For a motor purchased for a tank project, it claims that at 12V it produces 6000 RPM. This motor is attached to a 50:1 gear reduction gearbox. The output of this gearbox is attached to a 2.5" wheel. Determine the speed of the tank under these conditions in miles/hour (mph).

Answer:

ΔU = 53 BTU

Explanation:

In thermodynamics we consider heat entering a system as positive and heat leaving as negative.

Also, work performed by the system is positive and work applied to the system is negative.

The total heat applied to the system is:

Q = 72 - 7 = 65 BTU

The work is:

L = 12 BTU

The first law of thermodynamics states that:

Q = L + ΔU

Where ΔU is the change in internal energy.

Rearranging:

ΔU = Q - L

ΔU = 65 - 12 = 53 BTU

Answer:

The laminar flow is generally given in high viscosity fluids such as honey or oil, it has the characteristic of flowing in an orderly manner, the walls of the tube have a zero speed while in the center it has a maximum speed.

turbulent flow is characterized by fluid velocity vectors presenting themselves in a disorderly manner and in all directions.

I attached the drawings for the velocity profile in laminar and turbulent flow.

Answer:

W=-940.36 KJ

Explanation:

Given that

[tex]P_1=1\ bar,V_1=4.25 {m^3}[/tex]

[tex]P_2=6.8\ bar[/tex]

Process follows pv=constant

So this is the isothermal process and work in isothermal process given as

[tex]W=P_1V_1\ln \dfrac{P_1}{P_2}[/tex]

Now by putting the values                (1.4 bar =140 KPa)

[tex]W=P_1V_1\ln \dfrac{P_1}{P_2}[/tex]

[tex]W=140\times 4.25 \ln \dfrac{1.4}{6.8}[/tex]

W=-940.36 KJ

Negative sign indicates that this is a compression process and work will given to the system.

The question requires calculating expressions using metric units and prefixes, converting them into SI units with appropriate prefixes, and ensuring these calculations are accurate to three significant figures.

The question involves evaluating expressions with different units and converting them to SI units with appropriate prefixes, which requires a good understanding of metric prefixes and significant figures. Furthermore, examples of such conversions are provided to aid in this process.

Each conversion is an exercise in carefully applying metric prefixes and understanding how to manipulate them within calculations.

Answer:

Q=36444.11 Btu

Explanation:

Given that

Initial temperature = 60° F

Final temperature = 110° F

Specific heat of water = 0.999 Btu/lbm.R

Volume of water = 90 gallon

Mass = Volume x density

[tex]1\ gallon = 0.13ft^3[/tex]

Mass ,m= 90 x 0.13 x 62.36 lbm

m=729.62 lbm

We know that sensible heat given as

Q= m Cp ΔT

Now by putting the values

Q= 729.62 x 0.999 x (110-60) Btu

Q=36444.11 Btu

Answer:

7.05 Hz

Explanation:

The natural frequency of a mass-spring system is:

[tex]f = \frac{1}{2 \pi}\sqrt{\frac{k}{m}}[/tex]

To determine the constant k of the spring we use Hooke's law:

Δl  = F / k

k = F / Δl

In the first case the force was the weight of the 20 kg mass and the Δl was 20 mm.

F = m * a

F = 2 * 9.81 = 19.6 N

Then:

k = 19.6 / 0.02 = 980 N/m

Therefore:[tex]f = \frac{1}{2 \pi}\sqrt{\frac{980}{0.5}} = 7.05 Hz[/tex]

Answer:

b) zero

Explanation:

The horizontal component of acceleration during projectile motion is usually assumed to be zero.Because in projectile motion horizontal component of velocity will remain the constant and we know that rate of change of velocity with time is called acceleration.So when velocity is constant then acceleration will be zero.

In projectile motion ,gravitational acceleration will be in only vertical direction.

Explanation:

Superheater has two types of parts which are:

Primary super-heater is first heater which is passed by the steam after steam comes out of steam drum.

