Answer:
il(t) = e^(-100t)
Explanation:
The current from the source when the switch is closed is the current through an equivalent load of 15 + 50║50 = 15+25 = 40 ohms. That is, it is 80/40 = 2 amperes. That current is split evenly between the two parallel 50-ohm resistors, so the initial inductor current is 2/2 = 1 ampere.
The time constant is L/R = 0.20/20 = 0.01 seconds. Then the decaying current is described by ...
il(t) = e^(-t/.01)
il(t) = e^(-100t) . . . amperes
Answer:
Engineers can design a train with a regenerative braking system
Explanation:
Assuming the point of the question is that the engineers want to focus on using energy efficiently when starting and stopping, they would likely want to consider a regenerative braking system. Such a system can store energy during braking so that it can be used during starting, reducing the amount of energy that must be supplied by an outside power source.
Answer:
The elevation at the high point of the road is 12186.5 in ft.
Explanation:
The automobile weight is 2500 lbf.
The automobile increases its gravitational potential energy in
. It means the mobile has increased its elevation.
The initial elevation is of 5183 ft.
The first step is to convert Btu of potential energy to adequate units to work with data previously presented.
British Thermal Unit -
Now we have the gravitational potential energy in lbf*ft. Weight of the mobile is in lbf and the elevation is in ft. We can evaluate the expression for gravitational potential energy as follows:
Where m is the mass of the automobile, g is the gravity, W is the weight of the automobile showed in the problem.
is the final elevation and
is the initial elevation.
Replacing W in the Ep equation
Finally, the next step is to replace the variables of the problem.
The elevation at the high point of the road is 12186.5 in ft.
Answer:
Explanation:
Using the kinematics equation
to determine the velocity of car B.
where;
initial velocity
= constant deceleration
Assuming the constant deceleration is = -12 ft/s^2
Also, the kinematic equation that relates to the distance with the time is:

Then:

The distance traveled by car B in the given time (t) is expressed as:

For car A, the needed time (t) to come to rest is:

Also, the distance traveled by car A in the given time (t) is expressed as:

Relating both velocities:





t = 2.25 s
At t = 2.25s, the required minimum distance can be estimated by equating both distances traveled by both cars
i.e.



d + 104.625 = 114.75
d = 114.75 - 104.625
d = 10.125 ft