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Natalka [10]
3 years ago
15

A particle moves on a line away from its initial position so that after t hours it is s = 6t2 + 2t miles from its initial positi

on. find the average velocity of the particle over the interval [1, 4]. include units in your answer.
Physics
1 answer:
docker41 [41]3 years ago
6 0
<span>32 mph
   First, let's calculate the location of the particle at t=1, and t=4
 t=1
 s = 6*t^2 + 2*t
 s = 6*1^2 + 2*1
 s = 6 + 2
 s = 8 t = 4
 s = 6*t^2 + 2*t s = 6*4^2 + 2*4
 s = 6*16 + 8
 s = 96 + 8
 s = 104
   So the particle moved from 8 to 104 over the time period of 1 to 4 hours. And the average velocity is simply the distance moved over the time spent. So:
 avg_vel = (104-8)/(4-1) = 96/3 = 32
   And since the units were miles and hours, that means that the average speed of the particle over the interval [1,4] was 32 miles/hour, or 32 mph.</span>
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A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per un
Zarrin [17]

Answer:

steady state temperature =88.7deg C

t=time within  1 deg C of it steady state is 8.31s

Explanation:

A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per unit length of 0.01 Ω/m. If a current of 100 A flows through the wire and the convection coefficient is 500W/m2K, what is the steady state temperature of the wire? From the time the current is applied, how long does it take for the wire to reach a temperature within 1-degC of the steady state value? The density of the wire is 8,000kg/m3, its heat capacity is 500 J/kgK and its thermal condu

The diameter of the wire is known to be=1mm

properties=

The density of the wire is 8,000 kg/m3,

heat capacity is 500 J/kgK

themal conductivity is 20W/m.K

electrical resistance per unit length of 0.01 Ω/m

from lump capavity method

B_{i} =\frac{hr/2}{k}

500*(2.5*10^-4)/20

0.006<0.1

we know also, to find steady state temperature

\piDh(T-Tinf)=I^{2} R_{e}

make T the subject of the equation , we have

T=25+\frac{100^2*0.01}{\pi*0.001*500 }

T=88.7 degC

rate of chnage in temperature

dT/dt=\frac{I^2*Re}{rho*c*\pi*D^2/4 } -\frac{4h}{rho*c*D} (T-Tinf)

at t=o and integrating both sides\frac{T-Tinf-(I^2*Re/\pi*Dh) }{Ti-Tinf-(I^2*Re/\pi*Dh } =exp\frac{-4ht}{rho*c*D}

we have

\frac{87.7-25-63.7}{25-25-63.7} =exp\frac{4*500t}{8000*500*0.001}

t=8.31s

steady state temperature =88.7deg C

t=time within  1 degC of it steady stae is 8.31s

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