For the circuit shown below if Vp=26 V and resistor R dissipates 345.6 KJ
in 24 h. Find E1 and R.

Answer:

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Answer:

The temperature after a long time will return to 15°C

Explanation:

Determine the temperature of the slab after a very long time

First we calculate the heat flow for m^2 area normal to the surface

= q / A = 650°c - 15°C / ( 1 / h  + L / K )

           = 635°c  / ( 1 / 220 + 0.1 / 110 )  = 116.416 kw/m^2

Total heat content in the slab is calculated as

= m* c * ΔT

= 8530 * A * 0.1 * 380 * ( 650 - 15 )

= 205828.9  kJ/m^2

The temperature will return to 15°C after a long time

Answer:

(T²/D³)sys1 = (T²/D³)sys2

sys1 = earth-moon

sys2 = earth-sat

(27.33day)²/(3.8e8m)³ = (1day)²/D³

D = cbrt(7.3e22m³) = 4.2e7 m

Explanation:

exchange in capacity whilst a satellite tv for pc differences altitude How plenty paintings could be executed to flow the satellite tv for pc into yet another around orbit it is bigger above the outdoors of the Earth? satellite tv for pc exchange in capacity with top Assuming the satellite tv for pc is to be boosted to a clean top r? Gravitational capacity capacity (to center of earth) new orbit(2) has a greater robust PE than old one(a million), so exchange is helpful PE = G m?m?/r earth GM = 3.98e14 ?PE = (GM)(m)(a million/r? – a million/r?) KE additionally differences. Get speed at each and every top. New orbit(2) has decrease speed, so exchange is damaging v = ?(GM/R) V? = ?(GM/r?) V? = ?(GM/r?) ?KE = –½m(V?² – V?²) ?KE = –½mGM(a million/r? – a million/r?) including the two ?E = (GM)(m)(a million/r? – a million/r?)– ½mGM(a million/r? – a million/r?) ?E = ½mGM(a million/r? – a million/r?) ?E = ½(3.98e14)(7500) [a million/(0.5e7) –a million/(3.3e7) ] ?E = ½(3.98e14)(7500)(1e-7) [a million/(0.5) –a million/(3.3) ] ?E = ½(3.98e7)(7500) [2 – 0.303 ] ?E = ½(3.98e7)(7500)(a million.70) ?E = 2.04e11 Joules edit, corrected .

The radius of a circular orbit will be "[tex]\frac{V}{2 \pi} \ km[/tex]".

According to the question,

Time = 1 day

Distance = [tex]2 \pi r[/tex]

Let,

As we know,

→ [tex]Distance = Velocity\times time[/tex]

By substituting the values, we get

→          [tex]2 \pi r = V\times 1[/tex]

→              [tex]r = \frac{V}{2 \pi} \ km[/tex]

Thus the above is the right answer.

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This question is incomplete, The missing image is uploaded along this answer below;

Answer:

the maximum stress in the cylinder is 3.23 ksi

Explanation:

Given the data in the question and the diagram below;

First we determine the initial Kinetic Energy;

T = [tex]\frac{1}{2}[/tex]mv²

we substitute

⇒ T = [tex]\frac{1}{2}[/tex] × (550/32.2) × (2)²

T = 34.16149 lb.ft

T =  ( 34.16149 × 12 ) lb.in  

T = 409.93788 lb.in

Now, the volume will be;

V = [tex]\frac{\pi }{4}[/tex]d²L

from the diagram; d = 0.5 ft and L = 1.5 ft

so we substitute

V =  [tex]\frac{\pi }{4}[/tex] × ( 0.5 × 12 in )² × ( 1.5 × 12 in )

V = 508.938 in³

So by conservation of energy;

Initial energy per unit volume = Strain energy per volume

⇒ T/V = σ²/2E

from the image; E = 6.48(10⁶) kip

so we substitute

⇒ 409.93788 / 508.938 = σ²/2[6.48(10⁶)]

