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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.)
Physics
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

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A lowly high diver pushes off horizontally with a speed of 2.00 m/s from the platform edge 10.0 m above the surface of the water
VladimirAG [237]

Answer:

(a) 1.6m

(b) 6.86m

(c) 2.86s

Explanation:

This is a horizontal shooting problem, so we will use the following equations.

For horizontal distance x we use:

x(t)=x_{0}+v_{0}t

Where x(t) is the horizontal distance at a time t, x_{0} is the initial horizontal position which in this case will be zero: x_{0}=0. And v_{0} is the initial speed: v_{0}=2m/s

So to solve (a):

(a) At what horizontal distance from the edge is the diver 0.800 s after pushing off?

we have that the time is: t=0.8s so the horizontal distance is:

x(0.8)=(2m/s)(0.8s)1.6m

<u>The answer for (a) is 1.6m</u>

Now, to solve (b) we need the equation for vertical distance:

y(t)=y_{0}-\frac{1}{2}gt^2

Where y(t) is the vertical distance at a time t, y_{0} is the initial vertical distance: y_{0}=10m And g is the gravitational acceleration: g=9.81m/s^2 }

(b) At what vertical distance above the surface of the water is the diver just then?

The time is the same as in the last question: t=0.8s

y(0.8)=10m-\frac{1}{2}(9.81m/s^2)(0.8)^2

y(0.8)=10m-\frac{1}{2}(9.81m/s^2)(0.64s^2)

y(0.8)=10m-\frac{1}{2}(6.2784m)

y(0.8)=10m-(3.1392)

y(0.8)=6.86m

<u>the answer to (b) is 6.86m</u>

(c) At what horizontal distance from the edge does the diver strike the water?

To solve (c) we first need to know how long it took to reach the height of the water, that is, for what time y = 0

Using y(t)=y_{0}-\frac{1}{2}gt^2

if y(t)=0

0=10-\frac{1}{2}gt^2\\-10=-\frac{1}{2}gt^2\\20=gt^2\\\frac{20}{g}=t^2\\ \sqrt{\frac{20}{g}}=t

and since g=9.81m/s^2

t=\sqrt{\frac{20}{9.81} } =\sqrt{2.039}=1.43s

At t=1.43s the car hits the water, so the horizontal distance at that time is:

x(t)=x_{0}+v_{0}t

x(1.43)=(2m/s)(1.43s)=2.86m

<u>the answer to (c) is 2.86s</u>

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