Answer:
Power= Heat energy\ time
heat energy= mc∆T
specific heat of water = 4180
1.5*4180*(60-25)
=219450
time = 10× 60= 600 secs
power = 219450/600
Power= 365.75 Watt
=0.37 KW
Explanation:
the actual has to be bigger because the heater might be required to handle more ...also to accommodate extra energy.
The phasor technique is not valid if the frequencies of the sinusoids in the time domain are different. Part F - Use phasors to combine sinusoids The phasor technique makes it pretty easy to combine several sinusoidal functions into a single sinusoidal expression without using trigonometric identities. However, you cannot use the phasor technique in all cases. Select the expressions below for which the phasor technique cannot be used to combine the sinusoids into a single expression.
a. 45 sin(2500t – 50°) + 20 cos(1500t +20°)
b. 25 cos(50t + 160°) + 15 cos(50t +70°)
c. 100 cos(500t +40°) + 50 sin(500t – 120°) – 120 cos(500t + 60°) -100 sin(10,000t +90°) + 40 sin(10, 100t – 80°) + 80 cos(10,000t)
d. 75 cos(8t+40°) + 75 sin(8t+10°) – 75 cos(8t + 160°)
Answer:
a, c
Explanation:
As the problem statement tells you, the phasor technique cannot be used when the frequencies are different. The frequencies are different when the coefficients of t are different. The different ones are highlighted.
a. 45 sin(2500t – 50°) + 20 cos(1500t +20°)
b. 25 cos(50t + 160°) + 15 cos(50t +70°)
c. 100 cos(500t +40°) + 50 sin(500t – 120°) – 120 cos(500t + 60°) -100 sin(10,000t +90°) + 40 sin(10, 100t – 80°) + 80 cos(10,000t)
d. 75 cos(8t+40°) + 75 sin(8t+10°) – 75 cos(8t + 160°)
Answer:
A and CBy which the phaser technique cannot be used to combine the sinusoidal into a single expression.# derive is a kind of which dependency
. A fire hose with an inside diameter of 6.40 cm is connected to a water hydrant very close to the ground. The other end is attached to a nozzle with an inside diameter of 3.5 cm and is brought 8.0 m above the ground. The flow rate on the hose is 30.0 L/s. What is the pressure in the nozzle assuming the pressure inside the hose on the ground is 1.50 x 106 N/m2
Answer:
Solving
Explanation:
Trent is designing the grounds for a massive outdoor skate park. He has no experience with skateboarding at all, and although he is excited about this new project, he’s not sure where to begin. In planning this project, where should Trent concentrate his research? (Select all that apply.)
library research
skate-themed video games
interviews with skaters
documentaries on skater culture
Answer:
I’m sorry I don’t understand this is there more steps?♀️
Explanation:
Answer:
skate-themed video games
library research
interviews with skaters
documentaries on skater culture
(I got a 90% on the test so I dont know if this was the one that gave me the 90% but it said it was right)
Which option best describes a way engineers can provide maintenance, diagnoses, upgrades, or duplicates even for products they didn't design?
Answer:
Reverse Engineering
Explanation:
Just took the test
The compressed-air tank has an inner radius r and uniform wall thickness t. The gage pressure inside the tank is p and the centric axial load F is applied at the end cap. Use p = 1402 kPa, F= 13 kN, t 18 mm and r =306 mm.
Required:
Obtain the state of plane stress in the x - y coordinate system
Answer:
Explanation:
Given that:
The Inside pressure (p) = 1402 kPa
= 1.402 × 10³ Pa
Force (F) = 13 kN
= 13 × 10³ N
Thickness (t) = 18 mm
= 18 × 10⁻³ m
Radius (r) = 306 mm
= 306 × 10⁻³ m
Suppose we choose the tensile stress to be (+ve) and the compressive stress to be (-ve)
Then;
the state of the plane stress can be expressed as follows:
[tex](\sigma_ x) = \dfrac{Pd}{4t}+ \dfrac{F}{2 \pi rt}[/tex]
Since d = 2r
Then:
[tex](\sigma_ x) = \dfrac{Pr}{2t}+ \dfrac{F}{2 \pi rt}[/tex]
[tex](\sigma_ x) = \dfrac{1402 \times 306 \times 10^3}{2(18)}+ \dfrac{13 \times 10^3}{2 \pi \times 306\times 18 \times 10^{-3} \times 10^{-3}}[/tex]
[tex](\sigma_ x) = \dfrac{429012000}{36}+ \dfrac{13000}{34607.78467}[/tex]
[tex](\sigma_ x) = 11917000.38[/tex]
[tex](\sigma_ x) = 11.917 \times 10^6 \ Pa[/tex]
[tex](\sigma_ x) = 11.917 \ MPa[/tex]
[tex]\sigma_y = \dfrac{pd}{2t} \\ \\ \sigma_y = \dfrac{pr}{t} \\ \\ \sigma _y = \dfrac{1402\times 10^3 \times 306}{18} \ N/m^2 \\ \\ \sigma _y = 23.834 \times 10^6 \ Pa \\ \\ \sigma_y = 23.834 \ MPa[/tex]
When we take a look at the surface of the circular cylinder parabolic variation, the shear stress is zero.