After steam is heated on super primary heater, then the steam is passed on secondary super-heater so to be heated again. Thus, on secondary super-heater, the steam formed is hottest steam among others.

Steam from secondary super-heater which becomes the superheated steam, flow to rotate the High-Pressure Turbine.

Answer:

Chord ratio:

  It is the ratio of thickness to the chord.It is also knows as thickness ratio.

Chord ratio measure the performance of wing when it is operating at the transonic speed.

When the speed cross the speed of sound wave then that wave creates the shock wave and these shock leads to produce drag force on the aerofoil profile.When Mach number is one then it means that if Mach increase then it will leads increase then drag force.

Answer:

Explanation:

Solar energy is one of the Renewable sources of energy and it is sufficient to meet the entire needs of the world.

But it is not used extensively because of the inconsistency. Inconsistency means that sun rays are not consistent throughout the year. If the weather is not good or in winters earth does not receive sun rays continuously.

And another reason to not to set up solar energy is that it is very expensive not everyone can afford it.  

The modulus of elasticity for the aluminum bar is calculated using the relationship between stress and strain, given by Young's modulus formula. The given values are substituted in the formula to find that the modulus of elasticity for the aluminum is approximately 60 GPa.

Calculating the Modulus of Elasticity for Aluminum

To calculate the modulus of elasticity (also known as Young's modulus) for the aluminum bar, we can use the relationship described by Hooke's Law for elastic deformation, where stress is directly proportional to strain. The formula to find Young's modulus (Y) is:

Y = (F \/ A) \/ (\u0394L \/ L0)

Here, F is the force applied, A is the cross-sectional area, \u0394L is the change in length, and L0 is the original length of the bar.

Using the given values:

The calculation is as follows:

Y = (66,700 N / 272.25 x 10-6 m2) / (0.43 x 10-3 m / 0.125 m)

When the calculations are done, we find that the modulus of elasticity for the aluminum bar is approximately 60 GPa (Gigapascals), which is within the typical range for aluminum.

Calculate the modulus of elasticity of aluminum given specific values for force, length, cross-sectional area, and elongation during tension.

Young's modulus is a measure of a material's stiffness and is calculated using the formula: Y = (F * Lo) / (A * AL). Substituting the given values, we can find the modulus of elasticity of aluminum to be approximately 68 GPa.

Answer:

The Danner process is used to make glass tubes

Explanation:

In this process the molten glass passes from the feeder to a hollow ceramic cylinder that is tilted downward, so that the molten glass can "run"; This ceramic cylinder is spinning constantly.

As the molten glass passes through the Danner tube, compressed air is blown to prevent the glass from being damaged. On the opposite side of the feeder, i.e, on the other side of the tube a bubble is also formed so-called drawing onion, the glass tube is removed in a free buckling in a traction line from the onion.

When the stretching speed remains constant and an increase in the blowing pressure occurs, larger tube diameters are created and the wall thickness decreases.

This process allows tube diameters between 2 and 60 mm.

This is how glass bottles are made.

Answer:

a. The heat flux through the sheet of insulation is 19.92 W/m^2

b. The rate of heat transfer through the sheet of insulation is 125.28 W

c. The thermal resistance of the sheet due to the conduction is 0.86 Km^2/W.

Explanation:

From the heat conduction Fourier's law it can be state for a wall of width e and area A :

q = Q/ΔT  = k*A* (T2-T1)/e

Where q is the rate of heat transfer, k the conductivity constant, and T2 and T1 the temperatures on the sides of the wall. Replacing the values in the correct units, we obtained the rate of heat transfer:

q =  0.029 W/*mK * (3m*3m) * (12°K) / (0.025m)

(The difference in temperatures in Kelvin is the same than in Celcius degres).

q =  0.029 W/*mK * (9 m^2) * (12°K) / (0.025m)  = 125.28 W

The heat flux is calculated by dividing q by the area of the wall:

q/A = k* (T2-T1)/e = 0.029 W/*mK  * (12°K) / (0.025m) = 19.92 W/m^2

The thermal resistance of the sheet is defined as:

