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KatRina [158]
3 years ago
8

An ideal transformer has 75 turns in the primary coil and n turns in the secondary coil. a 120 v rms 60hz ac voltage source is c

onnected to the primary coil. a 10 ω resistor is connected to the secondary coil, forming a secondary circuit. the average power dissipated in the secondary circuit is 160 w. what is the number ns of turns in the secondary coil?
Physics
1 answer:
denpristay [2]3 years ago
4 0

think you messed up the symbol for resistor as resistors are measured in ohms where the symbol used for ohms is Greek omega

solving for average power in secondary coil:

average power =(current rms)^2*resistance⇒with a little algebra:

current rms=(√average power/resistance)

current rms=√160W/10Ω

current rms=4amps.

average power is also equal to current rms*voltage rms

with some algebra we can solve for voltage in the secondary wire:

voltage rms= average power/ current rms

voltage rms= 160W/4A

voltage rms=40Volts

now that we have voltage in the soecondary we can solve for the amount of turns in the secondary: Voltage secondary/voltage primary=number of turns in secondary/ number of turns in primary. using some algerbra we can solve for number of turns in secondary: (Voltage secondary/voltage primary)*number of turns in primary=number of turns in secondary

(40V/120V)*75turns=number of turns in secondary

number of turns in secondary=25turns

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With an initial velocity of 20km per hour a car accelerated at 8m/s2 for 10s,what is the position of the car at the end of 10s
Dima020 [189]
  • initial velocity=u=20km/h=5.5m/s
  • Acceleration=a=8m/s^2
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\\ \sf\longmapsto s=ut+\dfrac{1}{2}at^2

\\ \sf\longmapsto s=5.5(10)+\dfrac{1}{2}(8)(5.5)^2

\\ \sf\longmapsto s=55+4(5.5)^2

\\ \sf\longmapsto s=55+4(30.25)

\\ \sf\longmapsto s=55+121

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4 0
2 years ago
Rank the following objects by their accelerations down an incline (assume each object rolls without slipping) from least to grea
Alexxx [7]

Answer:

acceleration are

     hollow cylinder < hollow sphere < solid cylinder < solid sphere

Explanation:

To answer this question, let's analyze the problem. Let's use conservation of energy

Starting point. Highest point

          Em₀ = U = m g h

Final point. To get off the ramp

          Em_f = K = ½ mv² + ½ I w²

notice that we include the kinetic energy of translation and rotation

         

energy is conserved

        Em₀ = Em_f

        mgh = ½ m v² +1/2 I w²

angular and linear velocity are related

         v = w r

         w = v / r

we substitute

          mg h = ½ v² (m + I / r²)

          v² = 2 gh   \frac{m}{m+ \frac{I}{r^2} }

          v² = 2gh    \frac{1}{1 + \frac{I}{m r^2} }

this is the velocity at the bottom of the plane ,, indicate that it stops from rest, so we can use the kinematics relationship to find the acceleration in the axis ax (parallel to the plane)

         v² = v₀² + 2 a L

where L is the length of the plane

         v² = 2 a L

         a = v² / 2L

we substitute

         a = g \ \frac{h}{L} \  \frac{1}{1+ \frac{I}{m r^2 } }

let's use trigonometry

         sin θ = h / L

         

we substitute

         a = g sin θ   \ \frac{h}{L} \  \frac{1}{1+ \frac{I}{m r^2 } }

the moment of inertia of each object is tabulated, let's find the acceleration of each object

a) Hollow cylinder

      I = m r²

we look for the acerleracion

      a₁ = g sin θ    \frac{1}{1 + \frac{mr^2 }{m r^2 } }1/1 + mr² / mr² =

      a₁ = g sin θ    ½

b) solid cylinder

       I = ½ m r²

       a₂ = g sin θ  \frac{1}{1 + \frac{1}{2}  \frac{mr^2}{mr^2} } = g sin θ   \frac{1}{1+ \frac{1}{2} }

       a₂ = g sin θ   ⅔

c) hollow sphere

     I = 2/3 m r²

     a₃ = g sin θ   \frac{1}{1 + \frac{2}{3} }

     a₃ = g sin θ \frac{3}{5}

d) solid sphere

     I = 2/5 m r²

     a₄ = g sin θ  \frac{1 }{1 + \frac{2}{5} }

     a₄ = g sin θ  \frac{5}{7}

We already have all the accelerations, to facilitate the comparison let's place the fractions with the same denominator (the greatest common denominator is 210)

a) a₁ = g sin θ ½ = g sin θ      \frac{105}{210}

b) a₂ = g sinθ ⅔ = g sin θ     \frac{140}{210}

c) a₃ = g sin θ \frac{3}{5}= g sin θ       \frac{126}{210}

d) a₄ = g sin θ \frac{5}{7} = g sin θ      \frac{150}{210}

the order of acceleration from lower to higher is

   

     a₁ <a₃ <a₂ <a₄

acceleration are

     hollow cylinder < hollow sphere < solid cylinder < solid sphere

8 0
3 years ago
A current of 5.0 a flows through an electrical device for 10 seconds. how many electrons flow through this device during this ti
melisa1 [442]
1 Amp = 1 Coulomb/sec
1 Coulomb/sec = 6.25*10^18 electrons/sec

Therefore,
5.0 A = 5 C/s = 5*6.25*10^18 = 3.125*10^19 e/s

In 10 second, number of electrons are calculated as;
Number of electrons through the device = 3.125*10^19*10 = 3.125*10^20 electrons

4 0
3 years ago
Which objects would sink in honey, which has a density of 1.4 g/cm³? Check all that apply.
bazaltina [42]

Answer:

any object that has density more than 1.4

Explanation:

The object that has density more than 1.4 is denser than the honey

6 0
3 years ago
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