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Murrr4er [49]
3 years ago
15

A cubical furnace with external dimensions of 0:6 m is constructed from reclay brick. If the wall thickness is 40 mm, the inner

surface temp 630C , and the outer surface temperature is 70C, calculate the heat loss (in kW) from the furnace.
Physics
1 answer:
Alja [10]3 years ago
8 0

Answer:

Explanation:

surface area of furnace = 6 x .6²

=2.16 m²

Rate of heat loss = k A ( θ₂ -θ₁ ) / t where k is coefficient of thermal conductivity , A is surface involved , θ₂ -θ₁  is temperature difference and t is thickness .

For fire clay thermal conductivity

= .05 W / m . K

.05 x 2.16 x ( 630 - 70 ) / .04

= 1512 W

= 1.5  kW .

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A thin insulating rod is bent into a semicircular arc of radius a, and a total electric charge Q is distributed uniformly along
storchak [24]

Answer:

v = \frac{kQ}{a}  

Explanation:

We define the linear density of charge as:

\lambda = \frac{Q}{L}

     Where L is the rod's length, in this case the semicircle's length L = πr

The potential created at the center by an differential element of charge is:

dv = \frac{kdq}{r}

          where k is the coulomb's constant

                     r is the distance from dq to center of the circle

Thus.

v = \int_{}^{}\frac{kdq}{a}  

v = \frac{k}{a}\int_{}^{}dq

v = \frac{kQ}{a}     Potential at the center of the semicircle

4 0
3 years ago
What is the bending of a wave around a barrier?
Mamont248 [21]
That's wave 'diffraction'.
3 0
3 years ago
1<br> A truck increases its speed from 15 m/s to 60 m/s in 15 s. Its acceleration is
MAVERICK [17]
Acceleration = (Vf - Vi)/t
Since Vf= 60m/s
Vi= 15m/s
T= 15s
=> a= (60m/s - 15m/s)/15s
= 3
So the acceleration is 3m/s^2
3 0
3 years ago
A truck pushes a pile of dirt horizontally on a frictionless road with a net force of 20\, \text N20N20, start text, N, end text
meriva

Answer:

300 Nm ; 300 J

Explanation:

Given that:

Force (F) = 20 N

Distance (d) = 15 m

The kinetic energy (Workdone) = Force * Distance

Kinetic Energy = 20N * 15m

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K. E = 1/2

4 0
3 years ago
Read 2 more answers
A physics professor demonstrates the Doppler effect by tying a 600 Hz sound generator to a 1.0-m-long rope and whirling it aroun
cluponka [151]

Answer:Highest frequency  =618.89Hz

Lowest frequency=582.22Hz

Explanation:

 The linear velocity of a sound generator  is related to angular velocity and is given as

Vs = rω  where

r = the radius of circular path = 1.0 m

ω is the angular velocity of the sound generator. = 100 rpm

1 rev/min = 0.10472 rad/s

100rpm =10.472 rad/ s

Vs = rω

= 1m x 10.472rad/ s=  10.472m/s

A) Highest frequency  heard by a student in the classroom = Maximum frequency. Using the Doppler effect formulae,

f max = (v/ v-vs) fs

Where , v is the speed of the sound in air at 20 degrees celcius =

343 metres per second

vs is the linear velocity of the sound generator=10.472m/s

fs is the frequency of the sound generator= 600 Hz

f max = (343/ 343 - 10.472) x 600

=343/332.528) x600

=618.89Hz

B) Lowest frequency  heard by a student in the classroom = Minimum frequency

f min = (v/ v+vs) fs

(343/ 343 + 10.472) x 600

=343/353.472) x 600

=582.22hz

5 0
3 years ago
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