508.938σ² =  5,312,794,924.8

σ² = 10,438,982.5967  

σ = √10,438,982.5967

σ = 3230.9414  

σ = 3.2309 ksi  ≈ 3.23 ksi    { three significant figures }

Therefore, the maximum stress in the cylinder is 3.23 ksi

Answer:

a) the minimum sampling rate is 6 kHz

b) the minimum numbers of required samples are 120

c) the minimum length of the analog signal is 0.02 s

Explanation:

Given the data in the question;

(a) the minimum sampling rate;

band width of analog signal xₐ(t) is;

bandwidth B = 3kHz

Now, according to sampling theorem, minimum sampling rate F[tex]_s[/tex] must be twice the bandwidth of the signal.

so

F[tex]_s[/tex] = 2B

F[tex]_s[/tex] = 2( 3 kHz )

F[tex]_s[/tex] = 6 kHz

Therefore, the minimum sampling rate is 6 kHz

(b) the minimum number of required samples;

Let L represent the minimum number of samples required,

given that; required resolution of the spectrum of the signal is less than or equal to 50 Hz

F[tex]_s[/tex]/L ≤ 50

L ≥ F[tex]_s[/tex]/50

L ≥ ( 6 × 1000 Hz ) / 50

L ≥ 6000 / 50

L ≥ 120

Therefore, the minimum numbers of required samples are 120

(c) the minimum length of the analog signal record(in seconds).

minimum number of samples required is 120

T = L /  F[tex]_s[/tex]

T = 120 / ( 6 × 1000 Hz )

T = 120 / 6000

T = 0.02 s

Therefore, the minimum length of the analog signal is 0.02 s

Answer:

Hermetic compressors are ideal for small refrigeration systems, where continuous maintenance cannot be ensured.

Answer:

Option B

Explanation:

An evaporator along with  cold low pressure refrigerant absorbs heat from the air within the passenger compartment thereby supplying cool air for the occupants.

Hence, option B is correct

Answer:

Explanation:

From the information given:

original diameter [tex]d_o[/tex] = 10 mm

final diameter [tex]d_f =[/tex] 7.5 mm

Cold work tensile strength of brass = 380 MPa

Recall that;

[tex]\text {The percentage CW }= \dfrac{\pi (\dfrac{d_o}{2})^2 - \pi(\dfrac{d_f}{2})^2 }{\pi(\dfrac{d_o}{2})^2} \times 100[/tex]

[tex]\implies \dfrac{\pi (\dfrac{10}{2})^2 - \pi(\dfrac{7.5}{2})^2 }{\pi(\dfrac{10}{2})^2} \times 100[/tex]

[tex]\implies43.87\% \ CW[/tex]

→ At 43.87% CW, Brass has a tensile strength of around 550 MPa, which is greater than 380 MPa.

→ At 43.87% CW, the ductility is less than 5% EL, As a result, the conditions aren't met.

To achieve 15% EL, 28% CW is allowed at most

i.e

The lower bound cold work = 15%

The upper cold work = 28%

The average = [tex]\dfrac{15+28}{2}[/tex] = 21.5 CW

Now, after the first drawing, let the final diameter be [tex]d_o^'[/tex]; Then:

[tex]4.5\% \ CW = \dfrac{\pi (\dfrac{d_o^'}{2})^2 - (\dfrac{7.5}{2})^2}{\pi (\dfrac{d_o^'}{2})^2}\times 100[/tex]

By solving:

[tex]d_o^'} = 8.46 mm[/tex]

To meet all of the criteria raised by the question, we must first draw a wire with a diameter of 8.46 mm and then 21.5 percent CW on it.

Answer:

acceleration=change in velocity/ time

Explanation:

The velocity of an object is its speed in a particular direction. Velocity is a vector quantity because it has both a magnitude and an associated direction. To calculate velocity, displacement is used in calculations, rather than distance.

