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Salsk061 [2.6K]
3 years ago
10

A parallel RLC resonant circuit has a resistance of 200 Ω. If it is known that the bandwidth is 80 rad/s and the lower half-powe

r frequency is 800 rad/s, find the values of the parameters L and C.
Physics
1 answer:
d1i1m1o1n [39]3 years ago
5 0

Answer:

L= 3.6mH

C =9.9 microfarad

Explanation:

Resonant frequency fr

fr = fl + 1/2 BW

fr = 800+ 1/2×80

=800+40

=840 rad/s

Bandwidth (BW)

BW = fr/Q

Q = quality factor

Q= fr/ BW

Q = 840/80

Q= 10.5

Quality factor = R/Xl

Xl = inductive reactance

Xl = R/Q

Xl = 200/10.5

Xl = 19.05 ohms

Xl =2πfL

L= Inductance

L = Xl /2πf

L =19.05/5278.56

L= 3.6mH

Capacitor C

1= fr^2 × 4π^2LC

C = 1/fr^24π^2L

C = 1/100307.5

C= 9.9microfarad

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Answer:

-0.55m/s

Explanation:

Given that: For the boy

Weight = 745N

Velocity = +0.35 m/s

Mass of the boy = ?

g = 9.81m/s^2

W = mg

745 = m×9.81

m = 75.94kg

For the girl

Given that:

Weight = 477 N

g = 9.81m/s^2

m = ?

W = mg

477 = m×9.81m/s^2

m = 48.62kg

To solve for the v of the girl, the two has to add up

48.62kg×v + 75.94kg×+0.35 m/s = 0

48.62v + 26.579 = 0

48.62v = - 26.579

v = -26.579/48.62

v = -0.5466

v = -0.55m/s

Hence, the velocity of the girl is -0.55m/s.

The negative sign is as a result of the two of them moving is opposite direction.

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3 years ago
When looking at a graph what are the first 3 things you should do?
Valentin [98]

Answer: Look where the points are.

Explanation:

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2 years ago
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The bohr shift on the oxygen-hemoglobin dissociation curve is produced by changes in
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Answer:

THE BOHR SHIFT ON THE OXYGEN-HEMOGLOBIN DISSOCIATION CURVE IS PRODUCED BY CHANGES IN THE CONCENTRATION OF CARBON IV OXIDE.

Explanation:

The oxygen-hemoglobin dissociation curve shows the relationship between the saturated hemoglobin concentration and oxygen. It shows how the blood hold on to and releases oxygen. The Bohr shift can occur as a result of changes in concentration of carbon iv oxide and other factors such as acidity or pH, 2,3-bisphosphoglycerate, exercise, also temperature of the body. These factors contributes to the right or left shift on the curve. Carbon iv oxide prevents the binding of oxygen to the hemoglobin. The is because hemoglobin has the same binding site for both oxygen and carbon iv oxide. Carbon iv oxide increase also leads to a change in the pH of the blood through the formation of bicarbonate ion. Bicarbonate ion formation causes reduced acidity and therefore lead a shift in the dissociation curve for more of the carbon iv oxide to be excreted as hemoglobin's affinity for oxygen reduces. And when the concentration of carbon iv oxide is low in the plasma, acidity increases and this provides more affinity for oxygen by the hemoglobin.

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The motion of a car on a position-time graph is represented with a horizontal line. What does this indicate about the car's moti
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Answer: The answer is B

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2 years ago
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To meet a U.S. Postal Service requirement, employees' footwear must have a coefficient of static friction of 0.5 or more on a sp
motikmotik

Answer:

0.79 s

Explanation:

We have to calculate the employee acceleration, in order to know the minimum time. According to Newton's second law:

\sum F_x:f_{max}=ma_x\\\sum F_y:N-mg=0

The frictional force is maximum since the employee has to apply a maximum force to spend the minimum time. In y axis the employee's acceleration is zero, so the net force is zero. Recall that f_{max}=\mu N

Now, we find the acceleration:

\mu N=ma_x\\\mu mg=ma_x\\a_x=\mu g\\a_x=0.83(9.8\frac{m}{s^2})\\a_x=8.134\frac{m}{s^2}

Finally, using an uniformly accelerated motion formula, we can calculate the minimum time. The employee starts at rest, thus his initial speed is zero:

x=v_0t+\frac{1}{2}a_xt^2\\2x=a_xt^2\\t=\sqrt{\frac{2x}{a}}\\t=\sqrt{\frac{2(3.2m)}{8.134\frac{m}{s^2}}}\\t=0.79 s

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3 years ago
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