R = e / k

Replacing the values in the proper units:

R = 0.025 m /  0.029 W/*mK = 0.86 Km^2/W

Answer:

shunt resistor is 432µΩ

correct option is  432µΩ

Explanation:

given data

resistance R = 24.0 Ω

full scale deflection current Ig = 180 μA

galvanometer current I = 10.0 A

to find out

what shunt resistor is required

solution

we will apply here full scale deflection current  formula that is

Ig = I × [tex]\frac{r}{r + R}[/tex]    ...............1

here r is shunt current and R is resistance and I is galvanometer

put here all value

180 × [tex]10^{-6}[/tex] = 10 × [tex]\frac{r}{r + 24}[/tex]

r = 18 ×  [tex]10^{-6}[/tex]  × 24

r = 432 ×  [tex]10^{-6}[/tex] Ω

so shunt resistor is 432µΩ

correct option is  432µΩ

Shunt resistor required for 10.0 A ammeter: approximately 432 μΩ.

To construct an ammeter using a galvanometer, a shunt resistor is connected in parallel to the galvanometer. The shunt resistor diverts most of the current, allowing only a fraction of the current to pass through the galvanometer, thus enabling it to measure higher currents.

The relationship between the current passing through the galvanometer and the shunt resistor can be expressed using the formula:

[tex]\[I_{\text{total}} = I_{\text{g}} + I_{\text{s}}\][/tex]

Where:

[tex]- \(I_{\text{total}}\)[/tex] is the total current (10.0 A)

[tex]- \(I_{\text{g}}\)[/tex] is the current passing through the galvanometer (180 μA)

[tex]- \(I_{\text{s}}\)[/tex]  is the current passing through the shunt resistor

Given that the galvanometer resistance [tex]\(R_{\text{g}} = 24.0 Ω\)[/tex]  and full-scale deflection current \[tex](I_{\text{g}} = 180 μA\),[/tex] we can calculate the shunt resistance required using Ohm's Law:

[tex]\[I_{\text{g}} = \frac{V_{\text{g}}}{R_{\text{g}}}\][/tex]

Where:

- [tex]\(V_{\text{g}}\)[/tex] is the voltage across the galvanometer

Since the galvanometer is designed to deflect full-scale at 180 μA, the voltage across it is:

[tex]\[V_{\text{g}} = I_{\text{g}} \times R_{\text{g}}\]\[V_{\text{g}} = (180 \times 10^{-6}) \times 24\]\[V_{\text{g}} = 0.00432 \, \text{V}\][/tex]

The current passing through the shunt resistor can be calculated using:

[tex]\[I_{\text{s}} = \frac{V_{\text{total}}}{R_{\text{s}}}\][/tex]

Since the total current is 10.0 A and the voltage across the shunt resistor is the same as the voltage across the galvanometer (as they are in parallel), we have:

[tex]\[I_{\text{s}} = \frac{10.0}{R_{\text{s}}}\][/tex]

Now, the total current equals the sum of the currents through the galvanometer and shunt resistor:

[tex]\[10.0 = 180 \times 10^{-6} + \frac{0.00432}{R_{\text{s}}}\][/tex]

Solving for [tex]\(R_{\text{s}}\):[/tex]

[tex]\[R_{\text{s}} = \frac{0.00432}{10.0 - 180 \times 10^{-6}}\]\[R_{\text{s}} ≈ \frac{0.00432}{10.0}\]\[R_{\text{s}} ≈ 432 \, \text{μΩ}\][/tex]

So, the required shunt resistor is approximately 432 μΩ. Therefore, the closest answer is 432 μΩ.

