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BlackZzzverrR [31]
3 years ago
13

A bottle with a volume of 193 U. S. fluid gallons is filled at the rate of 1.9 g/min. (Water has a density of 1000 kg/m3, and 1

U.S. fluid gallon = 231 in.3.) How long does the filling take?
Physics
1 answer:
dlinn [17]3 years ago
8 0

Answer:

0.732 years

Explanation:

We are given that

Volume of fluid filled in bottle=193 U.S

1 U.S fluid  gallon=231 cubic inch

193 U.S gallon=231\times 193=44583in^3

1 in^3=1.63871\times 10^{-5}m^3

44583in^3=44583\times 1.63871\times 10^{-5}=0.731m^3

Density of water=1000kg/m^3

Mass=Volume\times density

Using the formula

Mass=0.731\times 1000=731kg

1kg =1000g

731kg=731\times 1000=731\times 10^3g

By using 1000=10^3

1.9 g of fluid takes time to fill in the bottle=1 min

1 g of fluid takes time to fill in the bottle=\frac{1}{1.9}min

731\times 10^3g of fluid takes time to fill in the bottle=\frac{1}{1.9}\times 731\times 10^3min

Time=384.7\times 10^3min

1min=\frac{1}{60\times 24\times 365}years

Time=384.7\times 10^3\times \frac{1}{60\times 24\times 365}years

Time =0.732 years

Hence, it takes to filling the 193 U.S fluid gallons in bottle=0.732 years

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Harlamova29_29 [7]

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A kangaroo jumps straight up to a vertical height of 1.66 m. How long was it in the air before returning to Earth? Express your
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Answer:

The kangaroo was 1.164s in the air before returning to Earth

Explanation:

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x = x_{0} + V_{0}t + \frac{1}{2}at^2

Where:

x = Final distance

xo = Initial point

Vo = Initial velocity

a = Acceleration

t = time

We have the following values:

x = 1.66m      

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Vo = 0 m/s (each jump starts from the floor and from a resting position)

a = 9.8 m/s^2 (the acceleration is the one generated by the gravity of earth)

t =This is just the time it takes to the kangaoo reach the 1.66m, we don't know the value.

Now replace the values in the equation

x = x_{0} + V_{0}t + \frac{1}{2}at^2

1.66 = 0 + 0t + \frac{1}{2}9.8t^2

1.66 = 4.9t^2

\frac{1.66}{4.9}  = t^2

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It takes to the kangaroo 0.582s to go up and the same time to go down then the total time it is in the air before returning to earth is

t = 0.582s + 0.582s

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Zina [86]
<span>The initial speed, u of plane in terms of velocity of sound which may be taken as U
u=142/331=0.429*U
It crosses the sound barrier after says t seconds then we have 331-142=23.1*t or t is given 8.18 s exactly t=9/11s.
After 18 seconds the plane will traveling with velocity V
V=142+18*23.1=557.8 m/s==1.685*U</span>
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Pavlova-9 [17]

Answer:

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