Answer:
1.42 KJ
Explanation:
solution:
power in beginning [tex]p_{0}[/tex]=(1.5 V).(9×[tex]10^{-3}[/tex] A)
= 13.5 mW
after continuous 37 hours it drops to
[tex]p_{37}[/tex]=(1 V).(9×[tex]10^{-3}[/tex] A)
=9 mW
When the voltage will drop energy will not remain the same but the voltage drop will always remain same if the voltage was drop to for example from 5 V to 4.5 V the drop will remain the same.
37 hours= 37.60.60
=133200 s
w=(9×[tex]10^{-3}[/tex] A×133200 )+[tex]\frac{1}{2}(13.5.10^{-3}-9.10^{-3})(133200)[/tex]
=1.42 KJ
NOTE:
There maybe a calculation error but the method is correct.
At a point in the flow over a high-speed missile, the local velocity and temperature are 300 ft/s and 500oR, respectively. Calculate the Mach number M and the characteristic Mach number, M* at this point.
Answer:
0.267
Explanation:
The Mach number is given by the following formula:
[tex]Ma = \frac{v}{c}[/tex]
where v = speed of the object
c= speed of air
Ma = Mach number
From the data:
Velocity of the object = 91.44 m/s
Velocity of air = 343 m/s
Therefore, the Mach number is given by the following formula:
Ma = [tex]\frac{91.44}{343}[/tex]
= 0.267
A rigid 14-L vessel initially contains a mixture of liquid water and vapor at 100°C with 12.3 percent quality. The mixture is then heated until its temperature is 180°C. The final state is superheated water and the internal energy at this state should be obtained by interpolation. Calculate the heat transfer required for this process. Use data from the steam tables.
Answer:
98.13kJ
Explanation:
Given that;
The rigid 10-L vessel initially contains a mixture of liquid water & vapor at
[tex]T_1 =100^0C\\T_2 = 180^0C[/tex]
We are to calculate the heat transfer required during the process by obtaining our data from the steam tables.
In order to do that, let start with our Energy Balance
So, Energy Balance for closed rigid tank system is given as:
[tex]\delta E_{system} = E_{in} - E_{out}[/tex]
Since the K.E and P.E are insignificant;
∴ K.E = P.E = 0
[tex]Q_{in}= \delta U + W\\Q_{in} = m(u_2-u_1)+ W[/tex]
Where;
m = mass flow rate of the mixture
[tex](u_2-u_1)[/tex] = corresponding change in the internal energy at state point 2 and 1
However, since we are informed that the vessel is rigid, then there is no work done in the system, then W turn out to be equal to zero .i,e
W = 0
we have our above equation re-written as:
[tex]Q_{in}= m (u_2-u_1)+0\\[/tex]
[tex]Q_{in}= m(u_2-u_1)[/tex]
We were told to obtain our data from the steam table, so were going to do just that
∴ At inlet temperature [tex]T_1 = 100^0C[/tex], the given quality of mixture of liquid water and vapor [tex](x_1)[/tex] = 123% = 0.123
Using the equation:
[tex]v_1 = v_f + x_1v_{fg}\\v_1 = v_f + x_1(v_g-v_f)[/tex]
where;
[tex]v_1[/tex] = specific volume at state 1
[tex]v_f[/tex] = specific volume of the liquid
[tex]v_g[/tex] = specific volume of the liquid vapor mixture
The above data from the steam table is given as;
[tex]v_f[/tex] = 0.001043 m³/kg
[tex]v_g[/tex] = 1.6720 m³/kg
so; we have
[tex]v_1 = 0.001043 + 0.123(1.6720-0.001043)[/tex]
[tex]v_1 = 0.001043+0.123(1.670957)[/tex]
[tex]v_1 =0.001043+0.205527711[/tex]
[tex]v_1= 0.206570711m^3/kg[/tex]
[tex]v_1=0.2065m^3/kg[/tex]
At [tex]T_1[/tex] = 100°C and [tex]x_1=0.123[/tex];
the following steam data from the tables were still obtained for the internal energy; which is given as:
Internal Energy [tex](u_1)[/tex] at the state 1
[tex]u_1= u_f + xu_{fg}[/tex]
where;
specific internal energy of the liquid [tex](u_f)[/tex] = 419.06 kJ/kg
The specific internal energy of the liquid vapor mixture [tex](u_{fg})[/tex] = 2087.0 kJ/kg
∴ since ; [tex]u_1= u_f + xu_{fg}[/tex]
[tex](u_1)[/tex] = 419.06 + (0.123 × 2087.0)
[tex](u_1)[/tex] = 675.761 kJ/kg
As the tank is rigid, so as the volume which is kept constant:
[tex]v_2=v_1\\=0.2065 m^3/kg[/tex]
Now, let take a look at when [tex]T_2 = 180^0C[/tex] from the data in the steam tables
Specific volume of the liquid [tex](v_f)[/tex] = 0.00113 m³/kg
specific volume of the liquid vapor mixture [tex](v_g)[/tex] = 0.19384 m³/kg
The quality of the mixture at the final state 2 can be determined by using the equation shown below:
[tex]v_2=v_f+x_2v_{fg}[/tex]
[tex]x_2=\frac{v_2-v_f}{v_{fg}}[/tex]
[tex]x_2=\frac{v_2-v_f}{v_g-v_f}[/tex]
[tex]x_2=\frac{0.2065-0.00113}{0.19384-0.00113}[/tex]
[tex]x_2=\frac{0.20537}{0.19271}[/tex]
= 1.0657
From our usual steam table; we still obtained data for the Internal Energy when [tex]T_2=180^0C[/tex]
Specific internal energy of the liquid [tex](u_f)[/tex] = 761.92 kJ/kg
Specific internal energy of the liquid vapor mixture [tex]u_{fg}[/tex] = 1820.88 kJ/kg
Calculating the internal energy at finsl state point 2 ; we have:
[tex]u_2=u_f+u_{fg}[/tex]
= 761.92 + (1.0657 × 1820.88)
= 761.92 + 1940.511816
= 2702.431816
[tex]u_2[/tex] ≅ 2702.43 kJ/kg
Furthermore, let us calculate the mass in the system; we have:
[tex]m= \frac{V_1}{v_1}[/tex]
where V₁ = the volume 10 - L given by the system and v₁ = specific volume at state 1 as 0.2065
V₁ = the volume 10 - L = 10 × ( 0.001 m³/L)
v₁ = 0.2065
∴
mass (m) = [tex]\frac{10(0.001m^3/L)}{0.2065}[/tex]
= 0.04842 kg
Now, we gotten all we nee do calculate for the heat transfer that is required during the process:
[tex]Q_{in}= m(u_2-u_1)[/tex]
[tex]Q_{in}= 0.04842(2702.43-675.761)[/tex]
[tex]Q_{in}= 0.04842(2026.669)[/tex]
[tex]Q_{in}= 98.13 kJ[/tex]
Therefore, the heat transfer that is required during the process = 98.13 kJ
There you have it!, I hope this really helps alot!