Answer: barries

Explanation:

Answer:

Relative roughness = 0.04

Explanation:

Given that:

Diameter = 5 cm

roughness = 2 mm

At inlet:

Minor coefficient loss [tex]k_{L1} = 0.4[/tex]

At exit:

Minor coefficient loss [tex]k_{L2} = 1[/tex]

Height h = 4m

Length = 5 m

To find the relative roughness:

Relative roughness is a term that is used to describe the set of irregularities that exist inside commercial pipes that transport fluids. The relative roughness can be evaluated by knowing the diameter of the pipe made with the absolute roughness in question. If we denote the absolute roughness as e and the diameter as D, the relative roughness is expressed as:

[tex]e_r = \dfrac{e}{D}[/tex]

[tex]e_r = \dfrac{0.2 }{5}[/tex]

[tex]\mathbf{e_r = 0.04}[/tex]

Answer:

the answer would be B motor

Answer: a accumulator

Explanation:

Answer:

I) Voltage - is the pressure from an electrical circuit's power source that pushes charged electrons (current) through a conducting loop, enabling them to do work such as illuminating a light. In brief, voltage = pressure, and it is measured in volts (V).

II) Current - is the movement of electrons through a wire. Electric current is measured in amperes (amps) and refers to the number of charges that move through the wire per second. If we want current to flow directly from one point to another, we should use a wire that has as little resistance as possible.

III) Electrical Power - is the rate, per unit time, at which electrical energy is transferred by an electric circuit. The SI unit of power is the watt, one joule per second. Electric power is usually produced by electric generators, but can also be supplied by sources such as electric batteries.

IV) Electrical Energy - is a form of energy resulting from the flow of electric charge. Energy is the ability to do work or apply force to move an object. In the case of electrical energy, the force is electrical attraction or repulsion between charged particles.

Explanation:

I hope ot helps to you a lot! Correct me if I'm wrong.

Answer:

car a is moving faster than the car b

Answer:

B: Car A and car B are moving in opposite directions.

C: Car A is moving faster than car B.

E: Car A and car B are moving toward each other until they cross over.

Explanation:

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Answer:

Hello your question is incomplete below is the complete question

Design a counter that counts the following sequence of 2-0-1-3 and repeat. Use the JK flip-flops given to you at the start of the semester. These values will be displayed on a seven-segment display like the one used in Lab 3

answer : attached below

Explanation:

Designing a counter that counts in a given sequence can be done using Logic gates that will be used to control the counter. we will design the counter using the 7 segment display

Note : The first Image is the segment display and the second image is the design of the counter

Answer:  is b

Explanation:

Answer:

72 years

Explanation:

The degree of consideration is the same for both bridges = 90%

Height  of first highway bridge( d1 ) = 5 ft

Time to consolidation ( t1 )= 4.5 years

Height of second bridge ( d2 ) = 20 ft

Time to consolidation ( t2 ) = ?

we will apply this relation below

Tv = Cv * t / d^v

Tv = constant

for a single drainage condition : t ∝ d^v hence; d = H

∴ [tex]\frac{t_{2} }{t_{1} } = (\frac{d_{2} }{d_{1} })[/tex]^2

t2 = t1 ( d2/d1 )^2

   = 4.5 ( 20 / 5 )^2

  = 72 years

Answer: its c

Explanation:

Solution :

import [tex]$\text{java}.$[/tex]util.*;

public [tex]$class$[/tex] currency{

  public static [tex]$\text{void}$[/tex] main(String[tex]$[]$[/tex] args) {

      Scanner input [tex]$=$[/tex] new Scanner(System[tex]$\text{.in}$[/tex]);

      System[tex]$\text{.out.}$[/tex]print("Enter [tex]$\text{number of}$[/tex] quarters:");

      int quarters = input.nextInt();

      System.out.print("Enter number of dimes:");

      int [tex]$\text{dimes =}$[/tex] input.nextInt();

      System[tex]$\text{.out.}$[/tex]print("Enter number of nickels:");

      int nickels = input.nextInt();

      System[tex]$\text{.out.}$[/tex]print("Enter number of pennies:");

      int [tex]$\text{pennies = }$[/tex] input.nextInt();

      // computing dollors

      double dollars = (double) ((quarters*0.25)+(dimes*0.10)+(nickels*0.05)+(pennies*0.01));

