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3 years ago

8

Two containers initially contain dry air at 40 °C. Water is added to both containers so that the relative humidity is 30% in con

tainer #1, and 80% in container #2. The wet bulb temperature in container #1 is _________ the wet bulb temperature in #2.
a. less than
b. greater than
c. the same as

1 answer:

enot [183]3 years ago

8 0

Answer:

A) Wet bulb temperature of #1 is less than that of #2

Explanation:

This can be gotten from pinpointing the states of the two containers on a psychometric chart.

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Find the resistance of a circuit that draws 4 amperes with 8 volts applied?

vodka [1.7K]

Answer:

2 ohms

Explanation:

V = I * R

8 = 4 * R

8 / 4 = R

R = 2 ohms

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Explain how smart materials can be used by manufacturers to improve health and safety for children's products and goods.

Ierofanga [76]

...simplify devices, reducing weight and the chance of failure.

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2 years ago

Which starting circuit uses fuses, switches, and smaller wires to energize a relay and solenoid?

zysi [14]

Answer:

Option B (Starter Control Circuit) is the right option.

Explanation:

This same switching is normally put upon this isolated side of something like the transmission Arduino microcontroller throughout the configuration that is using the ignition just to command the broadcast.
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All other given options are not related to the given instance. So the above option is correct.

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3 years ago

The particle travels along the path defined by the parabola y=0.5x2, where x and y are in ft. If the component of velocity along

JulsSmile [24]

Answer:

D=41.48 ft

$a=54.43\ ft/s^2$

Explanation:

Given that

y=0.5 x²

Vx= 2 t

We know that

$V_x=\dfrac{dx}{dt}$

At t= 0 ,x=0

$x=\int V_x.dt$

At t= 3 s

$x=\int_{0}^{3} 2t.dt$

$x=[t^2\left\right ]_0^3$

x= 9 ft

When x= 9 ft then

y= 0.5 x 9² ft

y= 40.5 ft

So distance from origin is

x= 9 ft ,y= 40.5 ft

$D=\sqrt{9^2+40.5^2} \ ft$

D=41.48 ft

$a_x=\dfrac{dV_x}{dt}$

Vx= 2 t

$a_x= 2\ ft/s^2$

At t= 3 s , x= 9 ft

y=0.5 x²

$a_y=\dfrac{d^2y}{dt^2}$

y=0.5 x²

$\dfrac{dy}{dt}=x\dfrac{dx}{dt}$

$\dfrac{d^2y}{dt^2}=\left(\dfrac{dx}{dt}\right)^2+x\dfrac{d^2x}{dt^2}$

Given that

$\dfrac{dx}{dt}=2t$

$\dfrac{dx}{dt}=2\times 3$

$\dfrac{dx}{dt}=6\ ft/s$

$a_y=\dfrac{d^2y}{dt^2}=6^2+9\times 2\ ft/s^2$

$a_y=54\ ft/s^2$

$a=\sqrt{a_x^2+a_y^2}\ ft/s^2$

$a=\sqrt{2^2+54^2}\ ft/s^2$

$a=54.43\ ft/s^2$

7 0

3 years ago

A hair dryer is basically a duct of constant diameter in which a few layers of electric resistors are placed. A small fan pulls

Inessa05 [86]

Answer:

the percent increase in the velocity of air is 25.65%

Explanation:

Hello!

The first thing we must consider to solve this problem is the continuity equation that states that the amount of mass flow that enters a system is the same as what should come out.

m1=m2

Now remember that mass flow is given by the product of density, cross-sectional area and velocity

(α1)(V1)(A1)=(α2)(V2)(A2)

where

α=density

V=velocity

A=area

Now we can assume that the input and output areas are equal

(α1)(V1)=(α2)(V2)

$\frac{V2}{V1} =\frac{\alpha1 }{\alpha 2}$

Now we can use the equation that defines the percentage of increase, in this case for speed

$i=(\frac{V2}{V1} -1) 100$

Now we use the equation obtained in the previous step, and replace values

$i=(\frac{\alpha1 }{\alpha 2} -1) 100\\i=(\frac{1.2}{0.955} -1) 100=25.65$

the percent increase in the velocity of air is 25.65%

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3 years ago