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pogonyaev
3 years ago
13

Before designing the filter, one must understand the relationship between the output voltage of the circuit and the frequency. F

or a series RLC circuit, how do the impedances of the circuit elements depend on the frequency?
a. The impedance magnitudes do not depend on frequency.
b. Both impedance magnitudes decrease when the frequency increases.
c. The magnitude of the impedance of the capacitor decreases when the frequency increases; the magnitude of the impedance of the inductor increases when the frequency increases.
d. The magnitude of the impedance of the capacitor increases when the frequency increases; the magnitude of the impedance of the inductor decreases when the frequency increases.
Physics
1 answer:
Kamila [148]3 years ago
5 0

Answer:

Option b. is correct

Explanation:

An RLC electrical circuit consists of constituent components: a resistor (R), an inductor (L), and a capacitor (C). A resistor, an inductor, and a capacitor are connected in series or parallel.

The impedances of the circuit elements depend on the frequency.

Both impedance magnitudes decrease when the frequency increases

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A car is 2.0 km west of a traffic light at t = 0 and 5.0 km east of the light at t = 6.0 min. Assume the origin of the coordinat
egoroff_w [7]

Answer:

(a) Position Vectors V₁= -2î km, V₂=5î km

(b) Displacement Δx=7 km

Explanation:

Given data

Distance=2 km west at t=0

Distance=5 km east at t=6 min

Positive x is the east direction

To find

(a)Car position vector at given times

(b)Displacement between 0 to 6.0 min

Solution

For Part (a) car position vector at given times

At t=0 the distance=2 km west so conclude that x₁=-2 because it is in negative side So vector V₁

V₁= -2î km

At t=6.0 the distance=5 km east so conclude that x₂=5 because it is in positive side So vector V₂

V₂=5î km

For (b) displacement between 0 to 6.0 min

According to following mathematical law we can conclude that

Δx=x₂-x₁

Δx=5-(-2)km

Δx=7 km

6 0
3 years ago
Two solid steel shafts are fitted with flanges that are then connected by bolts as shown. the bolts are slightly undersized and
Lelu [443]

Answer:

The maximum shear stress in shaft AB, T_{ABmax} is 15 MPa

The maximum shear stress in shaft CD,  T_{CDmax} is 45.9 MPa

Explanation:

The formula for a shaft polar moment of inertia, J is given by  

J = \pi \times \frac{D^4}{32} =\pi \times \frac{r^4}{2}

Therefore, we have

J_{AB} = \pi \times \frac{D_{AB}^4}{32} =\pi \times \frac{r_{AB}^4}{2}

Where:

D_{AB} = Diameter of shaft AB = 30 mm = 0.03 m

r_{AB} = Radius of shaft AB = 15 mm = 0.015 m

∴ J_{AB} = \pi \times \frac{0.03^4}{32} =\pi \times \frac{0.015^4}{2} = 7.95 × 10⁻⁸ m⁴

and

J_{CD} = \pi \times \frac{D_{CD}^4}{32} =\pi \times \frac{r_{CD}^4}{2}

Where:

D_{CD} = Diameter of shaft CD = 36 mm = 0.036 m

r_{CD} = Radius of shaft CD = 18 mm = 0.018 m

Therefore,

J_{CD} = \pi \times \frac{0.036^4}{32} =\pi \times \frac{0.018^4}{2} = 1.65 × 10⁻⁷ m⁴

Given that the shaft AB and CD are rotated 1.58 ° relative to each other, we have;

1.58 °= 1.58 \times \frac{2\pi }{360} rad = 2.76 × 10⁻² rad.

That is \phi_r = 2.76 × 10⁻² rad.

However  \phi_r =  \phi_{C/D} -  \phi_{B/A}  

Where:

\phi_{B/A} = \frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G} and

\phi_{C/D} = \frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G}

T_{AB} and T_{CD}= Torque on shaft AB and CD respectively

T_{AB}  = Required

T_{CD}= 500 N·m

L_{AB} and L_{CD} = Length of shafts AB an CD respectively

L_{AB}  = 600 mm = 0.6 m

L_{CD} = 900 mm = 0.9 m

G = Shear modulus of the material = 77.2 GPa

Therefore;

\phi_r =  \phi_{C/D} -  \phi_{B/A}  =\frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G} -\frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}

2.76 × 10⁻² rad =\frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G} -\frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}

=\frac{500\cdot 0.9}{1.65 \times 10^{-7} \cdot 77.2\times 10^9} -\frac{T_{AB}\cdot 0.6}{7.95\times 10^{-7} \cdot 77.2\times 10^9}

Therefore;

T_{AB} =  79.54 N.m

Where T = T_{AB} + T_{CD} =

Therefore T_{CD total } = 500 - 79.54 = 420.46 N·m

τ_{max} = \frac{T\times R}{J}

\tau_{ABmax} = \frac{T_{AB}\times R_{AB}}{J_{AB}} =  \frac{79.54\times 0.015}{7.95\times 10^{-8}} = 15 MPa

\tau_{CDmax} = \frac{T_{CD}\times R_{CD}}{J_{CD}} = \frac{420.46\times 0.018}{1.65\times 10^{-7}} = 45.9 MPa

7 0
3 years ago
Is there a way to approximate the distance from a black hole's singularity to it's event horizon ? If so, what formula is used t
strojnjashka [21]

Answer:

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4 0
3 years ago
5. Find the volume of the composite space figure to the nearest whole number.
Dmitry [639]
The shape is missing but let's consider it a semi-cylinder attached to the rectangular prism.
Given:
radius = 4.5 mm 
<span>Height = 11 mm </span>

<span>Volume of cylinder = (1/2)(pi)(4.5)^2(11)   (the shape is divided into half)
                           V = 349.89 mm cubed
Volume of prism = L x W x H
                           = 9 x 11 x 6
                           = 594 mm cubed
Total volume of the composite shape = 111.375 + 594 
                                                            = 943.89 mm cubed
Rounded answer = 944 mm cubed.</span>
6 0
4 years ago
The volume of vessel is 6 litres. convert it into ml​
dalvyx [7]
6000 ml
There are 1000ml in one liter
5 0
3 years ago
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