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Nookie1986 [14]

2 years ago

7

A baking dish is removed from a hot oven and placed on a cooling rack. As the dish cools down to 35 C from 175 C, its net radian

t power decreases to 12.0 W. What was the net radiant power of the baking dish when it was first removed from the oven? Assume that the temperature in the kitchen remains at 22 C as the dish cools down.

1 answer:

pogonyaev2 years ago

3 0

Answer:

net radiant power is 275.28 W

Explanation:

Given data

net radiant power P(F) = 12 W

initial temperature Ta = 175 °C = 175 + 273 = 448 K

final temperature Tb = 35 °C = 35 + 273 = 308 K

temperature in the kitchen T(f) = 22 °C = 22 + 273 = 295 K

to find out

net radiant power

solution

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Answer: 1 m/s

Explanation:

We have an object whose position $r$ is given by a vector, where the components X and Y are identified by the unit vectors $i$ and $j$ (where each unit vector is defined to have a magnitude of exactly one):

$r=[2 m + (2 m/s) t] i + [3 m - (1 m/s^{2})t^{2}] j$

On the other hand, velocity is defined as the variation of the position in time:

$V=\frac{dr}{dt}$

This means we have to derive $r$ :

$\frac{dr}{dt}=\frac{d}{dt}[2 m + (2 m/s) t] i + \frac{d}{dt}[3 m - (1 m/s^{2})t^{2}] j$

$\frac{dr}{dt}=(2 m/s) i - (\frac{1}{2} m/s^{2} t) j$ This is the velocity vector

And when $t=2s$ the velocity vector is:

$\frac{dr}{dt}=(2 m/s) i - (\frac{1}{2} m/s^{2} (2 s)) j$

$\frac{dr}{dt}=2 m/s i - 1m/s j$ This is the velocity vector at 2 seconds

However, the solution is not complete yet, we have to find the module of this velocity vector, which is the speed $S$ :

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Finally:

$S=1 m/s$ This is the speed of the object at 2 seconds

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2 years ago

A hockey stick strikes a hockey puck of mass 0.17 kg. If the force exterted on the hockey puck is 35.0 N and there is a force of

GREYUIT [131]

Answer:

$a=190\ m/s^2$

Explanation:

Mass of a hockey puck, m = 0.17 kg

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We need to find the acceleration of the hockey puck.

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Now, using second law of motion,

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$a=\dfrac{F}{m}\\\\a=\dfrac{32.3}{0.17}\\\\a=190\ m/s^2$

So, the acceleration of the hockey puck is $190\ m/s^2$ .

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2 years ago

The rear wheel on a clown’s bicycle has twice the radius of the front wheel. (a) When the bicycle is moving, is the linear speed

8_murik_8 [283]

Answer:

a). same as

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Explanation:

a). When a bicycle is moving, the linear speed at the top of the rear wheel is same as the linear speed at the top of the front wheel. Since the clown's bicycle is a rigid body, both the wheels that is the front wheel and the rear wheel will move with the same linear speed.

b). Since we know that angular speed varies inversely to the radius of the wheel.

That is ω = 1 / r

Since the rear wheel has twice the radius of that of the front wheel, therefore real wheel will have less angular speed than the front wheel.

Therefore, the angular speed of the rear wheel is less than the angular speed of the front wheel.

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1 year ago

At its lowest setting a centrifuge rotates with an angular speed of ω1 = 250 rad/s. When it is switched to the next higher setti

dalvyx [7]

Answer:

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Part(b): The angular displacement is $2629~rad$ .

Explanation:

Part(a):

If $\omega_{1},~\omega_{2}~and~\alpha$ be the initial angular speed, final angular speed and angular acceleration of the centrifuge respectively, then from rotational kinematic equation, we can write

$\alpha = \dfrac{\omega_{2} - \omega_{1}}{t}......................................................(I)$

where ' $t$ ' is the time taken by the centrifuge to increase its angular speed.

Given, $\omega_{i} = 250~rad~s^{-1}$ , $\omega_{f} = 750~rad~s^{-1}$ and $t = 9.5~s$ . From equation ( $I$ ), the angular acceleration is given by

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Part(b):

Also the angular displacement ( $\Delta \theta$ ) can be written as

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2 years ago

A long cylindrical rod of diameter 200 mm with thermal conductivity of 0.5 W/m⋅K experiences uniform volumetric heat generation

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Answer:

a, 71.8° C, 51° C

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Explanation:

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D(o) = 400 mm

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k(s) = 4 W/m.K

k(h) = 25 W/m².K

The expression for heat generation is given by

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q = π . 0.1² . L . 24000

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Thermal conduction resistance, R(conv) = 0.0318/L

Using energy balance equation,

Energy going in = Energy coming out

Which is = q, which is 754L

From the attachment, we deduce that the temperature between the rod and the sleeve is 71.8° C

At the same time, we find out that the temperature on the outer surface is 51° C

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