Answer:

natural frequency = 2.55 Hz

period of vibration = 0.3915 s

Explanation:

given data

weight = 20 lb

distance = 1 in = [tex]\frac{1}{12}[/tex] ft

weight = 30 lb

to find out

Determine the natural frequency and the period of vibration

solution

we first calculate here stiffness k by given formula that is

k = [tex]\frac{weight}{diatnace}[/tex]  ..........1

k = [tex]\frac{20}{1/12}[/tex]

k = 240 lb/ft

so

frequency = [tex]\sqrt{\frac{k}{m} }[/tex]   ..................2

put here value k and mass m = [tex]\frac{weight}{g}[/tex]

frequency = [tex]\sqrt{\frac{240}{30/32.2} }[/tex]  

frequency = 16.05 rad/s

and

period of vibration = [tex]\frac{2* \pi }{frequency}[/tex]

period of vibration = [tex]\frac{2* \pi }{16.05}[/tex]

period of vibration = 0.3915 s

and

natural frequency = [tex]\frac{1 }{period of vibration}[/tex]

natural frequency = [tex]\frac{1 }{0.3915}[/tex]

natural frequency = 2.55 Hz

Answer:

water=14.1 kg

air=0.1348kg

Explanation:

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties such as pressure and temperature.

To solve this exercise we must find the specific volume of water and air in the two states and then subtract them and multiply them by the volume of the tank to find the change in mass, the following equation is inferred

Δm=V(ρ2-ρ1)

Where V= tank volume=500L=0.5m^3

ρ2= density of fluid in state 2

ρ1=density of fluid in state 1

Δm=change of mass

for water

ρ1=971.8kg/m^3(80C)

ρ2=1000kg/m^3(5C)

Δm=0.5(1000-971.8)

Δm=14.1 kg

for air

ρ1=0.9994kg/m^3(80C)

ρ2=1.269kg/m^3(5C)

Δm=0.5(1.269-0.9994)

Δm=0.1348kg

Answer:

TRUE

Explanation:

Polymers can be natural as well as synthetic

The polymer which are found in nature are called natural polymer tease polymer are not synthesized, they are found in nature

Example of natural polymers is cellulose, proteins etc

On the other hand synthetic polymers are not found in nature they are synthesized in market

There are many example of synthetic polymer

Example : nylon, Teflon etc  

So it is a true statement

Answer:

Zero.

Explanation:

Lets take velocity V is given as in 2 dimensional flow

  V=u i +v j

V=f(x,y)

u=f(x,y)

v=f(x,y)

Rotational ability ω given as

[tex]\omega =\dfrac{1}{2}\left (\dfrac{\partial v}{\partial x}-\dfrac{\partial u} {\partial x}\right)[/tex]

If

ω=0 ,then flow will be  irrotational.

ω ≠0 ,then flow will be rotational.

So we can say that , ω will be zero every where for  irrotational  two dimensional flow .

Answer:

The coefficient of thermal expansion tells us how much a material can expand due to heat.

Explanation:

Thermal expansion occurs when a material is subjected to heat and changes it's shape, area and volume as a result of that heat. How much that material changes is dependent on it's coefficient of thermal expansion.

Different materials have different coefficients of thermal expansion (i.e. It is a material property and differs from one material to the next). It is important to understand how materials behave when heated, especially for engineering applications when a change in dimension might pose a problem or risk (eg. building large structures).

Answer: 78.89%

Explanation:

Given : Sample size : n=  1200

Sample mean : [tex]\overline{x}=2.45 [/tex]

Standard deviation : [tex]\sigma=0.07[/tex]

We assume that it follows Gaussian distribution (Normal distribution).

Let x be a random variable that represents the shaft diameter.