Thus;
[tex]\tau _{xy} =0[/tex]
Two identical bulbs in parallel in a radio create a total resistance of 15 ohms in the circuit. What's the resistance of each of the bulbs A. 10 ohms B. 7.5 ohms C. 45 ohms D. 30 ohms
Answer
letter B is the answer because 7.5+7.5=15
The resistance of each of the bulbs will be 30 ohms. The correct option is D.
What is resistance?To resist the passage of current, use a resistor. When a resistor is included in a circuit, the amount of current that flows through it drops.
Using ohms as the unit, resistance acts as a gauge to quantify how easily current will flow through a circuit.
When resistance falls, current rises, and when resistance rises, current falls. In order to make sure that current flows in circuits at the proper rate, resistors are crucial.
Here, it is given that:
Total resistance is 15 ohms.
We know that, two identical resistors in parallel have an equivalent resistance half the value of either resistor.
So, the total resistance of 15 ohms have the resistance of each of the bulbs as 30 ohms.
Thus, the correct option is D.
For more details regarding resistance, visit:
https://brainly.com/question/29427458
#SPJ2
Based on the concept that it is better to prevent falls happening in the first place, which of the following safety methods meets that criteria?
Answer:fall arrest harness
Explanation:cuz it’s just right
How many FastEthernet interfaces does a 2960 switch have?
Answer:
24 i believe
Explanation:
Iman wanted to find out at what temperature butter melts. She placed a cube of solid butter in a beaker. She put the beaker on a hot plate. She turned up the temperature dial one degree every two minutes. Then she recorded the temperature at which the butter started to melt.
Iman wanted to determine if her results were reliable. She decided to repeat the experiment. Which statement describes the main purpose of repeating the experiment?
to test a different substance
to see if the results are similar
to practice completing experiments
to determine if butter will melt twice
Answer:
Ans: To see if the results are similar
True or false, Increasing the spring force of the pressure plate that clamps the clutch disc to the flywheel increases torque capacity
but takes more foot pressure to operate the clutch pedal.
Answer: True
Explanation:
Una cuerda fuertemente estirada tiene sus puntos extremos fijos en x=0 y x=L. Si se le da un desplazamiento inicial F(x)=ax(L-x) desde la posición de equilibrio. Donde a es una constante y luego se suelta, encuentre el desplazamiento en cualquier tiempo t>0. Donde c=1/π.
Answer is in a photo. I couldn't attach it here, but I uploaded it to a file hosting. link below! Good Luck!
bit.[tex]^{}[/tex]ly/3a8Nt8n
Some wire of radius is 1.262mm has a resistance of 20Ω. Determine the resistance of a wire of the same length and material if the diameter is 1.6mm
Answer:
12.44Ω
Explanation:
For a question such as this, the assumption must be that the resistance is inversely proportional to the square of the diameter. The resistance will be ...
20Ω × (1.262/1.6)² ≈ 12.44Ω
Answer:
12.44[tex]ohms[/tex]
the resistance of a wire of the same length and material if the diameter is 1.6 mm of the given measuresExplanation:
hope it helps
The two major forces opposing the motion of a vehicle moving on a level road are the rolling resistance of the tires, Fr, and the aerodynamic drag force of the air flowing around the vehicle, Fd, given respectively by Fr, = fW, Fd= CdA1/2 rhoV2 where f and Cd are constants known as the rolling resistance coefficient and drag coefficient, respectively, W and A are the vehicle weight and projected frontal area, respectively, V is the vehicle velocity, and rho is the air density. For a passenger car with W = 3,550 lbf, A = 23.3 ft^2, and Cd = 0.34, and where f = 0.02 and rho = 0.08 lbm/ft^3.
