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Aleksandr [31]
3 years ago
13

A boy on the beach holds a spherical balloon filled with air. at 10:00am, the temperature on the beach is 20°c and the balloon h

as a diameter of 30 cm. two hours later, the balloon diameter is 30.5 cm. assume that the air is an ideal gas and that no air was lost or added, what is the temperature on the beach at noon?
Physics
1 answer:
Anon25 [30]3 years ago
6 0
For idea gases, volume is directly proportional to temperature. That is, an increase in temperature leads to increase in volume and vice versa.

Therefore,
V1/T1 = V2/T2 => T2 = (V2*T1)/V1

Assuming that the balloon is spherical in shape,

V= 4/3*pi*R^3.... In the formula for calculating T2, 4/3*pi cancels out.

R1 = 30/2 15 cm; R2 = 30.5/2 = 15.25 cm; T1 = 20+273.15 =293.15 K

Therefore,

T2 = (R2^3*T1)/R1^3 = (15.25^3*293.15)/15^3 = 308.05 K = 34.9 °C
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Answer:

The magnification is m  = 0.3674

Explanation:

From the question we are told that

  The  power of the lens is  P = -4.00 D(dioptre)

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 Generally the focal length is mathematically represented as

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converting to  cm  

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=>   v =  -15.8 \ cm

Generally the magnification is mathematically represented as

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=>    m  = \frac{- 15.8}{-43}

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</span>
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