      System[tex]$\text{.out.}$[/tex]format("You have : $%.2f",dollars);

  }

}

Answer:

true

Explanation:

Answer: its c

Explanation:

Answer:

Levi Ackerman  is the tritagonist of the anime/manga series Attack on Titan. He is a Captain in the Survey Corps , known to be the strongest soldier alive. He has a harsh and unsocial personality, but is well-regarded by his subordinates and he cares about their lives. Levi Ackerman spent his childhood in the Underground City. It was the slums of the Attack on Titan’s entire world

Explanation:

Answer:

mass flow rate = 0.0534 kg/sec

velocity at exit = 29.34 m/sec

Explanation:

From the information given:

Inlet:

Temperature [tex]T_1 = -16^0\ C[/tex]

Quality [tex]x_1 = 0.2[/tex]

Outlet:

Temperature [tex]T_2 = -16^0 C[/tex]

Quality  [tex]x_2 = 1[/tex]

The following data were obtained at saturation properties of R134a at the temperature of -16° C

[tex]v_f= 0.7428 \times 10^{-3} \ m^3/kg \\ \\ v_g = 0.1247 \ m^3 /kg[/tex]

[tex]v_1 = v_f + x_1 ( vg - ( v_f)) \\ \\ v_1 = 0.7428 \times 10^{-3} + 0.2 (0.1247 -(0.7428 \times 10^{-3})) \\ \\ v_1 = 0.0255 \ m^3/kg \\ \\ \\ v_2 = v_g = 0.1247 \ m^3/kg[/tex]

[tex]m = \rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ m = \dfrac{1}{0.0255} \times \dfrac{\pi}{4}\times (1.7 \times 10^{-2})^2\times 6 \\ \\ \mathbf{m = 0.0534 \ kg/sec}[/tex]

[tex]\rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ A_1 =A_2 \\ \\ \rho_1v_1 = \rho_2v_2 \\ \\ \implies \dfrac{1}{0.0255} \times6 = \dfrac{1}{0.1247}\times (v_2)\\ \\ \\\mathbf{\\ v_2 = 29.34 \ m/sec}[/tex]

Answer: c

Explanation:

The received power if the effective aperture area of the receiving antenna is 20 m2 is 177.77 m2.

In physics, power is referred to as the rate of energy conversion or transfer over time. The unit of power in the SI system, often known as the International System of Units, is the Watt (W). A single joule per second is one watt.

Power was formerly referred to as activity in some research. A scalar quantity is power. As power is always a function of labor done, it follows that if a person's output varies during the day depending on the time of day, so will his power.

A measure of the pace at which energy is transferred, power is a physical quantity. As a result, it can be described as the pace of job completion relative to time.

Therefore, The received power if the effective aperture area of the receiving antenna is 20 m2 is 177.77 m2.

To learn more about Power, refer to the link:

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Answer:

the elongation of the metal alloy is 21.998 mm

Explanation:

Given the data in the question;

K = σT/ (εT)ⁿ

given that metal alloy true stress σT = 345 Mpa, plastic true strain εT = 0.02,

strain-hardening exponent n = 0.22

we substitute

K = 345 / [tex]0.02^{0.22[/tex]

K = 815.8165 Mpa

next, we determine the true strain

(εT) = (σT/ K)^1/n

given that σT = 412 MPa

we substitute

(εT) = (412 / 815.8165 )^(1/0.22)

(εT) = 0.04481 mm

Now, we calculate the instantaneous length

[tex]l_i[/tex] = [tex]l_0e^{ET[/tex]

given that [tex]l_0[/tex] = 480 mm

we substitute

[tex]l_i[/tex] =[tex]480mm[/tex] × [tex]e^{0.04481[/tex]

[tex]l_i[/tex] =  501.998 mm

Now we find the elongation;

Elongation = [tex]l_i - l_0[/tex]

we substitute

Elongation = 501.998 mm - 480 mm

Elongation = 21.998 mm

Therefore, the elongation of the metal alloy is 21.998 mm