The heat transfer required for this process is 129.168 kilojoules.
According to steam tables, the initial state of the water inside the vessel is:
[tex]P = 101.42\,kPa[/tex], [tex]T = 100\,^{\circ}C[/tex], [tex]\nu = 0.207\,\frac{m^{3}}{kg}[/tex], [tex]u = 675.761\,\frac{kJ}{kg}[/tex], [tex]x = 0.123[/tex] (Liquid-Vapor Mixture)
Given that process is isochoric, that is, at constant volume, we know the following information:
[tex]T = 180\,^{\circ}C[/tex], [tex]\nu = 0.207\,\frac{m^{3}}{kg}[/tex], Superheated vapor
We need to know the pressure and the internal energy must be determined by using linear interpolation from steam tables twice. First, we use [tex]T = 180\,^{\circ}C[/tex] as our pivot, then we construct the following information by linear interpolation:
Lower bound
[tex]P = 800\,kPa[/tex], [tex]T = 180\,^{\circ}C[/tex], [tex]\nu = 0.24700\,\frac{m^{3}}{kg}[/tex], [tex]u = 2593.9\,\frac{kJ}{kg}[/tex]
Upper bound
[tex]P = 1000\,kPa[/tex], [tex]T = 180\,^{\circ}C[/tex], [tex]\nu = 0.19444\,\frac{m^{3}}{kg}[/tex], [tex]u = 2583.0\,\frac{kJ}{kg}[/tex]
Lastly, we find the expected pressure and internal energy by linear interpolation where specific volume is our pivot:
[tex]P = 952.207\,kPa[/tex], [tex]T = 180\,^{\circ}C[/tex], [tex]\nu = 0.207\,\frac{m^{3}}{kg}[/tex], [tex]u = 2585.605\,\frac{kJ}{kg}[/tex]
Once found all the required information, we can find the required heat transfer ([tex]Q[/tex]), in kilojoules, by means of this formula:
[tex]Q = \frac{V}{\nu}\cdot (u_{2}-u_{1})[/tex] (1)
Where:
[tex]V[/tex] - Volume of the vessel, in cubic meters,[tex]\nu[/tex] - Specific volume of water, in cubic meters per kilogram.[tex]u_{1}[/tex], [tex]u_{2}[/tex] - Initial and final internal energy, in kilojoules per kilogram.If we know that [tex]V = 0.014\,m^{3}[/tex], [tex]\nu = 0.207\,\frac{m^{3}}{kg}[/tex], [tex]u_{1} = 675.761\,\frac{kJ}{kg}[/tex] and [tex]u_{2} = 2585.605\,\frac{kJ}{kg}[/tex], then the heat transfer required for this process is:
[tex]Q = \left(\frac{0.014\,m^{3}}{0.207\,\frac{m^{3}}{kg} } \right)\cdot \left(2585.605\,\frac{kJ}{kg}-675.761\,\frac{kJ}{kg} \right)[/tex]
[tex]Q = 129.168\,kJ[/tex]
The heat transfer required for this process is 129.168 kilojoules.
To learn more on heat transfer, we kindly invite to check this verified question: https://brainly.com/question/12107378
The current entering the positive terminal of a device is i(t)= 6e^-2t mA and the voltage across the device is v(t)= 10di/dtV.
( a) Find the charge delivered to the device between t=0 and t=2 s.
( b) Calculate the power absorbed.
( c) Determine the energy absorbed in 3 s.
Answer:
a) 2,945 mC
b) P(t) = -720*e^(-4t) uW
c) -180 uJ
Explanation:
Given:
i (t) = 6*e^(-2*t)
v (t) = 10*di / dt
Find:
( a) Find the charge delivered to the device between t=0 and t=2 s.
( b) Calculate the power absorbed.
( c) Determine the energy absorbed in 3 s.
Solution:
- The amount of charge Q delivered can be determined by:
dQ = i(t) . dt
[tex]Q = \int\limits^2_0 {i(t)} \, dt = \int\limits^2_0 {6*e^(-2t)} \, dt = 6*\int\limits^2_0 {e^(-2t)} \, dt[/tex]
- Integrate and evaluate the on the interval:
[tex]= 6 * (-0.5)*e^-2t = - 3*( 1 / e^4 - 1) = 2.945 C[/tex]
- The power can be calculated by using v(t) and i(t) as follows:
v(t) = 10* di / dt = 10*d(6*e^(-2*t)) /dt
v(t) = 10*(-12*e^(-2*t)) = -120*e^-2*t mV
P(t) = v(t)*i(t) = (-120*e^-2*t) * 6*e^(-2*t)
P(t) = -720*e^(-4t) uW
- The amount of energy W absorbed can be evaluated using P(t) as follows:
[tex]W = \int\limits^3_0 {P(t)} \, dt = \int\limits^2_0 {-720*e^(-4t)} \, dt = -720*\int\limits^2_0 {e^(-4t)} \, dt[/tex]
- Integrate and evaluate the on the interval:
[tex]W = -180*e^-4t = - 180*( 1 / e^12 - 1) = -180uJ[/tex]
A +13 nC point charge is placed at the origin, and a +8 nC charge is placed on the x axis at x=5m. At what position on the x axis is the net electric field zero? (Be careful to keep track of the direction of the electric field of each particle.)
Answer:
Explanation:
Given
Charge [tex]q_1=13\ nC is placed at x=0[/tex]
Charge [tex]q_2=8\ nC is placed at x=5\ m[/tex]
Electric field because of both the charges will be away from them
Electric field because of charge [tex]q_1[/tex] at distance r from it
[tex]E_1=\frac{kq_1}{r^2}[/tex]
[tex]E_1=\frac{9\times 10^9\times 13\times 10^{-9}}{r^2}[/tex]
Electric Field due to charge [tex]q_2[/tex] at distance of 5-r from it
[tex]E_2=\frac{kq_2}{(5-r)^2}[/tex]
[tex]E_2=\frac{9\times 10^9\times 8\times 10^{-9}}{(5-r)^2}[/tex]
at this Point Net Electric field is zero i.e.