Required:
Determine the power required, in HP, to overcome rolling resistance and aerodynamic drag when V is 55 mph.
Answer:
The power required to overcome rolling resistance and aerodynamic drag is 19.623 h.p.
Explanation:
Let suppose that vehicle is moving at constant velocity. By Newton's Law of Motion, the force given by engine must be equal to the sum of the rolling resistance and the aerodynamic drag force of the air. And by definition of power, we have the following formula:
[tex]\dot W = \left(f\cdot W +\frac{\rho\cdot C_{D}\cdot A\cdot v^{2}}{2\cdot g_{c}} \right)\cdot v[/tex] (1)
Where:
[tex]\dot W[/tex]- Power, in pounds-force-feet per second.
[tex]f[/tex] - Rolling resistance coefficient, no unit.
[tex]W[/tex] - Weight of the passanger car, in pounds-force.
[tex]\rho[/tex] - Density of air, in pounds-mass per cubic feet.
[tex]C_{D}[/tex] - Drag coefficient, no unit.
[tex]A[/tex] - Projected frontal area, in square feet.
[tex]v[/tex] - Vehicle speed, in feet per second.
[tex]g_{c}[/tex] - Pound-mass to pound-force ratio, in pounds-mass to pound-force.
If we know that [tex]f = 0.02[/tex], [tex]W = 3,550\,lbf[/tex], [tex]\rho = 0.08\,\frac{lbm}{ft^{3}}[/tex], [tex]C_{D} = 0.34[/tex], [tex]A = 23.3\,ft^{2}[/tex], [tex]v = 80.685\,\frac{ft}{s}[/tex] and [tex]g_{c} = 32.174\,\frac{lbm}{lbf}[/tex], then the power required by the car is:
[tex]\dot W = \left(f\cdot W +\frac{\rho\cdot C_{D}\cdot A\cdot v^{2}}{2\cdot g_{c}} \right)\cdot v[/tex]
[tex]\dot W = 10901.941\,\frac{lbf\cdot ft}{s}[/tex]
[tex]\dot W = 19.623\,h.p.[/tex]
The power required to overcome rolling resistance and aerodynamic drag is 19.623 h.p.
A stainless steel ball (=8055 kg/m3, Cp= 480 J/kgK) of diameter D =15 cm is removed from theoven at a uniform temperature of 350oC. The ball is then subjectedto the flow of air at 1 atm pressure and 30oC with a velocityof 6 m/s. The surface temperature of the ball eventuallydrops to 250oC. Determine the average convection heat transfercoefficient during this cooling process and estimate howlong this process has taken.
Answer:
i) 25.04 W/m^2 .k
ii) 23.82 minutes = 1429.2 secs
Explanation:
Given data:
Diameter of steel ball = 15 cm
uniform temperature = 350°C
p = 8055 kg/m^3
Cp = 480 J/kg.k
surface temp of ball drops to 250°C
average surface temperature = ( 350 + 250 ) / 2 = 300°C
i) Determine the average convection heat transfer coefficient during the cooling process
Note : Obtain the properties of air at 1 atm at average film temp of 30°C from the table " properties of air " contained in your textbook
average convection heat transfer coefficient = 25.04 W/m^2 .k
ii) Determine how long this process has taken
Time taken by the process = 23.82 minutes = 1429.2 seconds
Δt = Qtotal / Qavg = 683232 / 477.92 = 1429.59 secs
attached below is the detailed solution of the given question
A cylindrical vessel is pressurized to 1.7 MPa. The structure on which the vessel rests becomes damaged, resulting in uneven support that causes a twisting moment T0 of 8 106 N-m on the vessel. The vessel has an outer diameter of 2.5 m and a wall thickness of 50 mm. The steel wall has a uniaxial yield strength of 600 MPa. Determine the factor of safety with respect to yielding (a) assuming a maximum shear stress yield criterion and (b) assuming a von Mises yield criterion.
