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2 years ago

5

During a 72-ms interval, a change in the current in a primary coil occurs. This change leads to the appearance of a 6.0-mA curre

nt in a nearby secondary coil. The secondary coil is part of a circuit in which the resistance is 12 Ω. The mutual inductance between the two coils is 3.2 mH. What is the change in the primary current?

1 answer:

Gwar [14]2 years ago

7 0

Answer:

so change in primary current is 1.620 A

Explanation:

Given data

current I =6.0-mA = 6 × 10^(-3) A

resistance R = 12 Ω

mutual inductance M = 3.2 mH = 3.2 × 10^(-3) H

dt = 72-ms = 72 × 10^-(3) s

to find out

change in the primary current

solution

we know that

Electric and magnetic fields in secondary coil = mutual inductance × change in primary current / dt ............1

and we know also that Electric and magnetic fields in secondary coil = resistance × current

so = 6 × 10^(-3) × 12 = 72 × 10^(-3) volts

so that we say

change in primary current from equation 1

change in primary current = Electric and magnetic fields in secondary coil × dt / mutual inductance

change in primary current = 72 × 10^(-3) × 72 × 10^(-3) / 3.2 × 10^(-3)

change in primary current = 1.620

so change in primary current is 1.620 A

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A hot air balloon of total mass M (including passengers and luggage) is moving with a downward acceleration of magnitude a. As i

LUCKY_DIMON [66]

Answer:

The fraction of mass that was thrown out is calculated by the following Formula:

M - m = (3a/2)/(g²- (a²/2) - (ag/2))

Explanation:

We know that Force on a moving object is equal to the product of its mass and acceleration given as:

F = ma

And there is gravitational force always acting on an object in the downward direction which is equal to g = 9.8 ms⁻²

Here as a convention we will use positive sign with acceleration to represent downward acceleration and negative sign with acceleration represent upward acceleration.

Case 1:

Hot balloon of mass = M

acceleration = a

Upward force due to hot air = F = constant

Gravitational force downwards = Mg

Net force on balloon is given as:

Ma = Gravitational force - Upward Force

Ma = Mg - F (balloon is moving downwards so Mg > F)

F = Mg - Ma

F = M (g-a)

M = F/(g-a)

Case 2:

After the ballast has thrown out,the new mass is m. The new acceleration is -a/2 in the upward direction:

Net Force is given as:

-m(a/2) = mg - F (Balloon is moving upwards so F > mg)

F = mg + m(a/2)

F = m(g + (a/2))

m = F/(g + (a/2))

Calculating the fraction of the initial mass dropped:

$M-m = \frac{F}{g-a} - \frac{F}{g+\frac{a}{2} }\\M-m = F*[\frac{1}{g-a} - \frac{1}{g+\frac{a}{2} }]\\M-m = F*[\frac{(g+(a/2)) - (g-a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{g+(a/2) - g + a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{(3a/2)}{g^{2}-\frac{a^{2}}{2}-\frac{ag}{2}} ]$

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2 years ago

In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at 64,000 km. P

tatiyna

Answer:

Time period for first satellites 24.46 days and for second satellites 37.67 days

Explanation:

Given :

Distance of first satellites $r_{sat1} = 48000 \times 10^{3}$ m

Distance of second satellites $r _{sat2} = 64000 \times 10^{3}$ m

Distance of charon $r_{c} = 19600 \times 10^{3}$ m

Time period of charon $T_{c} = 6.39$ days

From the kepler's third law,

Square of the time period is proportional to the cube of the semi major axis.

$T^{2} = r^{3}$

$\frac{T}{r^{\frac{3}{2} } } = constant$

For first satellites,

$\frac{T_{c} }{r_{c} ^{\frac{3}{2} } } = \frac{T_{sat1} }{r_{sat1} ^{\frac{3}{2} } }$

${T_{sat1} } = 6.39 \times \frac{(48000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}$

$T_{sat1} = 24.46$ days

For second satellites,

$\frac{T_{c} }{r_{c} ^{\frac{3}{2} } } = \frac{T_{sat2} }{r_{sat2} ^{\frac{3}{2} } }$

${T_{sat2} } = 6.39 \times \frac{(64000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}$

