All categories

2 years ago

ANSWER ASAP What is the point system? this is for driving

Engineering

1 answer:

lara31 [8.8K]2 years ago

3 0

Answer:

A point system assigns point values to traffic violations for each state. When the DMV receives a conviction notice from the court, the offense is entered on your driving record and points are assigned. These points are counted during a specified time period.

Explanation:

You might be interested in

Build a 3-input XOR gate using a 3x8 decoder and an Or gate.please Quickly i have final

lana [24]

Answer:

Take a 3 to 8 decoder with active low outputs

Assuming you are familiar with the functioning of decoders,

The three inputs of decoder of course are the first, second and the carry bit which you feed to the subtractor.

Next we examine the truth table of the full subtractor i formatted in the picture.

Then write the minterms for the difference output and borrow output from the given truth table picture I have mentioned before!!

Explanation:

hopi it to help you!!

7 0

2 years ago

What is the total inductance of a circuit that contains two 10 uh inductors connected in a parallel?

kolbaska11 [484]

Answer:

5 microhenries

Explanation:

The effective value of inductors in parallel "add" in the same way that resistors in parallel do. The value is the reciprocal of the sum of the reciprocals of the inductances that are in parallel.

10 uH ║ 10 uH = 5 uH

The effective inductance is 5 uH.

6 0

3 years ago

One - tenth kilogram of air as an ideal gas with k= 1.4 executes a carnot refrigeration cycle as shown i fig. 5,16, the isotherm

maria [59]

Answer:

Hello your question is incomplete attached below is the missing part

a) p1 = 454.83 kPa, p2 = 283.359 Kpa , p3 = 536.423 kpa , p4 = 860.959kPa

b) W12 = 3.4 kJ, W23 = -3.5875 KJ, W34 = -4.0735 KJ, W41 = 3.5875 KJ

c) 5

Explanation:

Given data:

mass of air ( m ) = 1/10 kg

adiabatic index ( k ) = 1.4

temperature for isothermal expansion = 250K

rate of heat transfer ( Q12 ) = 3.4 KJ

temperature for Isothermal compression ( T4 ) = 300k

final volume ( V4 ) = 0.01m ^3

a) Calculate the pressure, in Kpa, at each of the four principal states

from an ideal gas equation

P4V4 = mRT4 ( input values above )

hence P4 = 860.959kPa

attached below is the detailed solution

b) Calculate work done for each processes

attached below is the detailed solution

C) Calculate the coefficient of performance

attached below is detailed solution

6 0

2 years ago

Obtain a relation for the logarithmic mean temperature difference for use in the LMTD method?

kolezko [41]

Answer:

The log mean temperature difference is:

ΔT,lm=(ΔT1-ΔT2)/㏑(ΔT1/ΔT2)

Explanation:

To evaluate the equivalent average temperature difference between two fluids we consider a parallel-flow double-pipe heat exchanger (see attached diagram). The temperature of the hot and cold fluids is large at the inlet of the heat exchanger and decreases exponentially toward the outlet.

We can assume that the outer surface of the heat exchanger is well insulated and that heat transfer only occurs between the two fluids. We can also assume negligible kinetic and potential. The energy balance on each fluid can be written as the rate of heat loss from the hot fluid is equal to the rate of heat gained by the cold fluid in any section of the heat exchanger:

Q = -m,h×c,ph×dT,h (1)

where Q=rate of heat loss, m=mass flow rate, c,ph=heat capacity of the hot fluid, dT,h= differential temperature of the hot fluid

Q = m,c×c,pc×T.c (2)

where Q=rate of heat loss, m=mass flow rate, c,ph=heat capacity of the cold fluid, dT,h= differential temperature of the cold fluid

The temperature of the hot fluid change is negative and is added to make Q positive. Solving equations 1 and 2 in terms of dT:

dT.h = - Q/(m,h×c,ph)

dT.c = Q/(m,c×c,pc)

and taking the difference:

dT,h-dT,c= d(T,h - T,c) = -Q(1/(m,h×c,ph) + 1/(m,c×c,pc)) (3)

The heat transfer rate in the differential section of the heat exchanger can be expressed as:

Q = U(T,h-T,c)×dA,s (4)

where U=overall heat transfer coefficients, dA,s = differential sectional area. Substitute equation 4 into 3:

d(T,h - T,c)/(T,h - T,c) = -U×dA,s×(1/(m,h×c,ph) + 1/(m,c×c,pc)) (5)

Integrating equation 5:

㏑((T,h out - T,c out)/(T,h in - T,c in)) = -U×A,s×(1/(m,h×c,ph) + 1/(m,c×c,pc)) (6)

The first law of thermodynamics requires the rate of heat transfer from hot and cold fluid to be equal.

Q= m×c, pc×(T, c out-T, c in) (7)

Q= m×c, ph×(T,h out-T, h in) (8)

Solve equations 7 and 8 for m,c×c, pc and m,h×c, ph and substituting into equation 6:

Q = U×A,s×ΔT,lm

Where the log mean temperature difference is:

ΔT,lm=(ΔT1-ΔT2)/㏑(ΔT1/ΔT2)

Download pdf

8 0

3 years ago

A rectangular plate with a width of 19 m and a height of 12 m is located 4 m below a water surface. The plate is tilted and make

Kruka [31]

Answer:

F = 33,324,295.32N

Explanation:

We will first of all find the height at which the plate is inclined. We use sine rule in this case

SinΘ = Opp/hyp

Opp=h, hyp=19m, Θ=35°

h = 19xSin35 = 10.899m

Therefore height h=10.899+4 = 14.899m

We then Calculate Area of the plate

Area = 12x19 = 228m²

Finally, we use an online software to calculate the Hydrostatic pressure

The result from the online computation is attached.

The pressure is p = 146159.19Pa

But pressure p is

Pressure=Force/Area

Making Force the subject

Hydrostatic Force = Pressure x Area

F= 146159.19 x 228 = 33,324,295.32N

F = 33,324,295.32N

4 0

3 years ago