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3 years ago

14

A car increases its speed from 9.6 meters per second to 11.2 meters per second in 4.0 seconds. The average acceleration of the c

ar during this 4.0-second interval is

2 answers:

Nataly_w [17]3 years ago

6 0

A=(vf-vi)/t
a=(11.2-9.6)/4
a=0.4m/s^2

Gwar [14]3 years ago

5 0

The acceleration is the change of speed/velocity over time. Thus to calculate this you do (V1-V2)/T or (11.2-9.6)/4 or 0.4 m/s^2

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The pressure difference of the first bubble is $\Delta P _1 =10 J/m^3$

The pressure difference of the second bubble is $\Delta P _2 =20 J/m^3$

The pressure difference on the second bubble is higher than that of the first bubble so when the valve is opened pressure from second bubble will cause air to flow toward the first bubble making is bigger

Explanation:

From the question we are told that

The radius of the first bubble is $r_1 = 10 \ mm=0.01 \ m$

The radius of the second bubble is $r_2 = 5 \ mm = 0.005 \ m$

The surface tension of the soap solution is $s = 25 \ mJ/m^2 = 25*10^{-3} J/m^2$

Generally according to the Laplace's Law for a spherical membrane the pressure difference is mathematically represented as

$\Delta P = \frac{4 s}{R}$

Now the pressure difference for the first bubble is mathematically evaluated as

$\Delta P _1 = \frac{4 s}{r_1}$

substituting values

$\Delta P _1 = \frac{4 *25 *10^{-3}}{0.01}$

$\Delta P _1 =10 J/m^3$

Now the pressure difference for the second bubble is mathematically evaluated as

$\Delta P _2 = \frac{4 s}{r_1}$

$\Delta P _2 = \frac{4 *25 *10^{-3}}{0.005}$

$\Delta P _2 =20 J/m^3$

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