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2 years ago

11

Jake uses a fire extinguisher to put out a small fire. When he squeezes the handle, the flame rettardant is released from the ex

tinguisher with considerable force. As a result, Jake has to reinforce his grip on the extinguisher to hold it in place as it pushes back into his hands. The projectile in this scenario is the . The recoil in this scenario is the . The force on the flame rettardant is the force on the extinguisher. The mass of the flame rettardant is the mass of the extinguisher. The acceleration of the flame rettardant is the acceleration of the extinguisher.

2 answers:

Tpy6a [65]2 years ago

7 0

Can you attach a picture of the actual problem?

pogonyaev2 years ago

5 0

The projectile in this scenario is the flame r.

The recoil in this scenario is the extinguisher.

The force on the flame r is equal to the force on the extinguisher.

The mass of the flame r is less than the mass of the extinguisher.

The acceleration of the flame r is greater than the acceleration of the extinguisher.

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Water evaporating from a pond does so as if it were diffusing across an air film 0.15 cm thick. The diffusion coefficient of wat

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Answer:

The water level will drop by about 1.24 cm in 1 day.

Explanation:

Here Mass flux of water vapour is given as

$j_{H_2O}=\frac{D}{l} \bigtriangleup c$

where

$j_{H_2O}$ is the mass flux of the water which is to be calculated.

D is diffusion coefficient which is given as $0.25 cm^2/s$

l is the thickness of the film which is 0.15 cm thick.
$\bigtriangleup c$ is given as

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT}$

In this

$P_{sat}$ is the saturated water pressure, which is look up from the saturated water property at 20°C and 0.5 saturation given as 2.34 Pa

$P_a$ is the air pressure which is given as 0.5 times of $P_{sat}$

R is the universal gas constant as $8.314 kJ/kmol-K$

T is the temperature in Kelvin scale which is $20+273= 293K$

By substituting values in the equation

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT} \\ \bigtriangleup c= \frac{P_{sat}-0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5 \times 2.34}{8.314 \times 293} \\\bigtriangleup c= 0.48 mol/m^3$

Converting $\bigtriangleup c$ into $cm^3/cm^3$

As 1 mole of water 18 $cm^3$ so

$\bigtriangleup c= 0.48 mol/m^3 \\ \bigtriangleup c= 0.48 \times 18 \times 10^{-6} cm^3/cm^3 \\ \bigtriangleup c= 8.64 \times 10^{-6} cm^3/cm^3$

Putting this in the equation of mass flux equation gives

$j_{H_2O}=\frac{D}{l} \bigtriangleup c \\ j_{H_2O}=\frac{0.25}{0.15} \times 8.64 \times 10^{-6} \\ j_{H_2O}=14.4 \times 10^{-6} cm/s$

For calculation of water level drop in a day, converting mass flux as

$j_{H_2O}=14.4 \times 10^{-6} \times 24 \times 3600 cm/day\\ j_{H_2O}=1.24 cm/day$

So the water level will drop by about 1.24 cm in 1 day.

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Answer:

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0.56

Explanation:

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Answer:

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where

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m is the mass of the car

g is the gravitational acceleration

v is the speed of the car

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Re-arranging the equation,

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And by substituting the data of the problem, we find the speed at which the car begins to skid:

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BabaBlast [244]

I could help
If I know what I was reading I’m sorry I need to translate

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We know that acceleration is change in velocity by time taken for that change.

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So acceleration of frog is 61.66 m/ $s^2$

o it is evident that frog is capable of remarkable accelerations.

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