The process of a liquid becoming a gas is call what

the current approximately 7.0 ms after closing the switch is about 250 mA, which is option (a).

To find the current through the circuit 7.0 ms after the switch is closed, we can use the concept of an RL circuit. The current in an RL circuit follows an exponential growth equation, given by:

I(t) = (V/R)(1 - e^(-t/τ))

Where:

I(t) is the current at time t.

V is the voltage from the power supply (15 V in this case).

R is the resistance (60 Ω).

τ (tau) is the time constant of the circuit, given by L/R, where L is the inductance (45 mH = 0.045 H).

First, calculate the time constant τ:

τ = L/R = 0.045 H / 60 Ω = 0.00075 s.

Now, plug in the values into the equation to find I(7.0 ms):

t = 7.0 ms = 0.007 s.

I(0.007 s) = (15 V / 60 Ω) * (1 - e^(-0.007 s / 0.00075 s))

I(0.007 s) = (0.25 A) * (1 - e^(-9.333...))

Now, calculate the current:

I(0.007 s) ≈ (0.25 A) * (1 - e^(-9.333...))

I(0.007 s) ≈ (0.25 A) * (1 - 0.0000962) [Using e^(-9.333...) ≈ 0.0000962]

I(0.007 s) ≈ (0.25 A) * (0.9999038)

I(0.007 s) ≈ 0.24998 A

I(0.007 s) ≈ 250 mA

So, the current approximately 7.0 ms after closing the switch is about 250 mA, which is option (a).

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Initial current = 0

Final current = (15 V) / (60 ohms) = 0.25 Ampere

Current along the way = 0.25 · (1 - e^- time / time-constant)

"time-constant" = L/R = (0.045 / 60) =  7.5 x 10⁻⁴ second

Current = 0.25 · (1 - e^-10,000t/7.5)

When t = 7 ms,

Current = 0.25 · ( 1 - e^-70/7.5)

Current = 0.25 · (1 - e^-9.33)

Current = 0.25 · (1 - 8.84 x 10⁻⁵)

Current = 0.25 · (0.9999)

Current = so close to 250 mA that you can't tell the difference.

The reason is that 7.0 mS is 9.3 time-constants, and during EVERY time-constant, the current grows by 37% of the distance it still has left to go. So after 9.3 of these, it's practically AT the target.

I have a feeling that the time in the question is SUPPOSED TO BE 7 microseconds.  If that's true, then

Current = 0.25 · (1 - e^-[ 7 x 10⁻⁶ / 7.5 x 10⁻⁴ ]

Current = 0.25 · (1 - e^-0.00933)

Current = 0.25 · (1 - 0.9907)

Current = 0.25 · (0.0093)

Current = 2.32 mA  ?

No, that can't be it either.

Well !  Now, I'm going to determine the true and correct final answer in the only cheap and sleazy way I have left ... by looking at all the choices offered, and eliminating the absurd ones.

The effect of an inductor in the circuit is to resist any change in current.  The final current in this circuit is when it's not trying to change any more.  So the final current is just the battery with a resistor across it ... (12 V) / (60 ohms).  That's 0.25 Ampere, or 250 mA.  The current starts at zero when the switch closes, and it builds up and builds up to 250 mA.  It's never more than 250 mA.  

So look at the choices !  The only one that not more than 250 mA is choice-A .

THAT has to be it.  7.0 mS is a no-brainer.  It's 9.3 time-constants after the switch closes, the current has built up to 99.99% of its final value by then, it's not really trying to change much any more, the inductor is just about finished having any effect on the current, and the current is essentially at its final value of 250 mA.  The action is all over.

Now, I fully realize that Mister "Rishwait" is a bot and all, and nobody really needs the answer to this question.  But every cloud has a silver lining.  It's a numskull question, but it earned me 10 points, and it's been a truly fascinating trip down Memory Lane.

Answer:

-20.158ft-lb

Explanation:

Check the attached files for the explanation.

Alicia ran 15 miles at the faster pace.

Formulating the equations:

Let Alicia run with the speed of 12mph for A hours and with the speed of 8mph for B hours.

Total distance is 21 miles, so we can make the first equation as:

12A + 8B = 21      .............(i)

Now the total time taken is 2 hours, which means:

A + B = 2             ..............(ii)

Now multiply equation (ii) by 8 and subtract from equation (i)

then we get,   4A = 5

A = 5/4 hours

So Alicia run at the speed of 12mph for 5/4 hours, total distance covered is:

d = 12 × 5/4

d = 15 miles

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Answer:

15 miles

Explanation:

Alicia's speed in the first part of the race is:

where is the length of the first part and the time taken to cover this part.

Alicia's speed in the second part of the race is:

where is the length of the second part and the time taken to cover this part.

The total length of the race is

And the total time taken by Alicia to complete the race is

We have in total 4 equations with 4 unknown quantities. From the first 2 equations we find:

And substituting into the second group of equations:

Calling A the first equation and B the second equation, if we compute (A-8B) we get:

And so

So Alicia runs for 1.25 hours at 12 mph (second part of the race). Therefore, the distance travelled during this part is:

Therefore the answer is 15 miles.

Answer:

a. center of mass acceleration supposed to be acceleration due to gravity, 9.81 m/s^2,

b. T1 = 9.81m1 N; c. T3 =9.81(M1+M2) N; d. T3-T1, e. (T3-T1)/M2; f. (M1+M2)T3/M3

Explanation:

Final answer:

In this frictionless, massless pulley system, the center of mass accelerates downward with an acceleration equal to the acceleration due to gravity. The tension in the string connected to mass M1 is equal to M1g, and the tension in the string connected to mass M2 is equal to m1g + M2g. The magnitudes of the accelerations of M1 and M2 are equal, and the sum of the tensions T1 and T2 is equal to the tension T3.

Explanation:

a. The center of mass accelerates: When considering the system as a whole, the acceleration of the center of mass is determined by the net external force acting on the system. In this case, the only external force is the force due to gravity. Therefore, the center of mass accelerates downward with an acceleration equal to g, the acceleration due to gravity.

b. T1 is equal to M1g: The tension in the string connected to mass M1 is equal to the weight of M1, which is given by the formula T1 = M1g.

c. T3 is equal to m1g + M2g: The tension in the string connected to mass M2 is equal to the sum of the weights of M1 and M2, which is given by the formula T3 = m1g + M2g.

d. T1 is equal to T2: Since the pulley is assumed to be frictionless and massless, the tension in the string connected to mass M1 is the same as the tension in the string connected to mass M2.

e. The magnitude of the acceleration of M2 is equal to the magnitude of the acceleration on M1: This is due to the constraint imposed by the tension in the string. Since the tension in the string connecting M1 and M2 is the same, their accelerations must also be the same.

f. T1 + T2 is equal to T3: The sum of the tensions T1 and T2 is equal to the tension T3, as the total force acting on mass M2 is equal to the sum of the individual tensions.

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