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3 years ago

An explosion occurs 34 km away. The time it takes for its sound to reach your ears, traveling at 343 m/s, is

Physics

2 answers:

madam [21]3 years ago

7 0

Answer: 99.1 seconds

Explanation: You know that the total distance is 34km, and the speed of the sound is 343m/s.

Now, the total time that takes to something with a given velocity, to reach a distance D, is:

T = D/v

where D = 34km and v = 343m/s

Now, here we have different units, so we must convert the kilomeres into meters, this can be done as: 1km = 1000m then, 34km = 1000*34m = 34000m

Then the total time is:

T = 34000m/343m/s = 99.1 seconds

Ostrovityanka [42]3 years ago

6 0

Answer:

99 s

Explanation:

distance = rate × time

34000 m = (343 m/s) t

t = 99 s

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3 years ago

An object is moving along a straight line, and the uncertainty in its position is 1.90 m.

just olya [345]

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$2.78\times 10^{-35}\ \text{kg m/s}$

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$\Delta x$ = Uncertainty in position = 1.9 m

$\Delta p$ = Uncertainty in momentum

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m = Mass of object

From Heisenberg's uncertainty principle we know

$\Delta x\Delta p\geq \dfrac{h}{4\pi}\\\Rightarrow \Delta p\geq \dfrac{h}{4\pi\Delta x}\\\Rightarrow \Delta p\geq \dfrac{6.626\times 10^{-34}}{4\pi\times 1.9}\\\Rightarrow \Delta p\geq 2.78\times 10^{-35}\ \text{kg m/s}$

The minimum uncertainty in the momentum of the object is $2.78\times 10^{-35}\ \text{kg m/s}$

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$m=0.045\ \text{kg}$

Uncertainty in velocity is given by

$\Delta p\geq m\Delta v\geq 2.78\times 10^{-35}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{0.045}\\\Rightarrow \Delta v\geq 6.178\times 10^{-34}\ \text{m/s}$

The minimum uncertainty in the object's velocity is $6.178\times 10^{-34}\ \text{m/s}$

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$m=9.11\times 10^{-31}\ \text{kg}$

$\Delta v\geq \dfrac{\Delta p}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{9.11\times 10^{-31}}\\\Rightarrow \Delta v\geq 0.31\times 10^{-4}\ \text{m/s}$

The minimum uncertainty in the object's velocity is $0.31\times 10^{-4}\ \text{m/s}$ .

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