[tex]E_1=E_2[/tex]
[tex]\frac{9\times 10^9\times 13\times 10^{-9}}{r^2}=\frac{9\times 10^9\times 8\times 10^{-9}}{(5-r)^2}[/tex]
[tex]\frac{5-r}{r}=\sqrt{\frac{8}{13}}[/tex]
[tex]5-r=0.784 r[/tex]
[tex]5=1.784 r[/tex]
[tex]r=\frac{5}{1.784}[/tex]
[tex]r=2.80\ m[/tex]
Thus at [tex]x=2.8\ m[/tex] net electric field is zero
Water is the working fluid in an ideal Rankine cycle. The pressure and temperature at the turbine inlet are 1200 lbf /in^2 and 1100°F, respectively, and the condenser pressure is 140 lbf/in^2 The mass flow rate of steam entering the turbine is 1.4 x 10^6 lb/h. The cooling water experiences a temperature increase from 60 to 80°F, with negligible pressure drop, as it passes through the condenser. Determine for the cycle:
(a) the net power developed, in Btu/h
(b) the thermal efficiency.
(c) the mass flow rate of cooling water, in lb/h.
If a signal is transmitted at a power of 250 mWatts (mW) and the noise in the channel is 10 uWatts (uW), if the signal BW is 20MHz, what is the maximum capacity of the channel?
Answer:
C = 292 Mbps
Explanation:
Given:
- Signal Transmitted Power P = 250mW
- The noise in channel N = 10 uW
- The signal bandwidth W = 20 MHz
Find:
what is the maximum capacity of the channel?
Solution:
-The capacity of the channel is given by Shannon's Formula:
C = W*log_2 ( 1 + P/N)
- Plug the values in:
C = (20*10^6)*log_2 ( 1 + 250*10^-3/10)
C = (20*10^6)*log_2 (25001)
C = (20*10^6)*14.6096
C = 292 Mbps
On a given day, a barometer at the base of the Washington Monument reads 29.97 in. of mercury. What would the barometer reading be when you carry it up to the observation deck 500 ft above the base of the monument?
Answer:
The barometer reading will be 29.43 in
Explanation:
Using the formula of pressure variation
p2 - p1 = -yair * H
= 7.65 * [tex]10^{-2} \frac{lb}{ft^{3} } * 500 ft\\[/tex]
= 38.5 [tex]\frac{lb}{ft^{2} }[/tex]
According to the relationship between the pressure and the height of the mercury column
p = yHg * h --> where yHg and h is the barometer reading
yHg [tex](\frac{29.97}{12} ft)[/tex] - yHg * h1 = 38.5 [tex]\frac{lb}{ft^{2} }[/tex]
h1 = ([tex]\frac{29.97}{12} ft[/tex]) - [tex]\frac{38.5 \frac{lb}{ft^{2} } }{847 \frac{lb}{ft^{3} } }[/tex]
[tex][(\frac{29.97}{12} ft) - 0.0455 ft] - 12 \frac{in}{ft} \\\\h1 = 29.43 in[/tex]
The barometer reading be "29.43 in".
According to the question,
Barometer reading,
[tex]H' = 29.97 \ in[/tex]Pressure variation for incompressible and static fluid will be:
→ [tex]P_2-P_1 = V_{air} H[/tex]
[tex]= 7.65\times 10^{-2}\times 500[/tex]
[tex]= 38.5 \ lb/ft^2[/tex]
Pressure and height relationship for mercury will be:
→ [tex]P = V_{Hg} H'[/tex]
→ [tex]V.Hg\times \frac{29.97}{12} - V_{Hg} h_1 = 38.5[/tex]
→ [tex]h_1 = \frac{29.97}{12} - \frac{38.5}{847}[/tex]
[tex]= [\frac{29.97}{12} - 0.0455][/tex]
[tex]= 29.42 \ in[/tex]
Thus the above answer is right.
Learn more:
https://brainly.com/question/9572754
One the eight safety systems that failed was deemed to be ____________.
Select one:
a. political
b. misinterpreted pressure tests to determine whether the well had been sealed
c. the water temperature
d. the weather temperature
Answer: B. misinterpreted pressure tests to determine whether the well has been sealed.
Explanation:
Pressure test is a non-destructive test carried out to ensure there is integrity of the pressure shell on an equipment that is new or pressure and piping equipment installed earlier that has undergone some repairs.
After a crew carried out a variety of pressure tests to determine whether the well was sealed or not. The results of these tests were misinterpreted, so they thought the well was under control. This caused oil to spill.
The car travels along the circular path such that its speed is increased by at = (0.5et ) m/s2 , where t is in seconds. Determine the magnitudes of its velocity and acceleration after the car has traveled s =18 m starting from rest. Neglect the size of the carDetermine the magnitude of its acceleration after the car has traveled s = 18 m starting from rest
Answer:
acceleration = 24.23 ms⁻¹
Explanation:
Let's gather the data:
The acceleration of the car is given by a = 0.5 [tex]e^{t}[/tex]
The acceleration is the change in the speed in relation to time. In other words:
[tex]\frac{dv}{dt}[/tex] = a = a = 0.5 [tex]e^{t}[/tex] ...1
Solving the differential equation yields:
v = 0.5 [tex]e^{t}[/tex] + C₁
Initial conditions : 0 = 0.5 [tex]e^{0}[/tex] + C₁
C₁ = -5
at any time t, the velocity is: v= 0.5[tex]e^{t}[/tex]- 5
Solving for distance, s = 0. 5[tex]e^{t}[/tex] - 0.5 t - 0.5
18 = 0.5 [tex]e^{t}[/tex] - 0.5 t - 0.5
t = 3.71 s
Substitute t = 3.71 s
v= 0.5[tex]e^{t}[/tex]- 5
= 19.85 m/s
a = 0.5 [tex]e^{t}[/tex] ...1
= 20.3531
an = [tex]\frac{v^{2} }{p}[/tex]
= [tex]\frac{(19.853)^{2} }{30}[/tex]
= 13.1382
Magnitude of acceleration = [tex]\sqrt{ (a)^{2} + (an)^{2} }[/tex]
= [tex]\sqrt{(20.3531)^{2} +(13.1382)^{2} }[/tex]
= 24.23 ms⁻¹ Ans
A simple ideal Brayton cycle with air as the working fluid has a pressure ratio of 10. The air enters the compressor at 520 R and the turbine at 2000 R. Accounting for the variation of specific heats with temperature, determine (a) the air temperature at the compressor exit, (b) the back work ratio, and (c) the thermal efficiency.
