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4vir4ik [10]
3 years ago
8

A transformer is to be used to provide power for a computer disk drive that needs 6.4 V (rms) instead of the 120 V (rms) from th

e wall outlet. The number of turns in the primary is 405, and it delivers 500 mA (the secondary current) at an output voltage of 6.4 V (rms). Find the current in the primary.
Physics
1 answer:
Amanda [17]3 years ago
8 0

Answer:

The current in the primary is 0.026 A

Explanation:

Using the formula

I1 = (V1/V2)*I2

we have

I1 = (6.4/120)*0.500

I1 = 0.026 A

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For a specific volume of 0.2 m3/kg, find the quality of steam if the absolute pressure is (a) 40 kPa and (b) 630 kPa. What is th
ICE Princess25 [194]

Answer:

x=0.0498

x'=0.659

Explanation:

Specific Volume V=0.2m_3/kg

Absolute Pressure (a) P_a= 40kpa

Giving

T_a=75.87

v_f=1.265*10^{-3}m^3/kg

v_g=3.993m^3/kg

                               (b) P_a= 630kpa

Giving

T_b=160.13C

v_f'=1.10282*10^{-3} m^3/kg

v_g'=0.30286 m^3/kg

(a)

Generally the equation for quality of Steam X  is mathematically given by

x=\frac{v-v_f}{v_g-v_f}

x=\frac{0.2-1.0265*10^{-3}}{3.993-1.0265*10^{-3}}

x=0.0498

(b)

Generally the equation for quality of Steam X  is mathematically given by

x'=\frac{v-v_f'}{v_g'-v_f'}

x'=\frac{0.2-1.10*10^{-3}}{3.30-1.1*10^{-3}}

x'=0.659

4 0
3 years ago
If the sun had twice the mace how would that affect the gravitational force of the sun
daser333 [38]

Answer:  Gravity is the force that keeps planets in orbit around the Sun. Gravity alone holds us to Earth's surface.

Planets have measurable properties, such as size, mass, density, and composition. A planet's size and mass determines its gravitational pull.

A planet's mass and size determines how strong its gravitational pull is.

Models can help us experiment with the motions of objects in space, which are determined by the gravitational pull between them.

Explanation:

5 0
3 years ago
A rectangular loop of area A is placed in a region where the magnetic field is perpendicular to the plane of the loop. The magni
mina [271]

Answer:

Induced emf, \epsilon=-A\dfrac{B_{max}e^{-t/\tau}}{\tau}

Explanation:

The varying magnetic field with time t is given by according to equation as :

B=B_{max}e^{-t/\tau}

Where

B_{max}\ and\ t are constant

Let \epsilon is the emf induced in the loop as a function of time. We know that the rate of change of magnetic flux is equal to the induced emf as:

\epsilon=-\dfrac{d\phi}{dt}

\epsilon=-\dfrac{d(BA)}{dt}

\epsilon=-A\dfrac{d(B)}{dt}

\epsilon=-A\dfrac{d(B_{max}e^{-t/\tau})}{dt}

\epsilon=A\dfrac{B_{max}e^{-t/\tau}}{\tau}

So, the induced emf in the loop as a function of time is A\dfrac{B_{max}e^{-t/\tau}}{\tau}. Hence, this is the required solution.

7 0
3 years ago
1. The term that describes where the supply curve intersects the demand curve is known as
nadezda [96]
Equilibrium is the answer
6 0
2 years ago
(1.08×10-3)(9.3×10-4)
ira [324]
First you do the first parenthesis, (1.08 x 10 - 3) and you do it in the order of operations! (parenthesis, exponents, multiplication/division, add/subtract) to get 7.8. Then you take the second parenthesis (9.3 x 10 - 4) and do the same thing to get 89! You then times 7.8 by 89 to get 694.2! If it needs more elaboration just ask ^.^
4 0
3 years ago
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