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3 years ago

14

g As observed on earth, a certain type of bacteria is known to double in number every 24 hours. Two cultures of these bacteria a

re prepared, each consisting initially of one bacterium. One culture is left on earth and the other placed on a rocket that travels at a speed of 0.893c relative to the earth. At a time when the earthbound culture has grown to 256 bacteria, how many bacteria are in the culture on the rocket, according to an earth-based observer

1 answer:

tangare [24]3 years ago

8 0

Answer:

86.4 hrs

Explanation:

The amount of bacteria is initially 1

It doubles every 24 hrs.

After first 24 hrs, the amount = 2

After next 24 hrs = 4

After next 24 hrs = 8

After next 24 hrs = 16

After next 24 hrs = 32

After next 24 hrs = 64

After next 24 hrs = 128

After next 24 hrs = 256

Total time taken to reach 256 = 24 x 8 = 192 hrs

For the bacteria culture on the rocket that travels at a speed of 0.893c relative to the earth, this time is contracted by the relationship

t = t'(1 - ¥^2)^0.5

Where t is the contracted time =?

t' is the time on earth

¥ = v/c

Where v is the speed of the rocket

c is the speed of light

since v = 0.893c

¥ = 0.893

Substituting, we have

t = 192 x (1 - 0.893^2)^0.5

t = 192 x 0.2025^0.5

t = 192 x 0.45 = 86.4 hrs

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Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.

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Answer:0.061

Explanation:

Given

$T_C=300 k$

Temperature of soup $T_H=340 K$

heat capacity of soup $c_v=33 J/K$

Here Temperature of soup is constantly decreasing

suppose T is the temperature of soup at any instant

efficiency is given by

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$dW=Q(1-\frac{T_C}{T})$

$dW=c_v(1-\frac{T_C}{T})dT$

integrating From $T_H$ to $T_C$

$\int dW=\int_{T_C}^{T_H}c_v(1-\frac{T_C}{T})dT$

$W=\int_{T_C}^{T_H}33\cdot (1-\frac{300}{T})dT$

$W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}$

$W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]$

Now heat lost by soup is given by

$Q=c_v(T_C-T_H)$

Fraction of the total heat that is lost by the soup can be turned is given by

$=\frac{W}{Q}$

$=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}$

$=\frac{T_C-T_H-T_C\ln (\frac{T_C}{T_H})}{T_C-T_H}$

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Answer:

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