Give an example of one technology that is well matched to the needs of the environment, and one technology that is not.

A slight breeze is blowing over the hot tub above and yields a heat transfer coefficient h of 20 W/m2 -K. The air temperature is

patriot [66]

Answer:4050 W

Explanation:

Given

Heat transfer Coefficient(h)= $20 W/m^2-K$

Air temperature =75 F

surface area(A)= $7.5 m^2$

Temperature of hot tube is 102 F

We know heat transfer due to convection is given by

$Q=hA\left ( \Delta T\right )$

$Q=20\times 7.5\left ( 102-75\right )=4050 W$

7 0

3 years ago

A binary star system consists of two stars of masses m1m1m_1 and m2m2m_2. The stars, which gravitationally attract each other, r

d1i1m1o1n [39]

Answer:

$a_c_2=\dfrac{a_c_1\times m_1}{m_2}$

Explanation:

The question is: <em>Find the magnitude of the centripetal acceleration of the star with mass m₂</em>

The <em>centripetal acceleration</em> is the quotient of the centripetal force and the mass.

$a_c=\dfrac{F_c}{m}$

Thus, you can write the equations for each star:

$a_c_1=\dfrac{F_c_1}{m_1}$

$a_c_2=\dfrac{F_c_2}{m_2}$

As per Newton's third law, the centripetal forces are equal in magnitude. Then:

$a_c_1\times m_1=a_c_2\times m_2$

Now you can clear $a_c_2$ :

$a_c_2=\dfrac{a_c_1\times m_1}{m_2}$

6 0

4 years ago

Study the graphs below what type of trend and pattern in observed .

hichkok12 [17]

Answer:

negative and linear

Explanation:

its a straight line going down

8 0

3 years ago

Read 2 more answers

A steel with 1.0 % carbon is known as which one of the following: (a) eutectoid, (b) hypoeutectoid, (c) hypereutectoid, or (d) w

natta225 [31]

Answer:

Option c)

Explanation:

a.The carbon content of eutectoid steel, which is an carbon- iron alloy, is in between 0.80-0.83%

b. In hypoeutectoid steel, the carbon composition is less than 0.80% and is approximately 0.76%

c. In hypereutectoid steel, the carbon composition ranges from 0.80% - 2.0%

d. In wrought iron, the iron composition is quite low about 0.08%

3 0

3 years ago

A horizontal curve is being designed for a new two-lane highway (12-ft lanes). The PI is at station 250 + 50, the design speed i

kogti [31]

Answer

given,

Speed of vehicle = 65 mi/hr

= 65 x 1.4667 = 95.33 ft/s

e = 0.07 ft/ft

f is the lateral friction, f = 0.11

central angle,Δ = 38°

The PI station is

PI = 250 + 50

= 25050 ft

using super elevation formula

$e + f = \dfrac{v^2}{rg}$

$0.07 + 0.11 =\dfrac{95.33^2}{r\times 32.2}$

$r = \dfrac{95.33^2}{32.2\times 0.18}$

r = 1568 ft

As the road is two lane with width 12 ft

R = 1568 + 12/2

R = 1574 ft

Length of the curve

$L = \dfrac{\piR\Delta}{180}$

$L = \dfrac{\pi\times 1574\times 38}{180}$

L = 1044 ft

Tangent of the curve calculation

$T = R tan(\dfrac{\Delta}{2})$

$T = 1574 tan(\dfrac{38}{2})$

T = 542 ft

The station PC and PT are

PC = PI - T

PC = 25050 - 542

= 24508 ft

= 245 + 8 ft

PT = PC + L

= 24508 + 1044

=25552

= 255 + 52 ft

the middle ordinate calculation

$MO = R(1-cos\dfrac{\Delta}{2})$

$MO = 1574\times (1-cos\dfrac{38}{2})$

MO = 85.75 ft

degree of the curvature

$D = \dfrac{5729.578}{R}$

$D = \dfrac{5729.578}{1574}$

D = 3.64°

8 0

4 years ago