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pochemuha
3 years ago
14

Recall that the voltages VL(t) and VC(t) across the inductor and capacitor are not in phase with the respective currents IL(t) a

nd IC(t). In particular, which of the following statements is true for a sinusoidal current driver?
VL(t) and VC(t):
a) both lag their respective current
b) both lead their respective currents
c) VL(t) lags IL(t) and VC(t) leads IC(t)
d) VL(t) leads IL(t) and VC(t) lags IC(t)
Physics
1 answer:
Nikolay [14]3 years ago
7 0

Answer:

D) VL(t) leads IL(t) and VC(t) lags IC(t)

Explanation:

This is because The phase angle between voltage and current for inductors and capacitors is 90 degrees, or radians, also, this means that no power is dissipated in either the inductor or the capacitor, since the time average of current times voltage,

( I(t), V(t)), is zero.

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Ocean waves pass through two small openings, 20.0 m apart, in a breakwater. You're in a boat 70.0 m from the breakwater and init
Klio2033 [76]

Answer:

λ = 5.65m

Explanation:

The Path Difference Condition is given as:

δ=(m+\frac{1}{2})\frac{lamda}{n}  ;

where lamda is represent by the symbol (λ) and is the wavelength we are meant to calculate.

m = no of openings which is 2

∴δ= \frac{3*lamda}{2}

n is the index of refraction of the medium in which the wave is traveling

To find δ we have;

δ= \sqrt{70^2+(33+\frac{20}{2})^2 }-\sqrt{70^2+(33-\frac{20}{2})^2 }

δ= \sqrt{4900+(\frac{66+20}{2})^2}-\sqrt{4900+(\frac{66-20}{2})^2}

δ= \sqrt{4900+(\frac{86}{2})^2 }-\sqrt{4900+(\frac{46}{2})^2 }

δ= \sqrt{4900+43^2}-\sqrt{4900+23^2}

δ= \sqrt{4900+1849}-\sqrt{4900+529}

δ= \sqrt{6749}-\sqrt{5429}

δ=  82.15 -73.68

δ= 8.47

Again remember; to calculate the wavelength of the ocean waves; we have:

δ= \frac{3*lamda}{2}

δ= 8.47

8.47 = \frac{3*lamda}{2}

λ = \frac{8.47*2}{3}

λ = 5.65m

3 0
4 years ago
Scientific notation and graphing <br><br> 0.0004580 to scientific notation
Vlada [557]

Scientific Notation: 4.580 x 10^-4

Scientific e Notation: 4.580e-4

4 0
3 years ago
Read 2 more answers
5. How much heat is generated when an
Mila [183]

Answer:

I HOPE THIS IS CORRECT

Explanation:

Power of water =2 kw=2000w

Mass of water =200kg

difference in temperature ΔT=70−10=60oC

Concept

energy required to heat the water = energy given by water in time t=pt

energy required to increase tempeature of water by 60oC,Q=msΔT

S= specific heat =4200J/kgoC

              pt=msΔT

   2000×t=200×4200×60

      t=25200  

or   t=25.2×103sec.

6 0
3 years ago
A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Co
Tems11 [23]

(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

<h3>Center mass of the airplane</h3>

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>
  • The wheels under the wind do not pass through the center line.
  • The position of the front wheel is constant and it is zero mark (origin).
  • The rear wheels are at 21.7 m mark

Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

<h3>Mass of the plane at the position of the rear wheels</h3>

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}

18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg

<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN

<h3>Mass of the plane at the position of the front wheel</h3>

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

<h3>Force exerted by the ground on the front wheel</h3>

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

Learn more about center mass here: brainly.com/question/13499822

7 0
2 years ago
A sports car accelerates from 0 to 25 meters per second in 4 seconds. What is its acceleration?
WITCHER [35]

Answer:

6.25 ms²

Explanation:

..................

3 0
3 years ago
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