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1 year ago

The coils in a motor have a resistance of 0.14 Ω and are driven by an EMF of 62 V.

Physics

1 answer:

Brilliant_brown [7]1 year ago

7 0

The difference in the power required by the motor at start up and at operating speed is 9778.57

<h3>What is power? </h3>

This is defined as the rate in which energy is consumed. Electrical power is expressed mathematically as:

Power (P) = voltage (V) × current (I)

P = IV

Power = square voltage (V²) / resistance ®

P = V² / R

<h3>How to determine the power by the EMF</h3>

Voltage (V) = 37 V
Resistance (R) = 0.14 Ω
Power by EMF (P₁) =?

P = V² / R

P₁ = 62² / 0.14

P₁ = 27457.14 watts

<h3>How to determine the power by the back EMF</h3>

Voltage (V) = 62 V
Resistance (R) = 0.14 Ω
Power by back EMF (P₂) =?

P = V² / R

P₂ = 37² / 0.14

P₂ = 9778.57 watts

<h3>How to determine the power difference</h3>

Power by EMF (P₁) = 27457.14 watts
Power by back EMF (P₂) = 9778.57 watts
Difference in power =?

Difference = P₁ - P₂

Difference in power = 27457.14 - 9778.57

Difference in power = 17678.57 watts

Learn more about electrical power:

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rusak2 [61]

Answers:

a) $T=7.04(10)^{-10} s$

b) $5.11(10)^{12} cycles$

c) $2.06(10)^{26} cycles$

d) 46000 s

Explanation:

<h2>a) Time for one cycle of the radio wave</h2>

We know the maser radiowave has a frequency $f$ of $1,420,405,751.786 cycles/s$

In addition we know there is an inverse relation between frequency and time $T$ :

$f=\frac{1}{T}$ (1)

Isolating $T$ : $T=\frac{1}{f}$ (2)

$T=\frac{1}{1,420,405,751.786 cycles/s}$ (3)

$T=7.04(10)^{-10} s$ (4) This is the time for 1 cycle

<h2>b) Cycles that occur in 1 h</h2>

If $1h=3600s$ and we already know the amount of cycles per second $1,420,405,751.786 cycles/s$ , then:

$1,420,405,751.786 \frac{cycles}{s}(3600s)=5.11(10)^{12} cycles$ This is the number of cycles in an hour

<h2>c) How many cycles would have occurred during the age of the earth, which is estimated to be $4.6(10)^{9} years$ ?</h2>

Firstly, we have to convert this from years to seconds:

$4.6(10)^{9} years \frac{365 days}{1 year} \frac{24 h}{1 day} \frac{3600 s}{1 h}=1.45(10)^{17} s$

Now we have to multiply this value for the frequency of the maser radiowave:

$1,420,405,751.786 cycles/s (1.45(10)^{17} s)=2.06(10)^{26} cycles$ This is the number of cycles in the age of the Earth

<h2>d) By how many seconds would a hydrogen maser clock be off after a time interval equal to the age of the earth?</h2>

If we have 1 second out for every 100,000 years, then:

$4.6(10)^{9} years \frac{1 s}{100,000 years}=46000 s$

This means the maser would be 46000 s off after a time interval equal to the age of the earth

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3 years ago

A 220 kg crate hangs from the end of a rope of length L = 14.0 m. You push horizontally on the crate with a varying force F to m

kifflom [539]

<u>Answer</u>:

(a) magnitude of F = 797 N

(b)the total work done W = 0

(c)work done by the gravitational force = -1.55 kJ

(d)the work done by the pull = 0

(e) work your force F does on the crate = 1.55 kJ

<u>Explanation</u>:

<u>Given</u>:

