK [17]
By definition, the gain in PE (potential energy) is
ΔPE = m*g*h

Given:
mg = 40 N  (Note that m*g = weight)
h = 5 m

ΔPE = (40 N)*(5 m) =200 J

Answer: 200 J

5 0
3 years ago
Read 2 more answers
What type of relationship exists between the length of a wire and the resistance, if all other factors remain the same? A. Resis
ZanzabumX [31]
<h3>Answer</h3>

(A) Resistance is directly related to length.

<h3>Explanation</h3>

Formula for resistance

R = p(length) / A

where R = resistance

           p = resistivity(material of wire)

           A = cross sectional area

So it can be seen that resistance depends upon 3 factors that are length of wire , resistivity of wire and the cross sectional area of the wire.

If two of the factors, resistivity and cross sectional area, are kept constant then the resistance is directly proportional to the length of wire.

<h3> R ∝ length</h3>

This means that the resistance of the wire increases with the increase in length of the wire and decreases with the decrease of length of the wire.

5 0
3 years ago
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