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Ksju [112]
3 years ago
9

The voltage across a device and the current through it are:

Engineering
2 answers:
Strike441 [17]3 years ago
7 0

Answer: Charge, Q= 22.131 columbs

Power, P = 196.65 Watts

Explanation:

I(t) = 10( 1 - e(-0.5t)) A

I = dQ/dt

dQ = 10( 1 - e(-0.5t))dt

Integrating we have

Q = 10t - (20/(-0.5))e(-0.5t)

taking the value of t at 1s

we have,

Q = 10 + 20e(-0.5)

Q = 10 + 20/1.6487

Q = 10 + 12.131

Q = 22.131 Columbs

(b) Power

Power = voltage x current

Power = 5cos(2t) x 10(1 - e(-0.5t))

Power at t= 1, we substitute t=1

Power = 5cos(2) x 10(1 - e(-0.5))

Power = 5(0.9995) x 10(1 - 0.60654)

Power = 39.35 x 4.9975 = 196.65 Watts

Power = 196.65 Watts

butalik [34]3 years ago
3 0

Answer:

attached below

Explanation:

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kirill [66]

Answer:

The solution code is written in Java.

System.out.println(numItems);

Explanation:

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Charging method .Constant current method​
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Answer:

There are three common methods of charging a battery; constant voltage, constant current and a combination of constant voltage/constant current with or without a smart charging circuit.

Constant voltage allows the full current of the charger to flow into the battery until the power supply reaches its pre-set voltage.  The current will then taper down to a minimum value once that voltage level is reached.  The battery can be left connected to the charger until ready for use and will remain at that “float voltage”, trickle charging to compensate for normal battery self-discharge.

Constant current is a simple form of charging batteries, with the current level set at approximately 10% of the maximum battery rating.  Charge times are relatively long with the disadvantage that the battery may overheat if it is over-charged, leading to premature battery replacement.  This method is suitable for Ni-MH type of batteries.  The battery must be disconnected, or a timer function used once charged.

Constant voltage / constant current (CVCC) is a combination of the above two methods.  The charger limits the amount of current to a pre-set level until the battery reaches a pre-set voltage level.  The current then reduces as the battery becomes fully charged.  The lead acid battery uses the constant current constant voltage (CC/CV) charge method. A regulated current raises the terminal voltage until the upper charge voltage limit is reached, at which point the current drops due to saturation.

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How did studying pagodas help engineers create earthquake-proof structures in modern society?A. Engineers learned that the desig
Mila [183]

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3 years ago
Bind hole, 38 diameter, .50 deep
agasfer [191]

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4 0
2 years ago
Using a forked rod, a 0.5-kg smooth peg P is forced to move along the vertical slotted path r = (0.5 θ) m, whereθ is in radians.
-BARSIC- [3]

Answer:

N_c = 3.03 N

F = 1.81 N

Explanation:

Given:

- The attachment missing from the question is given:

- The given expressions for the radial and θ direction of motion:

                                       r = 0.5*θ

                                       θ = 0.5*t^2              ...... (correction for the question)

- Mass of peg m = 0.5 kg

Find:

a) Determine the magnitude of the force of the rod on the peg at the instant t = 2 s.

b) Determine the magnitude of the normal force of the slot on the peg.

Solution:

- Determine the expressions for radial kinematics:

                                        dr/dt = 0.5*dθ/dt

                                        d^2r/dt^2 = 0.5*d^2θ/dt^2

- Similarly the expressions for θ direction kinematics:

                                        dθ/dt = t

                                        d^2θ/dt^2 = 1

- Evaluate each at time t = 2 s.

                                        θ = 0.5*t^2 = 0.5*2^2 = 2 rad -----> 114.59°

                                        r = 1 m , dr / dt = 1 m/s , d^2 r / dt^2 = 0.5 m/s^2

- Evaluate the angle ψ between radial and horizontal direction:

                                        tan Ψ = r / (dr/dθ) = 1 / 0.5

                                        Ψ = 63.43°

- Develop a free body diagram (attached) and the compute the radial and θ acceleration:

                                        a_r = d^2r / dt^2 - r * dθ/dt

                                        a_r = 0.5 - 1*(2)^2 = -3.5 m/s^2

                                        a_θ =  r * (d^2θ/dt^2) + 2 * (dr/dt) * (dθ/dt)

                                        a_θ = 1(1) + 2*(1)*(2) = 5 m/s^2

- Using Newton's Second Law of motion to construct equations in both radial and θ directions as follows:

Radial direction:              N_c * cos(26.57) - W*cos(24.59) = m*a_r

θ direction:                      F  - N_c * sin(26.57) + W*sin(24.59) = m*a_θ

Where, F is the force on the peg by rod and N_c is the normal force on peg by the slot. W is the weight of the peg. Using radial equation:

                                       N_c * cos(26.57) - 4.905*cos(24.59) = 0.5*-3.5

                                       N_c = 3.03 N

                                       F  - 3.03 * sin(26.57) + 4.905*sin(24.59) = 0.5*5

                                       F = 1.81 N

4 0
3 years ago
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