Connect wires to make the correct logic outputs.

If the old radiator is replaced with a new one that has longer tubes made of the same material and same thickness as those in th

Nookie1986 [14]

Answer: hello some parts of your question is missing attached below is the missing information

The radiator of a car is a type of heat exchanger. Hot fluid coming from the car engine, called the coolant, flows through aluminum radiator tubes of thickness d that release heat to the outside air by conduction. The average temperature gradient between the coolant and the outside air is about 130 K/mm . The term ΔT/d is called the temperature gradient which is the temperature difference ΔT between coolant inside and the air outside per unit thickness of tube

answer : Total surface area = 3/2 * area of old radiator

Explanation:

we will use this relation

K = $\frac{Qd }{A* change in T }$

change in T = ΔT

therefore New Area ( A ) = 3/2 * area of old radiator

Given that the thermal conductivity is the same in the new and old radiators

3 0

3 years ago

An inductor has a 50.0-Ω reactance when connected to a 60.0-Hz source. The inductor is removed and then connected to a 45.0-Hz s

nignag [31]

Given:

$X_{L} = 50.0 \ohm$

frequency, f = 60.0 Hz

frequency, f' = 45.0 Hz

$V_rms} = 85.0 V$

Solution:

To calculate max current in inductor, $I_{L(max)$ :

At f = 60.0 Hz

$X_{L} = 2\pi fL$

$50.0 = 2\pi\times 60.0\times L$

L = 0.1326 H

Now, reactance $X_{L}$ at f' = 45.0 Hz:

$X'_{L} = 2\pi f'L$

$X'_{L} = 2\pi\times 45.0\times 0.13263 = 37.5\ohm$

Now, $I_{L(max)$ is given by:

$I_{L(max) = \sqrt {\frac{2V_{rms}}{X'_{L}}}$

$I_{L(max) = \sqrt {\frac{2\times 85.0}{37.5}} = 2.13 A$

Therefore, max current in the inductor, $I_{L(max)$ = 2.13 A

7 0

3 years ago

Hello I need some help with this please. Pick a problem in your school or community that you think could be solved with technolo

bezimeni [28]

Answer:

An AI operated automatic garbage collection system

Explanation:

There is always an issue in my neighbourhood with the garbagemen coming on time so having an automatic system will help in the overall efficiency in the task

7 0

3 years ago

Read 2 more answers

A cold air standard gas turbine engine with a thermal efficiency of 56.9 % has a minimum pressure of 100 kP

Aleks04 [339]

Answer:

a) 5.2 kPa

b) 49.3%

Explanation:

Given data:

Thermal efficiency ( л ) = 56.9% = 0.569

minimum pressure ( P1 ) = 100 kpa

<u>a) Determine the pressure at inlet to expansion process</u>

P2 = ?

r = 1.4

efficiency = 1 - [ 1 / (rp) $\frac{r-1}{r}$ ]

0.569 = 1 - [ 1 / (rp)^0.4/1.4

1 - 0.569 = 1 / (rp)^0.285

∴ (rp)^0.285 = 0.431

rp = 0.0522

note : rp = P2 / P1

therefore P2 = rp * P1 = 0.0522 * 100 kpa

= 5.2 kPa

b) Thermal efficiency

Л = 1 - [ 1 / ( 10.9 )^0.285 ]

= 0.493 = 49.3%

4 0

3 years ago

A sign erected on uneven ground is guyed by cables EF and EG. If the force exerted by cable EF at E is 46 lb, determine the mome

Vikki [24]

Answer:

M_AD = 1359.17 lb-in

Explanation:

Given:

- T_ef = 46 lb

Find:

- Moment of that force T_ef about the line joining points A and D.

Solution:

- Find the position of point E:

mag(BC) = sqrt ( 48^2 + 36^2) = 60 in

BE / BC = 45 / 60 = 0.75

Hence, E = < 0.75*48 , 96 , 36*0.75> = < 36 , 96 , 27 > in

- Find unit vector EF:

mag(EF) = sqrt ( (21-36)^2 + (96+14)^2 + (57-27)^2 ) = 115 in

vec(EF) = < -15 , -110 , 30 >

unit(EF) = (1/115) * < -15 , -110 , 30 >

- Tension T_EF = (46/115) * < -15 , -110 , 30 > = < -6 , -44 , 12 > lb

- Find unit vector AD:

mag(AD) = sqrt ( (48)^2 + (-12)^2 + (36)^2 ) = 12*sqrt(26) in

vec(AD) = < 48 , -12 , 36 >

unit(AD) = (1/12*sqrt(26)) * < 48 , -12 , 36 >

unit (AD) = <0.7845 , -0.19612 , 0.58835 >

$M_d = \left[\begin{array}{ccc}0.7845&-0.19612&0.58835\\36&96&27\\-6&-44&12\end{array}\right]$

M_AD = 1835.73 + 116.49528 - 593.0568

M_AD = 1359.17 lb-in

3 0

3 years ago