The question involves analyzing an ideal Brayton cycle to find the exit temperature from a compressor, the back work ratio, and the thermal efficiency of a gas turbine engine with air as the working fluid, accounting for the variation in specific heats with temperature.
Explanation:The student's question pertains to the ideal Brayton cycle, which is a thermodynamic cycle used to describe the workings of a gas turbine engine. Their question involves air with varying specific heats at different temperatures, typical in engineering thermodynamics analysis.
Although the appropriate formulas and calculations are not provided in the provided references, the typical approach would involve thermodynamic equations of state and specific heat relations to find the unknown temperatures, efficiency, and other parameters of interest. Analysis often includes using the Brayton cycle efficiency formula, temperature ratios, and the specific gas constant for air.
The back work ratio and thermal efficiency in particular can be calculated using the first and second laws of thermodynamics applied to each component of the cycle: the compressor and turbine. As specific heats vary with temperature, one typically uses mean values or functions that provide the specific heat values at the given temperatures.
Explanation of the Brayton cycle in gas turbines, calculation methods for compressor exit temperature, back work ratio, and thermal efficiency.
The Brayton cycle is a thermodynamic cycle that describes how gas turbines operate. It consists of four processes: two isentropic (reversible adiabatic) processes and two isobaric (constant-pressure) processes.
(a) The air temperature at the compressor exit can be calculated using the temperature ratios at the compressor and turbine, and the pressure ratio.
(b) The back work ratio is the ratio of the work required to drive the compressor to the work produced by the turbine.
(c) The thermal efficiency of the Brayton cycle can be determined by comparing the work output to the heat input.
A three-phase line with an impedance of (0.2 1 j1.0) Ω /phase feeds three balanced three-phase loads connected in parallel. Load 1: Absorbs a total of 150 kW and 120 kvar. Load 2: Delta connected with an impedance of (150 2 j48) Ω /phase. Load 3: 120 kVA at 0.6 PF leading. If the line-to-neutral voltage at the load end of the line is 2000 v (rms), determine the magnitude of the line-to-line voltage at the source end of the line.
Answer:
Source voltage (L-L) = 3479.50<2.13 V (in polar form)
Source voltage (L-L) = 3477.1 + j129.57 V (rectangular form)
Given Information:
A 3 phase source is feeding three loads connected in parallel.
Load voltage (L-N) = VLoad = 2000 V
Impedance of line = ZLine = 0.2 + j1.0 Ω
Load 1 = S1 = P + jQ = 150 + j120 kVA
Load 2 = S2 = Delta connected with Z2 = 150 - j48 Ω
Load 3 = S3 = 120 KVA at PF = 0.6 leading
Source voltage (L-L) = ?
Explanation:
The source voltage is = VLoad + total current*(ZLine)
Where total current is = I1 + I2 + I3
Lets first find current flowing in each of the loads
Load 1:
3 phase apparent power is given S1 = 150 + j120 kVA
Convert into per phase by diving by 3
S1 = (150 + j120)/3
S1 = 50 + j40 kVA
As we know, S = V1* (where * is the conjugate)
I1 = S1*/VLoad
I1 = (50,000 - j40,000)/2000 (notice minus sign due to conjugate)
I1 = 25 - j20 A
Load 2:
first convert delta impedance into wye by the relation
Zy = Zdelta/3
Zy = (150 - j48)/3
Zy = 50 - j16 Ω
I2 = V/Zy
I2 = 2000/(50 - j16)
I2 = 36.29 + j11.61 A
Load 3:
apparent power = 120 KVA
PF = cos(θ) = 0.6 leading
As we know, P = S*cos(θ) and Q = S*sin(θ)
θ = cos⁻¹(PF) = cos⁻¹(0.6) = 53.13°
P3 = 120*cos(53.13°) = 72 kW
Q3 = 120*sin(53.13°) = 96 kVAR
S3 = 72 + j96 kVA
Convert into per phase by diving by 3
S3 = (72 - 96)/3 (minus sign due to leading PF)
S3 = 24 - j32 kVA
I3 = S3*/VLoad
I3 = (24,000 + j32,000)/2000
I3 = 12 + j16 A
Total current = I1 + I2 + I3
Total current = (25 - j20) + (36.29) + j11.61) + (12 + j16)
Total current = 73.29 + j7.61
Source voltage (L-N) = VLoad + total current*(ZLine)
Source voltage (L-N) = 2000 + (73.29 + j7.61)*(0.2 + j1.0)
Source voltage (L-N) = 2007.05 + j74.81
Source voltage (L-L) = [tex]\sqrt{3}[/tex]*Source voltage (L-N)
Source voltage (L-L) = [tex]\sqrt{3}[/tex] (2007.05 + j74.81)
Source voltage (L-L) = 3477.1 + j129.57 V
Source voltage (L-L) = 3479.50<2.13 V (in polar form)
D *4.80 It is required to use a peak rectifier to design a dc power supply that provides an average dc output voltage of 12 V on which a maximum of ±1-V ripple is allowed. The rectifier feeds a load of 200. The rectifier is fed from the line voltage (120 V rms, 60 Hz) through a transformer. The diodes available have 0.7-V drop when conducting. If the designer opts for the half-wave circuit: (a) Specify the rms voltage that must appear across the transformer secondary. (b) Find the required value of the filter capacitor. (c) Find the maximum reverse voltage that will appear across the diode, and specify the PIV rating of the diode. (d) Calculate the average current through the diode during conduction. (e) Calculate the peak diode current.
Designing a DC power supply with a half-wave rectifier involves calculating the necessary transformer secondary RMS voltage, filter capacitor value, diode's peak inverse voltage, and average and peak diode currents considering the specific design requirements like output voltage and ripple.
Explanation:The question relates to designing a DC power supply using a half-wave rectifier, including calculations for transformer secondary voltage, filter capacitor value, peak inverse voltage (PIV), and diode current parameters. Solving such a design problem involves understanding and applying principles of electrical engineering, particularly those regarding rectification, filtering, and electrical power supply designs.
To calculate the RMS voltage required across the transformer secondary, one must consider the average output voltage required, the diode forward voltage drop, and the peak-to-peak voltage required to maintain the ripple within the specified limits.The filter capacitor value is crucial to minimize ripple and can be determined based on the allowable ripple voltage, load resistance, and the frequency of the rectified output.The PIV rating of the diode is determined by considering the maximum reverse voltage that the diode can withstand during operation, which is affected by the transformer's secondary RMS voltage and the rectifier configuration.Calculating the average and peak diode current involves understanding the load requirements and the electrical characteristics of the diode and the circuit.In the frequency domain, you view a signal whose highest frequency is fH=4.3kHz, and lowest frequency is fL=300Hz. What is the signal’s frequency bandwidth?