$T_{sat2} = 37.67$ days

Therefore, time period for first satellites = 24.46 days and for second satellites 37.67 days

8 0

2 years ago

If you are lying down and stand up quickly, you can get dizzy or feel faint. This is because the blood vessels don't have time t

sammy [17]

Complete Question

If you are lying down and stand up quickly, you can get dizzy or feel faint. This is because the blood vessels don’t have time to expand to compensate for the blood pressure drop. If your brain is 0.4 m higher than your heart when you are standing, how much lower is your blood pressure at your brain than it is at your heart? The density of blood plasma is about 1025 kg/m3 and a typical maximum (systolic) pressure of the blood at the heart is 120 mm of Hg (= 0.16 atm = 16 kP = 1.6 × 104 N/m2).

Answer:

The pressure at the brain is $P_b = 89.872 \ mm \ of \ Hg$

Explanation:

Generally is mathematically denoted as

$P = \rho gh$

Substituting $1025 kg/m^3$ for $\rho$ (the density) , $9.8 m/s^2$ for g (acceleration due to gravity) , 0.4m for h (the height )

We have that the pressure difference between the heart and the brain is

$P = 1025 * 9.8 *0.4$

$= 4018 N/m^2$

But the pressure of blood at the heart is given as

$P_h=120 mm of Hg =$ $120 * 133 = 1.59*10^3Pa$

Now the pressure at the brain is mathematically evaluated as

$P_b = P_h - P$

$= 1.596*10^4 - 4018$

$= 11982 N/m^2$

$P_b= \frac{11982}{133} = 89.872 \ mm \ of \ Hg$

3 0

2 years ago

An electron moving parallel to a uniform electric field increases its speed from 2.0 × 107 m/s to 4.0 × 107 m/s over a distance

jeka94

Answer:

$1.8\times 105 N/C$

Explanation:

We are given that

$u=2\times 10^7 m/s$

$v=4\times 10^7 m/s$

d=1.9 cm= $\frac{1.9}{100}=0.019 m$

Using 1m=100 cm

We have to find the electric field strength.

$v^2-u^2=2as$

Using the formula

$(4\times 10^7)^2-(2\times 10^7)^2=2a(0.019)$

$16\times 10^{14}-4\times 10^{14}=0.038a$

$0.038a=12\times 10^{14}$

$a=\frac{12}{0.038}\times 10^{14}=3.16\times 10^{16}m/s^2$

$q=1.6\times 10^{-19} C$

Mass of electron,m $=9.1\times 10^{-31} kg$

$E=\frac{ma}{q}$

Substitute the values

$E=\frac{9.1\times 10^{-31}\times 3.16\times 10^{16}}{1.6\times 10^{-19}}$

$E=1.8\times 105 N/C$

7 0

2 years ago

A 600g toy train completes 10 laps of its circular track in 1 min 20s. If the radius of the track is 1.2 m, Find the centripetal

Lynna [10]

Wow ! This will take more than one step, and we'll need to be careful
not to trip over our shoe laces while we're stepping through the problem.

The centripetal acceleration of any object moving in a circle is

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Notice that we won't need to use the mass of the train.

We know the radius of the track. We don't know the trains speed yet,
but we do have enough information to figure it out. That's what we
need to do first.

Speed = (distance traveled) / (time to travel the distance).

Distance = 10 laps of the track.   Well how far is that ? ? ?

1 lap = circumference of the track = (2π) x (radius) = 2.4π meters

10 laps = 24π meters.

Time = 1 minute 20 seconds = 80 seconds

The trains speed is (distance) / (time)

   = (24π meters) / (80 seconds)

   = 0.3 π meters/second .

NOW ... finally, we're ready to find the centripetal acceleration.

<span> (speed)² / (radius)

   = (0.3π m/s)² / (1.2 meters)

   = (0.09π m²/s²) / (1.2 meters)

   =    (0.09π / 1.2) m/s²

   =    0.236 m/s² . (rounded)

If there's another part of the problem that wants you to find
the centripetal FORCE ...

Well, Force = (mass) · (acceleration) .

We know the mass, and we ( I ) just figured out the acceleration,
so you'll have no trouble calculating the centripetal force. </span>

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