The speed of sound in air at 0'c is 332m/s If it increases at the rate of 0.6m/s per degree, what will be the temperature when the velocity has increased to 344m/s
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The speed of sound in air at 0℃ = 332m/s
If it increases at the rate of 0.6m/s per degree,
Speed at x℃ = (332 + 0.6x)℃
Thus 344 = 332 + 0.6x
344 - 332 = 0.6x
12 = 0.6x
x = 12/ 0.6 = 20℃
Temperature is 20℃
If it increases at the rate of 0.6m/s per degree,
Speed at x℃ = (332 + 0.6x)℃
Thus 344 = 332 + 0.6x
344 - 332 = 0.6x
12 = 0.6x
x = 12/ 0.6 = 20℃
Temperature is 20℃
abhaybiradar02:
But my friend said 200 'c
no
it's printed wrong in the 9th tb
the correct answer is 20*
I think answer is 7.2*
what are you say
ya you are right brother
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I hope you will understand it
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