Answer:
[tex]BW=\Delta f= f_H-f_L=4300Hz-300Hz=4000Hz=4kHz[/tex]
Explanation:
Bandwidth is the difference between the upper and lower frequencies in a continuous set of frequencies. The bandwidth may be determined by use of the following formula:
[tex]BW=f_H-f_L\\\\Where:\\\\f_H=Upper\hspace{3}cutoff\hspace{3}frequency=4.3kHz=4300Hz\\f_L=Lower\hspace{3}cutoff\hspace{3}frequency=300Hz[/tex]
Hence the signal’s frequency bandwidth is:
[tex]BW=4300-300=4000Hz=4kHz[/tex]
Select the correct verb form to complete each of the following sentences.Members of the committee ____ issued a mobile phone and one roll of stamps. Neither the CEO nor the CFO _____any recollection of the illegal funds transfer. Mr. Adams and his assistant______ planning to attend the annual technology showcase. The finance committee _______planning their vacations around the annual June meeting.Everything in my home office _____ to stop functioning when I need it the most.
The correct verb forms to complete the given sentences are:
Members of the committee issued a mobile phone and one roll of stamps.
Neither the CEO nor the CFO has any recollection of the illegal funds transfer.
Mr. Adams and his assistant are planning to attend the annual technology showcase.
The finance committee is planning their vacations around the annual June meeting.
Everything in my home office seems to stop functioning when I need it the most.
In the first sentence, the verb "issued" is correct because it is in the past tense and agrees with the subject of the sentence, "Members of the committee."
In the second sentence, the verb "has" is correct because it is in the present tense and agrees with the subject of the sentence, "Neither the CEO nor the CFO." The verb "has" is also used when the subject is a singular noun that refers to two or more people or things joined by "neither" or "nor."
In the third sentence, the verb "are" is correct because it is in the present tense and agrees with the subject of the sentence, "Mr. Adams and his assistant." The verb "are" is also used when the subject is a plural noun.
In the fourth sentence, the verb "is" is correct because it is in the present tense and agrees with the subject of the sentence, "The finance committee." The verb "is" is also used when the subject is a singular noun that refers to a group of people or things.
In the fifth sentence, the verb "seems" is correct because it is in the present tense and agrees with the subject of the sentence, "Everything in my home office." The verb "seems" is also used to express an opinion or belief.
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A bus travels the 100 miles between A and B at 50 mi/h and then another 100 miles between B and C at 70 mi/h.
The average speed of the bus for the entire 200-mile trip is:
a. More than 60 mi/h.
b. Equal to 60 mi/h.
c. Less than 60 mi/h.
Answer:
c. less than 60 mi/h
Explanation:
To calculate the average speed of the bus, we need to calculate the total distance traveled by the bus, as well as the total time of travel of the bus.
Total Distance Traveled = S = 100 mi + 100 mi
S = 200 mi
Now, for total time, we calculate the times for both speeds from A to b and then B to C, separately and add them.
Total Time = t = Time from A to B + Time from B to C
t = (100 mi)/(50 mi/h) + (100 mi)(70 mi/h)
t = 2 h + 1.43 h
t = 3.43 h
Now, the average speed of bus will be given as:
Average Speed = V = S/t
V = 200 mi/3.43 h
V = 58.33 mi/h
It is clear from this answer that the correct option is:
c. less than 60 mi/h
The average speed of the bus for the entire 200-mile trip is less than 60 mi/h (Option C)
How to determine the time from A to BDistance = 100 miles Speed = 50 mi/hTime =?Time = Distance / speed
Time = 100 / 50
Time = 2 h
How to determine the time from B to CDistance = 100 miles Speed = 70 mi/hTime =?Time = Distance / speed
Time = 100 / 70
Time = 1.43 h
How to determine the average speed Total distance = 200 milesTime from A to B = 2 hTime from B to C = 1.43 hTotal time = 2 + 1.43 = 3.43 hAverage speed =?Average speed = Total distance / total time
Average speed = 200 / 3.43
Average speed = 58.31 mi/h
Thus, we can conclude that the average speed is less than 60 mi/h
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For a given set of input values, a NAND gate produces the opposite output as an OR gate with inverted inputs.A. True
B. False
Answer:
B.
Explanation:
For a given set of input values, A NAND gate produces exactly the same values as an OR gate with inverted inputs.
The truth table for a NAND gate with 2 inputs is as follows:
0 0 1
0 1 1
1 0 1
1 1 0
The truth table for an OR gate, is as follows:
0 0 0
0 1 1
1 0 1
1 1 1
If we add two extra columns for inverted inputs, the truth table will be this one:
0 0 1 1 1
0 1 1 0 1
1 0 0 1 1
1 1 0 0 0
which is the same as for the NAND gate, not the opposite, so the statement is false.
This means that the right choice is B.
A small manufacturing plant is located 2 km down a transmission line, which has a series reactance of 0.5 Ω/km. The line resistance is negligible. The line voltage at the plant is 480/08 V (rms), and the plant consumes 120 kW at 0.85 power factor lagging. Determine the voltage and power factor at the sending end of the transmission line by using.a. Complex power approachb. Circuit analysis approach.
Answer:
Complex analysis = 682.2 V
Explanation:
1. Using a complex power approach:
Data:
Power drawn by the load, Q load = 120 kW
Power factor lagging = 0.85
The reactive power is solved as follows:
tan [arccos (power factor)] = [tex]\frac{Qload}{Pload}[/tex]
tan (cos⁻¹ [0.85]) = [tex]\frac{Qload}{120}[/tex]
solving the equation above gives Qload = 74. 4 kVar
The complex power drawn in by the load is given as:
S load = P load + jQload
= 120 + j74.4 kVA
using the complex analysis above, we can solve for the current into the load like this: I = [tex]\frac{Sload}{Vload}[/tex]
= [tex]\frac{120 + j74.4}{480/8}[/tex]
= 294 (-31.8⁰) A
The power factor will be 682.2
2. Circuit power approach:
using the KVL:
Vsource = Zline + Vload
= j 1(294 (-31.8⁰) + 480
= 682.4
The power factor will be cos (21.5 - 31.79) = 0.598 lagging
In the kitchen of your house you would like to run a 360 W blender, a 60 W phone charger, a 840 W toaster oven, and four 120 W lights. Assuming your house is wired for 120 Volts, how much current will be pulled by these electrical devices
The kitchen appliances, including a blender, phone charger, toaster oven, and lights, will pull a total current of 15 Amperes when operating on a 120-Volt household electrical system.
