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3 years ago

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Consider a thin circular disk that has been heated up to 400 °C and then left inside a chamber to cool down. The chamber surface

is kept at 40 °C and the air inside is maintained at 25 °C. Assume that the disk is held at the center of the chamber, and we can ignore heat conduction effects in this problem. The disk surface property is known as ε=0.65. For the diameter of D=200 mm, what is the total rate of heat transferfrom the disk?

1 answer:

kogti [31]3 years ago

7 0

Answer:

hello your question lacks the required diagram attached below is the complete question with the required diagram

answer : Qtotal = 807.4 Mw

Explanation:

Given Data :

disk properties :

∈ = 0.65

D = 200 mm

Ts = 400⁰c

attached below is the detailed solution

The total rate of Heat transferred from the disk

Qtotal = 807.4 Mw

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Stan is driving north on his scooter at 8m/s, accelerates 11m/s (North) in 4s, drives a constant velocity for the next 15s, and

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A) Acceleration: $a_1 = 0.75 m/s^2, a_2 = 0, a_3 = -1.57 m/s^2$

B) The total displacement is 209.5 m north

C) The average velocity is 8.06 m/s north

Explanation:

A)

Acceleration is defined as:

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time taken for the velocity to change from u to v

Here we have:

- In the first segment,

u = 8 m/s north

v = 11 m/s north

t = 4 s

So the acceleration is

$a_1 = \frac{11-8}{4}=0.75 m/s^2$ (north)

- In the second segment, Stan drives at a constant velocity: so the final velocity is equal to the initial velocity,

u = v

Therefore, the acceleration is zero: $a_2 = 0$

- In the third segment,

u = 11 m/s (north)

v = 0 (he comes to a stop)

t = 7 s

So the acceleration is

$a=\frac{0-11}{7}=-1.57 m/s^2$

And the negative sign means the acceleration is south, opposite to the direction of motion.

B)

In a uniformly accelerated motion, the displacement can be calculated as:

$s=ut+\frac{1}{2}at^2$

where

u is the initial velocity

a is the acceleration

t is the time

- For the first segment, we have

$u = 0\\a = 0.75 m/s^2\\t=4 s$

So the displacement is

$s_1 = 0+\frac{1}{2}(0.75)(4)^2=6 m$

- For the second segment, we have

$u = 11 m/s\\a = 0\\t=15 s$

So the displacement is

$s_2 = (11)(15)+0=165 m$

- For the third segment, we have

$u = 11\\a = -1.57 m/s^2\\t=7 s$

So the displacement is

$s_3 = (11)(7)+\frac{1}{2}(-1.57)(7)^2=38.5 m$

So the total displacement is:

s = 6 m + 165 m + 38.5 m = 209.5 m

In the north direction (positive direction)

C)

The average velocity is given by:

$v=\frac{d}{t}$

where

d is the total displacement

t is the total time

Here we have:

d = 209.5 m

t = 26 s

Therefore, the average velocity is

$v=\frac{209.5}{26}=8.06 m/s$ (north)

Learn more about accelerated motion:

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2 years ago

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A wire is wrapped 200 times around a hollow cylinder of cross-sectional area 12 cm2 and length 10 cm. Assuming the turns are spr

Sonja [21]

Answer:

The answer is 10.857mJ

Explanation:

The energy stored in this solenoid is given by the below mentioned equation,

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Plugging this into the energy equation you obtain the equation for the total energy stored in the magnetic field of the solenoid, given by,

$U = u_{0} *N^{2}*A*I^{2}/(2*L)$

where $u_{0}$ is the permeability of free space which equals to $u_{0} = 4*\pi *10^{-7} T.m/A$ . Plugging all the quantities into the above equation from the data in the question after converting to standard units. of meters instead of centimeters, we get for the energy stored in the coil,

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A 103 kg physics professor has fallen into the Grand Canyon. Luckily, he managed to grab a branch and is now hanging 93 m below

siniylev [52]

Answer:

125.83672 seconds

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Power is given by

$P=\frac{W}{t}\\\Rightarrow t=\frac{W}{P}\\\Rightarrow t=\frac{93874.2}{746}\\\Rightarrow t=125.83672\ seconds$

The time taken by the horse to pull the professor is 125.83672 seconds

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3 years ago