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Studentka2010 [4]
3 years ago
12

An AC voltage source of amplitude 10V is supplied across an inductor and a resistor wired in series. The voltage amplitude acros

s the inductor is 5V. The voltage amplitude across the resistor is closest to:___________.
Physics
1 answer:
marin [14]3 years ago
4 0

Answer:

8.66 V

Explanation:

Vo = 10 v

VL = 5 V

VR = ?

According to the equation of LR circiut

V_{0}^{2}=V_{R}^{2}+ V_{L}^{2}

10 x 10 = VR² + 5 x 5

100 - 25 = VR²

VR² = 75

VR = 8.66 V

Thus, the voltage amplitude across the resistor is 8.66 v.

Explanation:

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A plane designed for vertical takeoff has a mass of 8.0 × 10³ kg. Find the net work done by all forces on the plane as it accele
artcher [175]

Answer:

<em>the net work done after starting from rest is =  2.4 × 10⁵ J</em>

Explanation:

Work: Work can be defined as the product of force and distance. The fundamental unit of work is Joules (J),  The unit of Energy is Joules (J), as such Energy and work are interchangeable during calculation, This is illustrated below

E = W = 1/2mv².......................... Equation 1

Where m = mass of the plane, v = velocity of the plane, E = Energy, W = work done.

v² = u² + 2as ................................. Equation 2.

Where v = final velocity of the plane, u = initial velocity of the plane, a = acceleration of the plane, distance of the plane.

<em>Given: a = 1.0 m/s², s = 30 m, u = 0 m/s (at rest)</em>

<em>Substituting these values into equation 2</em>

<em>v² = 0² +2×1×30</em>

<em>v² = 60</em>

<em>v = √60</em>

<em>v = 7.75 m/s</em>

Also given: m = 8.0 × 10³ kg, and v = 7.75 m/s

<em>Substituting these values into equation 1,</em>

<em>W = 1/2(8.0×10³)(7.75)²</em>

<em>W = (4.0×10³)(60)</em>

W = 240 × 10³ J

<em>W = 2.4 × 10⁵ J</em>

<em>Therefore the net work done after starting from rest is =  2.4 × 10⁵ J</em>

4 0
3 years ago
A boy is playing with a ball of mass 72g attached to a string. He is moving it at constant speed in a horizontal circle of radiu
irina [24]

Answer:

Explanation:

radius of circle r = 0.9 m.

(a ) In a motion on circular path , work done is zero because force ( centripetal force ) acts perpendicular to displacement .

( b )

Tension in string T = m ω²r

Putting the values

60 = .072 x ω² x 0.9

ω² = 926

ω = 30.4 rad /s

angle made in 20 revolutions θ = 20 x 2π = 126.6 rad

time taken = θ / ω

= 126.6 / 30.4

= 4.16 s .

3 0
3 years ago
Which part of a wind-powered system ultimately produces the electricity? A. nacelle B. blade C. turbine D. generator
djverab [1.8K]
<span>The correct answer is: (D) Generator

Explanation:
In wind-powered systems, the wind energy turns the blades around the rotor of a wind turbine. That rotor is connected to a generator that generates electricity. In other words, the kinectic energy of the wind is converted into electrical energy by using the generator in the wind-powered systems.</span>
5 0
3 years ago
Read 2 more answers
You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh
spayn [35]

To solve this problem we will apply the concepts related to Reyleigh's criteria. Here the resolution of the eye is defined as 1.22 times the wavelength over the diameter of the eye. Mathematically this is,

\theta = \frac{1.22 \lambda }{D}

Here,

D is diameter of the eye

D = \frac{1.22 (539nm)}{5.11 mm}

D= 1.287*10^{-4}m

The angle that relates the distance between the lights and the distance to the lamp is given by,

Sin\theta = \frac{d}{L}

For small angle, sin\theta = \theta

sin \theta = \frac{d}{L}

Here,

d = Distance between lights

L = Distance from eye to lamp

For small angle sin \theta = \theta

Therefore,

L = \frac{d}{sin\theta}

L = \frac{0.691m}{1.287*10^{-4}}

L = 5367m

Therefore the distance is 5.367km.

4 0
3 years ago
In a transformer, energy is carried from the primary coil to the secondary coil by:________
likoan [24]

In a transformer, energy is carried from the primary coil to the secondary coil by magnetic field in the iron core.

To find the answer, we have to know more about the transformer.

<h3>How transformer works?</h3>
  • An item utilized in the transfer of electric energy is a transformer.
  • AC current is used for transmission.
  • It is frequently used to modify the supply voltage between circuits without altering the AC frequency.
  • The fundamentals of mutual and electromagnetic induction govern how the transformer operates.
  • Magnetic field through the primary coil changes when primary coil current varies. the iron core of the secondary coil likewise has a magnetic field.
  • EMF is therefore generated in the secondary coil.

Thus, we can conclude that, in a transformer, energy is carried from the primary coil to the secondary coil by magnetic field in the iron core.

Learn more about the transformer here:

brainly.com/question/26787198

#SPJ4

5 0
2 years ago
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