Explanation:To calculate the total current pulled by various electrical devices in the kitchen, you add up their power ratings and divide the sum by the voltage of the electrical system. These devices, which include a 360 W blender, a 60 W phone charger, an 840 W toaster oven, and four 120 W lights, all operate on a standard household voltage of 120 Volts. The formula for current (I) when power (P) and voltage (V) are known is I = P/V.
First, calculate the total power consumed by all devices:
360 W (blender) + 60 W (phone charger) + 840 W (toaster oven) + (4 × 120 W) (lights) = 1800 W.
Then, to find the current, we use the formula with the total power and the household voltage: I = 1800 W / 120 V = 15 A. Therefore, these appliances will pull a total current of 15 Amperes.
A turntable A is built into a stage for use in a theatrical production. It is observed during a rehearsal that a trunk B starts to slide on the turntable 10 s after the turntable begins to rotate. Knowing that the trunk undergoes a constant tangential acceleration of 0.28 m/s2, determine the coefficient of static friction between the trunk and the turntable.
Answer:
The coefficient of static friction, μₛ, between the trunk and turntable = 0.32
Explanation:
For this motion of the trunk B,
Initial velocity, v₀ = 0
Tangential Acceleration, a = 0.28 m/s²
Time taken, t = 10s
Using equations of motion,
v = v₀ + at
v = 0 + 0.28 × 10 = 2.8 m/s
Frictional force, Fᵣ = μₛN
μₛ = coefficient of static friction,
N = Normal reaction exerted on the trunk B as a result of its weight = mg
Doing a force balance on the trunk B,
Force keeping the trunk B in circular motion must balance the frictional forces.
Force keeping the trunk B in circular motion, F = mv²/r
Fᵣ = F
μₛN = mv²/r but N = mg
μₛmg = mv²/r
μₛg = v²/r
μₛ = v²/gr
μₛ = 2.8²/(9.8 × 2.5) = 0.32
Hope this helps!!!
The coefficient of static friction between the trunk and the turntable is 1.
Explanation:To determine the coefficient of static friction between the trunk and the turntable, we can use the equation:
fs = m * at
where fs is the static friction, m is the mass of the trunk, and at is the tangential acceleration of the trunk.
Using the given values, we have:
0.28 = m * 0.28
Solving for m, we find:
m = 1 kg
Therefore, the coefficient of static friction between the trunk and the turntable is 1.
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Assume that the conversion of energy into mechanical work (at the wheel) in an internal combustion engine is 20%. Calculate gallons of gasoline required to deliver 30 horsepower at the wheel, for one hour. 1 HP = 746 Watts
1 HP for 1 hour is 0.746 kWh
1 Kwh = 3 412 BTU
To deliver 30 horsepower at the wheel for one hour with a 20% efficient internal combustion engine, approximately 2.88 gallons of gasoline are required.
Explanation:The question at hand involves calculating the quantity of gasoline required to deliver 30 horsepower (HP) at the wheel for one hour, given that the energy conversion efficiency of an internal combustion engine is 20%. First, understand that 1 HP is equivalent to 746 Watts, and therefore, 30 HP equals 22,380 Watts or 22.38 Kilowatts. Since 1 HP for 1 hour is 0.746 kWh, 30 HP for 1 hour is 22.38 kWh. Knowing the energy content of gasoline and the engine's efficiency will allow us to calculate the required gallons of gasoline.
Given that a gallon of gasoline releases about 140 MJ (or 38.89 kWh) of energy when burned and considering the 20% efficiency, the useful energy per gallon of gasoline is approximately 7.778 kWh. To find out how many gallons of gasoline are needed to produce 22.38 kWh at 20% efficiency, we divide the total required energy output (22.38 kWh) by the useful energy per gallon (7.778 kWh), resulting in approximately 2.88 gallons of gasoline needed.
Consider a composite wall that includes an 8-mm-thick hardboard siding, 40-mm by 170-mm hardwood studs on 0.65-m centers with glass fiber insulation (paper faced, 28 kg/m3 ), and a 12-mm layer of gypsum (vermiculite) wall board. What is the thermal resistance associated with a wall that is 2.5 m high by 6.5 m wide (having 10 studs, each 2.5 m high)?
Answer:
dimensions and materials information that are provided about composite wall (2.5mx6.5m 10 studs each 2.5m high)
to find:wall thermal resistance=??
properties:
Ka=0.094W/m.K
Kb=0.16W/m.K
Kc=0.17W/m.K
Kd=0.038W/m.K
using isothermal surface assumption:
(La/KaAa)=0.008/0.094(0.65x2.5)
=0.0524K/W
(LB/KbAb)=0.13/0.16(0.04x2.5)
=8.125K/W
(Ld/KdAd)=0.13/0.38(0.6x2.5)
=2.243K/W
core's equivalent resistance is:
Req=(1/Rb+1/Rd)⁻¹
=(1/8.125+1/2.243)⁻¹
=1.758K/W
For total resistance we will find sum of
Rtot=Ra+Req+Rc
=1.854K/W
Whereas 10 studs are used so:
Rtotal=(10x1/Rtot)⁻¹
=0.1854K/W
Match each situation with the type of material (conductor or inductor) you would want to use in it. You need to connect a recently landed plane to the Earth in order to ground it and remove the built-up precipitation static. You would want to use this kind of material: You need to move a live power line out of a puddle of water. There is a lot of charge moving through this line, and if any makes it to your hands it's going to hurt. You would want to use this kind of material: You need a smooth sphere to put a sensitive piece of equipment inside that will minimize any sparks between the sphere and pieces of equipment outside the sphere. You would want to use this kind of material:
Answer: for the following process, I will explain each process and where the material is best to be used.
1. You need to connect a recently landed plane to the Earth in order to ground it and remove the built-up precipitation static. You would want to use this kind of material:
Answer: Conductor
Explanation: for you to ground the plane, you need a conductor that can be able to direct the current down to the earth. Because electron can only flow freely in a conductor.
2. You need to move a live power line out of a puddle of water. There is a lot of charge moving through this line, and if any makes it to your hands it's going to hurt. You would want to use this kind of material:
Answer: Inductor
Explanation: an Inductor resist the flow of electric current through it. You have to use an inductor, to avoid been electrocuted by the live wire. If a conductor is used current will flow through it, which may lead to electrocutions, as the water is also a conductor of electricity.