Answer:
a) 15.075 , b ) 12.9
Explanation:
pressure on cylindrical vessel = 1.7 MPa
Twisting moment due to pressure = To of 8( 10^6 ) N-m
outer diameter of vessel = 2.5 m
wall thickness of vessel = 50 mm
Steel wall uniaxial yield strength = 600 MPa
Calculate the factor of safety
a) assuming a max shear stress yield criterion
safety factor = 15.075
b) assuming a von Mises yield criterion
safety factor = 12.9
attached below is the detailed solution of the problem
(TCO 3) Linear-phase filters are desirable in many applications because Group of answer choices they have a sharper frequency response. they do not cause phase-distortion of an input signal. they have better stop-band performance by eliminating side-lobes. they have the smallest number of filter coefficients compared to nonlinear filters.
Answer:
They do not cause phase-distortion of an input signal. ( option 2 )
Explanation:
Linear-phase filters are desirable in many applications simply because they do not cause phase-distortion of an input signal.
The signals inputted into a Linear-phase filter ( FIR type ) are delayed equally as frequency is delayed in the pass band but they are are distorted.
The main special function of the Linear-phase filter is to reduce signal distortion.
(TCO 3) Below is the frequency response of the IIR realization of an analog band-pass filter. What type of filter is this?
Answer:
Hello the frequency response is missing attached below is the missing image
answer : Elliptic Filter
Explanation:
The filter with such frequency response that is an analog band-pass filter as well is called an Elliptic filter.
And this is because an Elliptic filter posses a constant ripple in the stop band and a unity gain is seen in the pass band
Write a program to play the Card Guessing Game. Your program must give the user the following choices: - Guess only the face value of the card. - Guess only the suit of the card. - Guess both the face value and the suit of the card. Before the start of the game, create a deck of cards. Before each guess, use the function random_shuffle to randomly shuffle the deck.
how am I going to do this, I have a friend that might be able to help I will check
If it takes 35 s for the 50-Mg tugboat to increase its speed uniformly to 25 km/h, starting from rest, determine the force of the rope on the tugboat. The propeller provides the propulsion force F which gives the tugboat forward motion, whereas the barge moves freely. Also, determine F acting on the tugboat. The barge has a mass of 75 Mg.
Answer:
- the force of the rope on the tugboat is 14.87 kN
- Force acting on the tugboat is 24.79 kN
Explanation:
Given the data in the question;
tugboat increases its speed uniformly to 25 km/h
v₁ = 25 km/h = (25 × 1000) / ( 1 × 60min × 60sec )
= 25000m / 3600s
= 6.94 m/s
Now, lets determine the force on the rope using the following relation;
T = [tex]m_b[/tex]v₁ / t₁
[tex]m_b[/tex] is mass of barge( 75 Megagram = 75 × 10³ Kilogram ), time t₁ is 35 s and v₁ is 6.94 m/s
so we substitute
T = [(75 × 10³) × 6.94 ] / 35
T = 520500 / 35
T = 14871.43 N
T = 14871.43 / 1000
T = 14.87 kN
Therefore, the force of the rope on the tugboat is 14.87 kN
Now, to determine F acting on the tugboat;
[tex]Ft_1[/tex] = ( [tex]m_t[/tex] + [tex]m_b[/tex] )[tex]v_1[/tex]
we solve for F
F = ( [tex]m_t[/tex] + [tex]m_b[/tex] )[tex]v_1[/tex] / [tex]t_1[/tex]
where [tex]m_t[/tex] is mass of tugboat (50 Megagram = 50 × 10³ Kilogram )
so we substitute
F = [( (50 × 10³) + (75 × 10³) )6.94] / 35
F = [ 125000 × 6.94 ) / 35
F = 867500 / 35
F = 24785.7 N
F = 24785.7 / 1000
F = 24.79 kN
Therefore, Force acting on the tugboat is 24.79 kN
Switching to first order kinetics, enzymes are immobilized in 6 mm radius (R) beads. Beads are immersed in a solution containing substrate at a concentration (CAs) of 22 gmol/m3. The observed reaction rate (rA,obs) of the heterogeneous reaction is 4.3 x 10-6 gmol/sec. The first order rate constant (k1) is 0.24 /sec. Calculate the hypothetical rate, r*As of the reaction in this system.
Two HVAC/R technicians are discussing the soldering of copper pipes. Technician A says that when soldering, it's important to apply flux all the way to the end of the pipe. Technician B says that when soldering, you should twist the pipe inside the fitting to help spread the flux around the joint area. Which one of the following statements is correct?
Answer:
B is correct
Explanation:
name the process by which mild steel can be converted into high carbon steel and explain it briefly ?
