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Orlov [11]
3 years ago
14

A water tank has a square base of area 8 square meters. Initially the tank contains 70 cubic meters. Water leaves the tank, star

ting at t=0, at the rate of 2 + 4 t cubic meters per hour. Here t is the time in hours. What is the depth of water remaining in the tank after 3 hours?
Physics
1 answer:
OLEGan [10]3 years ago
8 0

Answer:

Explanation:

Given

base of water tank A=8\ m^2

Initial volume of water V=70\ m^3

Water leaves the tank at the rate of

\frac{\mathrm{d} V}{\mathrm{d} t}=-(2+4t^3)

Integrating rate to get desired volume

at t=0,V=70\ m^3

at t=3\ hr,V=V

therefore

\int_{70}^{V}dV=\int_{0}^{3}-\left ( 2+4t\right )dt

V-70=2t+2t^2|_0^3

V-70=-24

V=46\ m^3

and volume=Area\times h

h=\frac{V}{A}

h=\frac{46}{8}

h=5.75\ m

Thus height of water remaining is 5.75\ m      

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ANEK [815]

Answer:

t = 13.7 s or t = 14 s with proper significant figures

Explanation:

The initial speed is 0 m/s since the car starts from rest, acceleration is 5.5 m/s2 and distance is 523 m.

Since we have initial speed, acceleration and distance we can use the following formula to find the time. We can now use algebra to work out our answer.

d = vt + \frac{1}{2}at²

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iVinArrow [24]

Answer:

Decreases/Reduces

Explanation:

Fill in the blank:

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When a gas is rapidly compressed (say, by pushing down a piston) its temperature increases. When a gas expands against a piston,
shusha [124]

Answer:

Explained in explanation

Explanation:

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