Ask question

Ask question

All categories

Norma-Jean [14]

2 years ago

9

Calculate the net convective heat transfer loss rate from both vertical surfaces of the plate to the air. (hint: h values should

be the same for both surfaces of the vertical plate.)

1 answer:

pychu [463]2 years ago

7 0

Net convective heat transfer loss rate (gnet)

Gnet = 2h, A (Ts - T∞)

=2(4.51) (0.001) (25)

Gnet = 0.2250 W.

There are two types of convection: natural convection and forced convection. Natural convection is caused by differences in the density of liquids due to differences in temperature (eg "hot air rises"). Global atmospheric circulation and local weather phenomena (including wind) are due to convective heat transfer.

Newton's Law of Cooling is a widely used equation for calculating both forced and natural net convective. The formula for Newton's Law of Cooling is Q = h A ΔT. Equation parameters and their typical US and SI.

Learn more about Net convective at

brainly.com/question/9382711

#SPJ4

You might be interested in

The swinging pendulum has 10 joules of potential energy at its maximum height at points (1) and (5). If the mass of the pendulum

SVEN [57.7K]

The speed of the pendulum at point 3 is 1.4 m/s

Explanation:

We can solve this problem by using the law of conservation of energy. In fact, the mechanical energy of the pendulum (which is the sum of his potential energy + his kinetic energy) must be conserved. So we can write:

$U_1 +K_1 = U_3 + K_3$

where

$U_1$ is the initial potential energy, at the highest position

$K_1$ is the initial kinetic energy, at the highest position

$U_3$ is the final potential energy, at the lowest position

$K_3$ is the final kinetic energy, at the lowest position

We are told that:

$U_1 = 10 J$ is the potential energy of the pendulum at the maximum height

$K_1 = 0$ (when the pendulum is at maximum height, the speed is zero, so the kinetic energy is zero)

$U_3 = 0$ (potential energy is zero at the lowest position)

Therefore,

$K_3 = U_1 = 10 J$

Kinetic energy can be rewritten as

$K_3 = \frac{1}{2}mv^2$

where

m = 10 kg is the mass of the pendulum

v is its speed at point 3

Solving for v,

$v=\sqrt{\frac{2K_3}{m}}=\sqrt{\frac{2(10)}{10}}=1.4 m/s$

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

4 0

3 years ago

On a spacecraft, two engines are turned on for 789 s at a moment when the velocity of the craft has x and y components of v0x =

ddd [48]

Answer:

Explanation:

Given

time for which spacecraft turned on is 789 s

$v_{0x}=5410 m/s$

$v_{0y}=8100 m/s$

$x=2.78\times 10^6 m$

$y=4.73\times 10^6 m$

we know $s=ut+\frac{at^2}{2}$

$2.78\times 10^6=5410\times 789+\frac{a_x(789)^2}{2}$

$2.78=4.268+a_x\times 0.622$

$a_x=-2.392 m/s^2$

For y component

$s=ut+\frac{at^2}{2}$

$4.73\times 10^6=8100\times 789+\frac{a_y(789)^2}{2}$

$4.73=6.39+a_y\times 0.622$

$a_y=-2.66 m/s^2$

4 0

4 years ago

PLEASE HELP!!!!

Aleks [24]

The answer is step by step 65

4 0

3 years ago

Read 2 more answers

A parallel-plate capacitor with only air between its plates is charged by connecting the capacitor to a battery. The capacitor i

Delicious77 [7]

Answer:

3.91

Explanation:

Given that

Final reading of the voltmeter, V2 = 45 v

Initial reading of the voltmeter, V1 = 11.5 v

The dielectric constant k, of a material is usually given as

k = V2/V1

k = 45 / 11.5

k = 3.91

Therefore, the dielectric constant of the material as we've calculated above is sure to be 3.91.

I hope that helps you understand

5 0

4 years ago

I am a bit confused about this question.

gavmur [86]

How do you know when something is moving ? You ALWAYS have to compare it to something else. If the object in question changes its distance or direction from your house, or from your big toe, or from a stake in the ground in your front yard, then you say it's moving. The thing is: There's ALWAYS something else to compare it to.

I assume you're sitting on the couch now, staring at the TV, or at your computer, or at your phone. Compared to the couch, or to the tree in your front yard, or to somebody sitting on top of Mt. Everest, or to downtown Jerusalem, you're NOT moving. Your distance and direction from the reference point isn't changing.

BUT ... what if you compare yourself to somebody sitting at the North pole of the Sun ? He has to keep turning his eyes to watch you (because the Earth including you is in orbit around the sun). So your direction from him keeps changing, and 'relative' to him (compared to him), you're definitely moving.

Now let's go a little farther:

You're sitting in a comfy seat, reading a book that's in your lap. Maybe you're even getting sleepy. You're sitting still in the seat, and the book in your lap isn't moving.

SURPRISE ! Your comfy seat is in Row-27 of a passenger jet, and you're flying to Seattle to visit your Grandma. right now, you're just passing over Casper, Wyoming, and there's somebody down on the ground playing with a telescope. He looks at your airplane, and HE says that you, the seat you're sitting in, and your book are ALL moving at almost 500 miles an hour.

The difference is: YOU're comparing your book to the seat in front of you, and YOU say the book is not moving. The guy with the telescope is comparing the book to the ground he's standing on, and HE says your book is moving west at 500 miles an hour.

You're BOTH correct. The description of ANY motion always depends on what you're comparing to. If you're about to ask "What's the REAL motion of the book ?", then I'm sorry. There's NO SUCH THING as 'REALLY'. It always depends on what you're comparing to. Nine people can be watching the same object, and they can have nine different descriptions of its motion, and they're ALL correct. They're just comparing the object to different things in their own neighborhood, and the nine things are all moving in different ways.

The bottom line: MOTION IS ALWAYS RELATIVE (to something else).

8 0

3 years ago