Answer:
please give me brainlist and follow
Explanation:
Mild steel can be converted into high carbons steel by which of the following heat treatment process? Explanation: Case hardening, also referred as carburizing increases carbon content of steel, thus, imparting hardness to steel.
A magician claims that he/she has invented a novel, super-fantastic heat engine. This engine operates between two reservoirs of temperatures 250 K and 750 K, respectively. Please verify and explain if this is possible for any of the following QH (heat received from the hot reservoir), QL (heat rejected to the cool reservoir), and W (mechanical work produced):
a. (3a). (3p) QH= 900 kJ, Wmech = 400 kJ, QL = 600 kJ.
b. (3b). (3p) Qu= 900 kJ, W mech = 400 kJ, QL = 500 kJ.
c. (3c). (3p) Qh= 900 kJ, Wmech = 600 kJ, QL = 300 kJ.
d. (3b). (3p) Qh= 900 kJ, Wmech = 800 kJ, QL = 100 kJ.
Answer:
A) Not possible, B) Posible, C) Possible, D) Not possible.
Explanation:
The maximum theoretical efficiency for any thermal engine is defined by Carnot's cycle, whose energy efficiency ([tex]\eta[/tex]), no unit, is expressed below:
[tex]\eta = 1-\frac{T_{L}}{T_{H}}[/tex] (1)
Where:
[tex]T_{L}[/tex] - Cold reservoir temperature, in Kelvin.
[tex]T_{H}[/tex] - Hot reservoir temperature, in Kelvin.
If we know that [tex]T_{L} = 250\,K[/tex] and [tex]T_{H} = 750\,K[/tex], then the maximum theoretical efficiency for the thermal engine is:
[tex]\eta = 1-\frac{T_{L}}{T_{H}}[/tex]
[tex]\eta = 0.667[/tex]
For real thermal engines, the following inequation is observed:
[tex]0 \le \eta_{r} \le \eta[/tex] (2)
Where [tex]\eta_{r}[/tex] is the efficiency of the real heat engine, no unit.
There are two possible criteria to determine if a given heat engine is real:
Efficiency
[tex]\eta_{r} = 1 - \frac{Q_{L}}{Q_{H}}[/tex] (3)
Where:
[tex]Q_{L}[/tex] - Heat rejected to the cold reservoir, in kilojoules.
[tex]Q_{H}[/tex] - Heat received from the hot reservoir, in kilojoules.
Power output
[tex]W = Q_{H}-Q_{L}[/tex] (4)
Where [tex]W[/tex] is the power output, in kilojoules.
Now we proceed to verify each case:
A) [tex]Q_{H} = 900\,kJ[/tex], [tex]Q_{L} = 600\,kJ[/tex], [tex]W_{m} = 400\,kJ[/tex]
[tex]\eta_{r} = 0.333[/tex]
[tex]0 \le \eta_{r} \le \eta[/tex]
[tex]W = 300\,kJ[/tex]
[tex]W \ne W_{m}[/tex]
This engine is not possible.
B) [tex]Q_{H} = 900\,kJ[/tex], [tex]Q_{L} = 500\,kJ[/tex], [tex]W_{m} = 400\,kJ[/tex]
[tex]\eta_{r} = 0.444[/tex]
[tex]0 \le \eta_{r} \le \eta[/tex]
[tex]W = 400\,kJ[/tex]
[tex]W = W_{m}[/tex]
The engine is possible.
C) [tex]Q_{H} = 900\,kJ[/tex], [tex]Q_{L} = 300\,kJ[/tex], [tex]W_{m} = 600\,kJ[/tex]
[tex]\eta_{r} = 0.667[/tex]
[tex]0 \le \eta_{r} \le \eta[/tex]
[tex]W = 600\,kJ[/tex]
[tex]W = W_{m}[/tex]
The engine is possible.
D) [tex]Q_{H} = 900\,kJ[/tex], [tex]Q_{L} = 100\,kJ[/tex], [tex]W_{m} = 800\,kJ[/tex]
[tex]\eta_{r} = 0.889[/tex]
[tex]\eta_{r} > \eta[/tex]
[tex]W = 800\,kJ[/tex]
[tex]W = W_{m}[/tex]
The engine is possible.
A gas mixture containing 3 moles CO2, 5 moles H2 and 1 mole water is undergoing the following reactions CO2+3H2 →cH3OH + H2O Develop the expressions for the mole fraction of the species in terms of the extent of the reaction?