3. You need a smooth sphere to put a sensitive piece of equipment inside that will minimize any sparks between the sphere and pieces of equipment outside the sphere. You would want to use this kind of material:
Answer: Inductor
Explanation: to avoid spark, an inductor should be used, because when they is a friction between a conductors and an electric current, they will be a spark. So an inductor should be used to avoid spark. Inductors does not give a spark when in friction with an electric current
AN INDUCTOR IS ANY MATERIAL THAT RESIST THE FLOW OF ELECTRIC CURRENT THROUGH IT.
A CONDUCTOR IS ANY MATERIAL THAT ALLOWS THE FLOW ELECTRIC CURRENT THROUGH IT.
There are different kinds of material that conduct electricity. The answers are below;
When one wants to connect a recently landed plane to the Earth in order to ground it and remove the built-up precipitation static. You would use a Conductor.When there is a lot of charge moving through this line, and if any makes it to your hands it's going to hurt. You would use Inductor.When one need a smooth sphere to put a sensitive piece of equipment inside that will minimize any sparks between the sphere and pieces of equipment outside the sphere. You would use an Inductor.A key difference between a conductor and an inductor is that a conductor is known to be against an adjustment of voltage but an inductor only is against an adjustment of the current.
The inductor is known to stores energy because it has an attractive field. The conductor is said to stores energy because it serves as an electric field.
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print('Predictions are hard.") print(Especially about the future.) user_num = 5 print('user_num is:' user_num)
Answer:
The correct code is given below:-
print("Predictions are hard.")
print("Especially about the future.")
user_num = 5
print("user_num is:", user_num)
Define a named tuple Player that describes an athlete on a sports team. Include the fields name, number, position, and team.
Answer:
Explanation:
we would be analyzing this question with the important code given
#code :
from collections import namedtuple
#creating a named tuple named 'Player' with field names name, number, position and team
Player = namedtuple('Player',['name','number','position', 'team'])
cam = Player('Cam Newton','1','Quarterback','Carolina Panthers')
lebron = Player('Lebron James','23','Small forward','Los Angeles Lakers')
print(cam.name+'(#'+cam.number+')'+' is a '+cam.position+' for the '+cam.team+'.')
print(lebron.name+'(#'+lebron.number+')'+' is a '+lebron.position+' for the '+lebron.team+'.')
NB:
Lebron James (#23) rep. Small forward for the LA lakers
Cam Newton(#1) rep. a Quaterback for the Carolina Panthers
cheers i hope this helps
Two large, nonconducting plates are suspended 8.25 cm apart. Plate 1 has an area charge density of + 86.8 μC/m2 , and plate 2 has an area charge density of + 23.6 μC/m2 . Treat each plate as an infinite sheet. Two parallel vertical lines are horizontally separated from each other. Each line represents a nonconducting plate. The plate on the left is labeled plate 1, and the plate on the right is labeled plate 2. The region of space to the left of plate 1 is labeled region A. The region of space between the two plates is labeled region B. The region to the right of plate 2 is labeled region C. How much electrostatic energy U E is stored in 2.29 cm3 of the space in region A?
The electrostatic energy in region A can be found by first calculating the electric field due to Plate 1, using the given charge density, and then determining the energy density. This energy density is then multiplied by the volume in region A, converted to SI units, to find the total electrostatic energy stored.
Explanation:To find the electrostatic energy UE stored in a certain volume of region A, we must first determine the electric field in that region. For two infinite charged plates as described, we can use the principle of superposition. Each plate generates an electric field which is independent of the other. Plate 1, with a positive charge density, will generate an electric field directed away from itself, while Plate 2 will also generate an electric field directed away since it also has a positive charge density.
However, the student's scenario only involves calculating the energy in region A, where only the electric field due to Plate 1 exists since Plate 2's field does not extend into that region. The electric field E due to an infinite plate with charge density σ is E = σ / (2ε0) where ε0 is the permittivity of free space (ε0 ≈ 8.85 x 10-12 C2/N·m2). In this case, E for Plate 1 is calculated using σ = +86.8 μC/m2.
The electrostatic energy density u in a region with an electric field E is given by u = ε0E2/2. The energy density can then be multiplied by the volume to find the total energy UE. The volume in region A is 2.29 cm3. Hence, UE = u × Volume.
It's important to convert all measurements into SI units before performing the calculations. The area charge density has to be converted from μC/m2 to C/m2, and the volume from cm3 to m3.
A 10-V-emf battery is connected in series with the following: a 2-µF capacitor, a 2.0-Ω resistor, an ammeter, and a switch, initially open; a voltmeter is connected in parallel across the capacitor. At the instant the switch is closed, what are the current and capacitor voltage readings, respectively?
Answer:
I = 5.0 A V=0
Explanation:
Assuming that the ammeter is an ideal one (which means that the internal resistance is negligible compared with other resistors in the circuit), as the voltage through the capacitor can't change instantaneously, just after the switch is closed, will behave like a short, so applying Ohm's law, the current in the circuit will be as follows:
[tex]I =\frac{V}{R} = \frac{10 V}{2.0 \Omega} = 5.0 A[/tex]
As the voltage in the capacitor, can't change instantaneously, assuming an ideal voltmeter (infinite resistance) , the lecture on the voltmeter across the capacitor will be just 0.
As time goes by, the voltage measured will follow the following equation:
[tex]V = V0*( 1-e^{-\frac{t}{R*C}} )[/tex]
We see that when t=0, V=0.
A microwave transmitter has an output of 0.1 W at 2 GHz. Assume that this transmitter is used in a microwave communication system where the transmitting and receiving antennas are parabolas, each 1.2 m in diameter. a. What is the gain of each antenna in decibels
Answer:
The gain of each antenna in decibels is 353 .33 dB.
Explanation:
Given:
Frequency = 2 GHz
Diameter = 1.2 m
To Find:
The gain of each antenna in decibels = ?
