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Nuetrik [128]
2 years ago
8

A tank, in the shape of a cube with each side measuring 2 meters, is half-full of water. Above the water is air that is pressuri

zed to 5 kPa. Calculate the resultant force on one side of the tank due to the air and water (ignore atmospheric pressure) and determine the height of this force above the bottom of the tank.
Physics
1 answer:
slavikrds [6]2 years ago
5 0

Answer:

Explanation:

The tank is half full so height of water level is at 1 m . Centre of gravity of water will be at height of .5 m . Pressure will act at this point .

Pressure = h d g where h is height of centre of mass of water column , d is density of water and g is acceleration due to gravity .

Pressure of water column  = .5 x 10³ x 9.8 = 4.9 k Pa .

Air is pressurized to 5 kPa so

resultant pressure  on one side of the tank due to the air and water

= 4.9 + 5 kPa = 9.9 kPa .

Total force on one face = pressure x area of one face under water

= 9.9 x 10³ x .5 x 2²

= 19.8 kN .

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A steel wire of length 4.7 m and cross section 3 x 103 m2 stretches by the same amount as a copper wire of length 3.5 m and cros
EleoNora [17]

Answer:

The ratio of the young's modulus of steel and copper is 1.79\times10^{-2}

(c) is correct option.

Explanation:

Given that,

Length of steel wire = 4.7 m

Cross sectionA = 3\times10^{-3}\ m^2

Length of copper wire = 3.5 m

Cross sectionA = 4\times10^{-5}\ m^2

We need to calculate the ratio of young's modulus of steel and copper

Using formula of young's modulus for steel wire

Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta l}{l}}

Y_{s}=\dfrac{Fl_{s}}{A_{s}\Delta l}....(I)

The young's modulus for copper wire

Y_{c}=\dfrac{Fl_{c}}{A_{c}\Delta l}....(II)

From equation (I) and (II)

The ratio of the young's modulus of steel and copper

\dfrac{Y_{s}}{Y_{c}}=\dfrac{\dfrac{Fl_{s}}{A_{s}\Delta l}}{\dfrac{Fl_{c}}{A_{c}\Delta l}}

\dfrac{Y_{s}}{Y_{c}}=\dfrac{A_{c}\times l_{s}}{A_{s}\times l_{c}}

\dfrac{Y_{s}}{Y_{c}}=\dfrac{4\times10^{-5}\times4.7}{3\times10^{-3}\times3.5}

\dfrac{Y_{s}}{Y_{c}}=1.79\times10^{-2}

Hence, The ratio of the young's modulus of steel and copper is 1.79\times10^{-2}

5 0
3 years ago
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GrogVix [38]

Answer:

Explanation:

a) series resistors carry the same current

A = V/Re = 6/(16 + 6) = 0.2727272... ≈ 27 mA

b) V = V₀(R/Re) = 6(16/(16 + 6)) = 4.363636 ≈ 4.4 V

c) V = V₀(R/Re) = 6(6/(16 + 6)) = 1.636363 ≈ 1.6 V

 or V = 6 - 4.4 = 1.6 V

3 0
3 years ago
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We have: v = d/t
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Hope this helps!
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3 years ago
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r was thirsty and decided to mix up a pitcher of lemonade. She put lemon juice, water, and sugar into a pitcher and stirred it t
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2 years ago
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Answer:

Neither technician

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