Sasha goes back through her architectural design books for inspiration and designs the new country club to have a Romanesque feel, using rounded arches for the doorways and windows and constructing the walls with extra thickness to give a feeling of imposing massiveness to the building. The client is mystified by Sasha’s approach. He says he can’t visualize it working at all and demands she rethink the entire design. What would be best for Sasha to do in response to this criticism?
She should create a computer animated view of the design to walk the client through it.
She should get a better idea for what the client wants and expects.
She should try a different architectural style for her design.
She should build a model of her design to show the client how it will look when finished.
What is the function and role of product tear down charts, and how do engineers utilize them in the reverse engineering process?
Answer:
Product Teardown 28 pieces (1) Plastic packaging: protect and display product for purchase. (4) Exterior screws: hold case halves together. (1) Right case half: acts as part of a handle and contains the rest of the parts. (1) Left case half: acts as part of a handle and contains the rest of the parts.
Explanation:
A product teardown process is an orderly way to know about a particular product and identify its parts, system functionality to recognize modeling improvement and identify cost reduction opportunities. Unlike the traditional costing method, tear down analysis collects information to determine product quality and price desired by the consumers.
Answer:
?
Explanation:
A group of filmmaking students have produced four different short movies,
presentations about each one. Even though most of the content will be diffe
be the same for each production,
Air enters the combustor of a jet engine at Pressure=10 atm, T=1000R and M=0.2. Fuel is injected and burned with fuel-air ratio (by mass) of 0.06. The heat released during the combustion is 4.50000000 ft-lb per slug of fuel. Assuming one-dimensional frictionless flow with r=1.4 for the fuel-air mixture, calculate M2, p2 and T2 at the exit of combustor?
The rate of heat transfer between a certain electric motor and its surroundings varies with time as , Q with dot on top equals negative 0.2 open square brackets 1 minus e to the power of negative (0.05 t )end exponent close square brackets , where t is in seconds and Q with dot on top is in kW. The shaft of the motor rotates at a constant speed of omega equals 100 space r a d divided by s (about 955 revolutions per minute, or RPM) and applies a constant torque of tau equals 18 space N m to an external load. The motor draws a constant electric power input equal to 2.0 kW. For the motor, provide an expression to show the change in total energy, increment E, as a function of time.
Answer:
the required expression is [tex]E_2 - E_1[/tex] = 4[ 1 - [tex]e^{(-0.05t)}[/tex] ]
Explanation:
Given the data in the question;
Q = -0.2[ 1 - [tex]e^(-0.05t)[/tex] ]
ω = 100 rad/s
Torque T = 18 N-m
Electric power input = 2.0 kW
now, form the first law of thermodynamics;
dE/dt = dQ/dt + dw/dt = Q' + w'
dE/dt = Q' + w' ------ let this be equation 1
w' is the net power on the system
w' = [tex]w_{elect[/tex] - [tex]w_{shaft[/tex]
[tex]w_{shaft[/tex] = T × ω
we substitute
[tex]w_{shaft[/tex]= 18 × 100
[tex]w_{shaft[/tex] = 1800 W
[tex]w_{shaft[/tex] = 1.8 kW
so
w' = [tex]w_{elect[/tex] - [tex]w_{shaft[/tex]
w' = 2.0 kW - 1.8 kW
w' = 0.2 kW
hence, from equation 1, dE/dt = Q' + w'
we substitute
dE/dt = -0.2[ 1 - [tex]e^{(-0.05t)[/tex] ] + 0.2
dE/dt = -0.2 + 0.2[tex]e^{(-0.05t)[/tex] ] + 0.2
dE/dt = 0.2[tex]e^{(-0.05t)[/tex]
Now, the change in total energy, increment E, as a function of time;
ΔE = [tex]\int\limits^t_0}\frac{dE}{dt} . dt[/tex]
ΔE = [tex]\int\limits^t_0} 0.2e^{(-0.05t)} dt[/tex]
ΔE = [tex]\int\limits^t_0} \frac{0.2}{-0.05} [e^{(-0.05t)}]^t_0[/tex]
[tex]E_2 - E_1[/tex] = 4[ 1 - [tex]e^{(-0.05t)}[/tex] ]
Therefore, the required expression is [tex]E_2 - E_1[/tex] = 4[ 1 - [tex]e^{(-0.05t)}[/tex] ]