Solution:
Relationship between antenna gain and effective area
[tex]= \frac{4 \pi f^2 A_e}{c^2}[/tex]-----------------------------------(1)
Where
f is frequency
[tex]A_e[/tex] is effective area
c is the speed of light
[tex]A_e[/tex] = effective area of a parabolic antenna with a face area of A is 0.56A
[tex]A_e = 0.56(\pi r^2)[/tex]
r is the radius
r = [tex]\frac{1.2}{2}[/tex]
r = 0.6
[tex]A_e = 0.56(\pi (0.6)^2)[/tex]
[tex]A_e = 0.56( 0.36 \pi)[/tex]
[tex]A_e = 0.2016 \pi[/tex]
Substituting the values in eq(1)
[tex]= \frac{4 \pi (2 \times 10^9)^2 \times 0.2016\pi}{(3 \times 10^8)^2}[/tex]
[tex]= \frac{4 \pi (4 \times 10^{18} ) \times 0.2016\pi}{(9 \times 10^{16})}[/tex]
[tex]= \frac{4 \pi (4 \times 10^{2} ) \times 0.2016\pi}{9}[/tex]
= [tex]\frac{31.80 \times 10^{2}}{9}[/tex]
= 353 .33 dB
An annular aluminum fin of rectangular profile is attached to a circular tube having an outside diameter of 25 mm and a surface temperature of 250°C. The fin is 1 mm thick and 10 mm long, and the temperature and the convection coefficient associated with the adjoining fluid are 25°C and 25W/m2 .K, respectively.
a) What is the heat loss per fin? What is the fin efficiency?
b) If 200 such fins are spaced at 5-mm increments along the tube length, what is the heat loss per meter of tube length?
c) If the tube had no fins, what would be the heat loss per meter of tube length? d) Using this result and that from part (b), what is the "fin array effectiveness"?
Answer:
See attachment below
Explanation:
A double-pane insulated window consists of two 1 cm thick pieces of glass separated by a 1.8 cm layer of air. The window measures 4 m in width and 3 m in height. The inside air is at 27ºC with a convection coefficient of 12 W/m2K, and the outer surface of the glass is at -10ºC. (a) Sketch the thermal circuit. (b) Find the temperature of the glass surface inside the room. (c) Calculate the heat loss through the window. (d) Which, if any, thermal resistances are negligible (less than 1% of the total)?
Answer:
(b). T = 22.55 ⁰C
(c). q = 557.8 W
Explanation:
we take follow a step by step process to solving this problem.
from the question, we have that
The two glass pieces is separated by a 1.8 cm distance layer of air.
the thickness of glass piece is 1 cm
width = 4 m
the height = 3 m
(a). the sketch of the thermal circuit is uploaded in the picture below.
(b). the thermal resistance due to the conduction in the first glass plane is given thus;
R₁ = Lg / Kg A ................(1)
given that Kg rep. the thermal conductivity of the glass plane
A = conduction surface area
Lg = Thickness of glass plane4
taking the thermal conductivity of glass plane as Kg = 0.78 w/mk
inputting values into equation (1) we have,
R₁ = [1 (cm) ˣ 1 (m)/100 (cm)] / [(0.78 w/mk)(4m ˣ 3m)]
R₁ = 1.068 ˣ 10 ⁻³ k/w
Being that we have same thermal resistance in the first and second plane,
therefore R₁ = R₃ = 1.068 ˣ 10 ⁻³ k/w
⇒ Also the thermal resistance between air and glass as a result of the conduction by the layer is given thus
R₂ = La/KaA .....................(2)
given Ka = thermal conductivity of air
A = surface area
La = thickness of air
substituting values into the equation we have
R₂ = [1.8 (cm) ˣ 1 (m)/100 (cm)] / [(0.0262 w/mk)(4m ˣ 3m)]
R₂ = 5.73 ˣ 10⁻² k/w
Given the thermal resistance on the outer surface due to convection, we have
R₄ = 1/hA
inputting value gives R₄ = 1 / (12 w/m² ˣ 12m) = 6.94 ˣ 10⁻³k/w
R₄ = 6.94 ˣ 10⁻³k/w
Finally the sum total of thermal resistance = R₁ + R₂ + R₃ + R₄
R-total = 0.0663 kw
From this we can calculate the rate of heat loss
using q = Ti - To / R-total ..............(3)
given Ti and To is the inside and outside temperature i.e. 27⁰C and -10⁰C
from equation (3),
q = 27- (-10) / 0.0063 = 557.8 W
q = 557.8 W
⇒ Applying the heat transfer formula for inside surface glass temperature gives;
q = Ti - T₂ / R₃ + R₄
T₂ = Ti - q (R₃ + R₄)
T₂ = 27 - 557.8 (1.068ˣ10⁻³ + 6.94ˣ10⁻³ ) = 22.55°C
T₂ = 22.55°C
cheers i hope this helps
The break mechanism used to reduce recoil in certain types of guns consists essentially of a piston attached to the barrel and moving in a fixed cylinder filled with oil. As the barrel recoils with an initial velocity vo , the piston moves and oil is forced through orifices in the piston, causing the piston and the barrel to decelerate at a rate proportional to their velocity, i.e.a=-kv. Express:a) v in terms of tb) x in terms of tc) v in terms of x
Answer:
a) v = v₀(e^-kt)
b) x = -v₀(e^-kt)/k
c) v = -kx
Explanation:
a) a = dv/dt
a = -kv = dv/dt
dv/v = -kdt
Integrating the left hand side from v₀ to v and the right hand side from 0 to t
In (v/v₀) = -kt
v/v₀ = e^(-kt)
v = v₀(e^-kt)
b) v = dx/dt
dx/dt = v = v₀(e^-kt)
dx/dt = v₀(e^-kt)
dx = v₀(e^-kt)dt
Integrating the right hand side from 0 to t and the left hand side from 0 to x,
x = -v₀(e^-kt)/k
c) Since v = v₀(e^-kt)
x = -v₀(e^-kt)/k
x = -v/k
v = -kx
Hope this helps!
The velocity (v) expressed in terms of t is; v = v₀(e^(-kt))
The distance expressed in terms of t is; x = -v₀(e^-kt)/k
The velocity expressed in terms of x is; v = -kx
What is the velocity in terms of other parameters?
a) We know that acceleration is simply change in speed with respect to time. Thus;
a = dv/dt
We are told that a = -kv. Thus;
a = dv/dt = -kv
dv/v = -kdt
Integrating both sides with respective boundary conditions gives;
v₀ to v∫dv/v = 0 to t∫-kdt
⇒ In (v/v₀) = -kt
v/v₀ = e^(-kt)
v = v₀(e^(-kt))
b) We know that velocity is change in distance with respect to time. Thus;
v = dx/dt
From a above, we saw that v = v₀(e^(-kt))
Thus;
dx/dt = v₀(e^(-kt))
dx = v₀(e^(-kt))dt
Integrating both sides with respective boundary conditions gives;
0 to x∫dx = 0 to t∫v₀(e^(-kt))dt
x = -v₀(e^-kt)/k
c) Earlier we saw that v = v₀(e^-kt)
Since x = -v₀(e^-kt)/k, then;
x = -v/k
Thus, velocity is;
v = -kx
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