Using formula, [tex]z=\dfrac{x-\mu}{\sigma}[/tex], the z-value corresponds to 2.39 will be :-

[tex]z=\dfrac{2.39-2.45}{0.07}\approx-0.86[/tex]

z-value corresponds to 2.60 will be :-

[tex]z=\dfrac{2.60-2.45}{0.07}\approx2.14[/tex]

Using the standard normal table for z, we have

P-value = [tex]P(-0.86<z<2.14)=P(z<2.14)-P(z<-0.86)[/tex]

[tex]=P(z<2.14)-(1-P(z<0.86))=P(z<2.14)-1+P(z<0.86)\\\\=0.9838226-1+0.8051054\\\\=0.788928\approx0.7889=78.89\%[/tex]

Hence, the percentage of the diameter of the total shipment of shafts will fall between 2.39 inch and 2.60 inch = 78.89%

The Reynolds number is approximately 14,067, indicating turbulent flow of MEK through the pipe, as Re > 2100 denotes turbulence.

let's calculate the Reynolds number for the flow of liquid methyl ethyl ketone (MEK) through the pipe.

Given:

- Pipe diameter (D) = 2.067 inches = 0.0512 meters (converted from inches to meters)

- Fluid velocity (u) = 0.48 ft/s = 0.1463 meters/s (converted from feet per second to meters per second)

- Fluid density (ρ) = 0.805 g/cm³ = 805 kg/m³ (converted from grams per cubic centimeter to kilograms per cubic meter)

- Fluid viscosity (μ) = 0.43 centipoise = 0.43 x 10²  kg/(m*s)(converted from centipoise to kg/(m*s))

The Reynolds number (Re) is calculated as:

[tex]\[ Re = \frac{(D \cdot u \cdot \rho)}{\mu} \][/tex]

Plugging in the values:

[tex]\[ Re = \frac{(0.0512 \, \text{m} \cdot 0.1463 \, \text{m/s} \cdot 805 \, \text{kg/m³})}{0.43 \times 10^{-3} \, \text{kg/(m*s)}} \][/tex]

[tex]\[\text{Re} = \frac{(0.0074991 \, \text{m}^2/\text{s}^2 \cdot 805 \, \text{kg/m}^3)}{0.43 \times 10^{-3} \, \text{kg/(m*s)}}\]\[\text{Re} = \frac{6.0498755 \, \text{kg/(m*s)}}{0.43 \times 10^{-3} \, \text{kg/(m*s)}}\]\[\text{Re} \approx \frac{6.0498755}{0.00043}\]\[\text{Re} \approx 14,067.35\][/tex]

Since the calculated Reynolds number is greater than 2100 (Re > 2100), the flow of MEK through the pipe is turbulent.

Answer:

Piston movement of TDC to BDC is called stroke.

Explanation:

Step1

In any engine fuel is burned inside the cylinder and forces the piston to move from top dead center to bottom dead center. First piston is moved from top dead center to bottom dead center to allow the suction of fuel inside the cylinder. Then piston move toward top dead center and compresses the fuel. This process is called compression of fuel. Then fuel is burned with the spark inside the cylinder and the piston moves toward bottom dead center. This process is called expansion process. Last process is the expansion process that allows the exhaust out of the cylinder.

Step2

The piston reciprocates from bottom position of the cylinder to top position of the cylinder and vice versa. So, the motion of piston from top center to bottom center is called stroke. One movement of piston from TDC to BDC is called one stroke. There are total four strokes in petrol engine and diesel engine. So, two strokes are equal to one cycle or one rotation of crank and four strokes are equal to two rotation of crank.

Thus, piston movement of TDC to BDC is called stroke.

Answer:

The return stroke

Explanation:

Motion from top dead center to bottom dead center is called the return stroke because motion from the bottom to the top is the forward stroke.

Answer:

The volume up to cylindrical portion is approx  32355 liters.

Explanation:

The tank is shown in the attached figure below

The volume of the whole tank is is sum of the following volumes

1) Hemisphere top

Volume of hemispherical top of radius 'r' is

[tex]V_{hem}=\frac{2}{3}\pi r^3[/tex]

2) Cylindrical Middle section

Volume of cylindrical middle portion of radius 'r' and height 'h'

[tex]V_{cyl}=\pi r^2\cdot h[/tex]

3) Conical bottom

Volume of conical bottom of radius'r' and angle [tex]\theta [/tex] is

[tex]V_{cone}=\frac{1}{3}\pi r^3\times \frac{1}{tan(\frac{\theta }{2})}[/tex]