Mass of the crate, m = 220 kg

Length of the rope, L = 14.0m

Distance, d = 4.00m

<u>(a) What is the magnitude of F when the crate is in this final position</u>

Let us first determine vertical angle as follows

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

=> $\theta = Sin^{-1} \frac{d}{L}$ =

Now substituting thje values

=> $\theta = Sin^{-1} \frac{4}{12}$ =

=> $\theta = Sin^{-1} \frac{1}{3}$

=> $\theta = Sin^{-1}(0.333)$

=> $\theta = 19.5^{\circ}$

Now the tension in the string resolve into components

The vertical component supports the weight

=> $Tcos\theta = mg$

=> $T = \frac{mg}{cos\theta}$

=> $T = \frac{230 \times 9.8 }{cos(19.5)}$

=> $T = \frac{2254 }{cos(19.5)}$

=> $T = \frac{2254 }{0.9426}$

=>T =2391N

Therefore the horizontal force

$F = TSin(19.5)$

F = 797 N

b) The total work done on it

As there is no change in Kinetic energy

The total work done W = 0

<u>c) The work done by the gravitational force on the crate</u>

The work done by gravity

Wg = Fs.d = - mgh

Wg = - mgL ( 1 - Cosθ )

Substituting the values

= $-230 \times 9.8\times 12 ( 1 - cos(19.5) )$

= $-230 \times 9.8\times 12 ( 1 - 0.9426) )$

= $-230 \times 9.8\times 12 (0.0574)$

= $-230 \times 9.8\times 0.6888$

= $-230 \times 6.750$

= -1552.55 J

The work done by gravity = -1.55 kJ

<u>d) the work done by the pull on the crate from the rope</u>

Since the pull is perpendicular to the direction of motion,

The work done = 0

e)Find the work your force F does on the crate.

Work done by the Force on the crate

WF = - Wg

WF = -(-1.55)

WF = 1.55 kJ

<u>(f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)</u>

Here the work done by force is not equal to F*d

and it is equal to product of the cos angle and F*d

So, it is not equal to the product of the horizontal displacement and the answer to (a)

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3 years ago

Electromagmetic radiation comes in different forms which pair does not appear next to one an other in the electromagmetic spectr

vivado [14]

Answer:

A) microwaves and ultraviolet

Explanation:

this is the spectrum in order:

radio waves

microwaves

infra red

visible light

ultraviolet

X rays

gamma rays

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2 years ago

The aqueduct passes under Johnson Road in Lancaster through a siphon. The maximum capacity of the aqueduct is 350 m3/s. The heig

Mariulka [41]

Answer:

D ≈ 8.45 m

L ≈ 100.02 m

Explanation:

Given

Q = 350 m³/s (volumetric water flow rate passing through the stretch of channel, maximum capacity of the aqueduct)

y₁ - y₂ = h = 2.00 m (the height difference from the upper to the lower channels)

x = 100.00 m (distance between the upper and the lower channels)

We assume that:

the upper and the lower channels are at the same pressure (the atmospheric pressure).
the velocity of water in the upper channel is zero (v₁ = 0 m/s).
y₁ = 2.00 m (height of the upper channel)
y₂ = 0.00 m (height of the lower channel)
g = 9.81 m/s²
ρ = 1000 Kg/m³ (density of water)

We apply Bernoulli's equation as follows between the point 1 (the upper channel) and the point 2 (the lower channel):

P₁ + (ρ*v₁²/2) + ρ*g*y₁ = P₂ + (ρ*v₂²/2) + ρ*g*y₂

Plugging the known values into the equation and simplifying we get

Patm + (1000 Kg/m³*(0 m/s)²/2) + (1000 Kg/m³)*(9.81 m/s²)*(2 m) = Patm + (1000 Kg/m³*v₂²/2) + (1000 Kg/m³)*(9.81 m/s²)*(0 m)

⇒ v₂ = 6.264 m/s

then we apply the formula

Q = v*A ⇒ A = Q/v ⇒ A = Q/v₂

⇒ A = (350 m³/s)/(6.264 m/s)

⇒ A = 55.873 m²

then, we get the diameter of the pipe as follows

A = π*D²/4 ⇒ D = 2*√(A/π)

⇒ D = 2*√(55.873 m²/π)

⇒ D = 8.434 m ≈ 8.45 m

Now, the length of the pipe can be obtained as follows

L² = x² + h²

⇒ L² = (100.00 m)² + (2.00 m)²

⇒ L ≈ 100.02 m

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Answer:

Solution

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