Applying the given values we obtain the volume of the container up to cylinder is

[tex]V=\pi 1.5^2\times 4.0+\frac{1}{3}\times \frac{\pi 1.5^{3}}{tan30}=32.355m^{3}[/tex]

Hence the capacity in liters is [tex]V=32.355\times 1000=32355Liters[/tex]

Answer:

W  = 12.8 KW

Explanation:

given data:

mass flow rate = 0.2 kg/s

Engine recieve heat from ground water at 95 degree ( 368 K)  and reject that heat to atmosphere  at 20 degree (293K)

we know that maximum possible efficiency is given as

[tex]\eta = 1- \frac{T_L}{T_H}[/tex]

[tex]\eta = 1 - \frac{ 293}{368}[/tex]

[tex]\eta = 0.2038[/tex]

rate of heat transfer is given as

[tex]Q_H = \dot m C_p \Delta T[/tex]

[tex]Q_H = 0.2 * 4.18 8(95 - 20)[/tex]

[tex]Q_H = 62.7 kW[/tex]

Maximuim power is given as

[tex]W = \eta Q_H[/tex]

W = 0.2038 * 62.7

W  = 12.8 KW

The angular speed of a flywheel rotating at 300 revolutions per minute is calculated by converting revolutions to radians and minutes to seconds, resulting in an angular speed of 10π rad/s.

The question asks to determine the angular rate of rotation of a searchlight on a boat when targeting an automobile that is 3000 ft away and moving at 80 ft/s. This involves understanding the relationship between linear velocity, radius, and angular velocity, which is a crucial concept in trigonometry and physics.

To find the angular speed of the flywheel rotating at 300 revolutions per minute, we first convert revolutions per minute (rpm) to radians per second, knowing that one revolution is 2π radians and there are 60 seconds in a minute. Therefore:

Angular speed = 300 rpm × (2π radians/revolution) × (1 minute/60 seconds) = 300 × 2π / 60 rad/s = 10π rad/s.

Answer:

C.Cooling of thermal energy

Explanation:

In evaporative cooling liquid water convert into vapor by using thermal energy.In this out side air is used to evaporate the water.

Hot air enters from outside and this hot air gets contact with wet matrix .In wet matrix temperature of air will reduce and temperature of water will increases .After wet matrix heat transfer take place between air and water .After that by using force convection air pushed out and produce cooling.

Answer:

a)[tex]v=25.56\ kg/m^3[/tex]

b) [tex]v=38.8\ kg/m^3[/tex]

Explanation:

Given that

Pressure P = 15 MPa

Temperature = 1000 C = 1273 K

a)If assume as ideal gas

The gas constant for super heated steam R is 0.461 KJ/kg.K.

We know that ideal gas equation

P v  =RT

15 x 1000 x v =0.461 x 1273

[tex]v=25.56\ kg/m^3[/tex]

b) By using steam table

From steam table we can see that volume of super heated vapot at 15 MPa and 1273 K .

[tex]v=38.8\ kg/m^3[/tex]

Answer:

401.3 kg/s

Explanation:

The power plant has an efficiency of 36%. This means 64% of the heat form the source (q1) will become waste heat. Of the waste heat, 85% will be taken away by water (qw).

qw = 0.85 * q2

q2 = 0.64 * q1

p = 0.36 * q1

q1 = p /0.36

q2 = 0.64/0.36 * p

qw = 0.85 *0.64/0.36 * p

qw = 0.85 *0.64/0.36 * 600 = 907 MW

In evaporation water becomes vapor absorbing heat without going to the boiling point (similar to how sweating takes heat from the human body)

The latent heat for the vaporization of water is:

SLH = 2.26 MJ/kg

So, to dissipate 907 MW

G = qw * SLH = 907 / 2.26 = 401.3 kg/s

Answer:

m = 367.753 kg.s

Explanation:

GIVEN DATA:

Power plant capacity W=600 MW

Plant efficiency is [tex]\eta = 36\%[/tex]

[tex]efficiency = \frac{W}{QH}[/tex]

[tex]0.36 = \frac{600}{QH}[/tex]

QH = 1666.6 MW

from first law of thermodynamics we hvae

QH -QR = W

Amount of heat rejection is QR = 1066.66 MW

As per given information we have 15% heat released to atmosphere

[tex]QR = 0.15 \tiimes 1066.66 = 159.99 MW[/tex]

AND 85% to cooling water

[tex]Q = 0.85 \times 1066.66 = 906.66 MW[/tex]

from saturated water table

at temp 150 degree c we have Hfg = 2465.4 kJ/kg

rate of cooiling water is given as  =  mhfg

[tex]906.66 \times 1000  KW = m \times 2465.4[/tex]

m = 367.753 kg.s

where m is rate of makeup water that is added to offset

Answer:

W=-280.67 KJ

Explanation:

Given that

Initial pressure = 150 KPa

Final pressure = 600 KPa

Temperature T= 285 K

Mass m=2.4 Kg

We know that ,work in isothermal process given as

[tex]W=mRT\ ln\dfrac{P_1}{P_2}[/tex]

Gas constant for nitrogen gas R=0.296 KJ/kgK

Now by putting the values

[tex]W=mRT\ ln\dfrac{P_1}{P_2}[/tex]

[tex]W=2.4\times 0.296\times 285\ ln\dfrac{150}{600}[/tex]

W=-280.67 KJ

Negative sign indicates that it is compression process and work is done on the gas.

Answer:

a) 926 μJ

b) 3.802 mC

c) 8.61 A

d) 0.0721

e) 3.2137

Explanation:

The energy in the inductor is

[tex]El = \frac{1}{2}*L*I^2[/tex]

[tex]El = \frac{1}{2}*25*10^{-3}*(9.2*10^{-3})^2 = 1.06*10^{-6} J = 1.06 \mu J[/tex]

The energy store in a capacitor is

[tex]Ec = \frac{1}{2}*C*V^2[/tex]

The voltage in a capacitor is

V = Q/C

[tex]V = \frac{3.8*10^{-3}}{7.8*10^{-3}} = 0.487 V[/tex]

Therefore:

[tex]Ec = \frac{1}{2}*7.8*10^{-3}*0.487^2 = 9.256*10^{-4} J = 925.6 \mu J[/tex]

The total energy is Et = 925.6 + 1.1 = 926.7 μJ

At a certain point all the energy of the circuit will be in the capacitor, at this point it will have maximum charge

[tex]Ec = \frac{1}{2}*C*V^2[/tex]

V = Q/C

[tex]Ec = \frac{1}{2}*C*(\frac{Q}{C})^2[/tex]

[tex]Ec = \frac{1}{2}*\frac{Q^2}{C}[/tex]

[tex]Q^2 = 2*Ec*C[/tex]

[tex]Q = \sqrt{2*Ec*C}[/tex]

[tex]Q = \sqrt{2*926*10{-6}*7.8*10^{-3}} = 3.802 * 10{-3} C = 3.802 mC[/tex]

When the capacitor is completely empty all the energy will be in the inductor and current will be maximum

[tex]El = \frac{1}{2}*L*I^2[/tex]

[tex]I^2 = 2*\frac{El}{L}[/tex]

[tex]I = \sqrt{2*\frac{El}{L}}[/tex]

[tex]I = \sqrt{2*\frac{926.7*10^{-3}}{25*10^{-3}}} = 8.61 A[/tex]

At t = 0 the capacitor has a charge of 3.8 mC, the maximum charge is 3.81 mC

q = Q * cos(vt + f)

q(0) = Q * cos(v*0 + f)

3.8 = 3.81 * cos(f)

cos(f) = 3.8/3.81

f = arccos(3.8/3.81) = 0.0721

If the capacitor is discharging it is a half cycle away, so f' = f + π = 3.2137