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Nuetrik [128]
3 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]3 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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3 years ago
A 0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters
Artist 52 [7]

Answer:

Part a)

F = 7.76 N

Part b)

\theta = -137.7 degree

Part c)

\theta = -127.7 degree

Explanation:

As we know that acceleration is rate of change in velocity of the object

So here we know that

x = -19 + t - 3t^3

y = 20 + 7t - 9t^2

Part a)

differentiate x and y two times with respect to time to find the acceleration

a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)

a_x = \frac{d}{dt}(0 +1 - 9t^2)

a_x = -18t

a_y = \frac{d^2}{dt^2}(20 + 7t - 9t^2)

a_y = \frac{d}{dt}(0 +7 - 18t)

a_y = -18

Now the acceleration of the object is given as

\vec a = (-18t)\hat i + (-18)\hat j

at t= 1.1 s we have

\vec a = -19.8 \hat i - 18 \hat j

now the net force of the object is given as

\vec F = m\vec a

\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)

\vec F = -5.74 \hat i - 5.22 \hat j

now magnitude of the force will be

F = \sqrt{5.74^2 + 5.22^2} = 7.76 N

Part b)

Direction of the force is given as

tan\theta = \frac{F_y}{F_x}

tan\theta = \frac{-5.22}{-5.74}

\theta = -137.7 degree

Part c)

For velocity of the particle we have

v_x = \frac{dx}[dt}

v_x = (0 +1 - 9t^2)

v_y = \frac{dy}{dt}

v_y = (0 +7 - 18t)

now at t = 1.1 s

\vec v = -9.89\hat i - 12.8 \hat j

now the direction of the velocity is given as

\theta = tan^{-1}(\frac{v_y}{v_x})

\theta = tan^{-1}(\frac{-12.8}{-9.89})

\theta = -127.7 degree

7 0
3 years ago
A bike travels 30m/s for 3 seconds. How far did it travel?
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He traveled 90 meters
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What is the angle of reflection?
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Light travels at 3 × 108 m/s, and it takes about 8 min for light from the sun to travel to Earth. Based on this, the order of ma
N76 [4]

Answer:

The order of magnitude of the distance from the sun to Earth is 10⁸ km.

Explanation:

The order of magnitude of the distance from the sun to Earth can be calculated as follows:

c = \frac{x}{t}

Where:

c: is the speed of light = 3x10⁸ m/s

t: is the time = 8 min

Hence, the distance is:

x = c*t = 3 \cdot 10^{8} m/s*8 min*\frac{60 s}{1 min} = 1.44 \cdot 10^{11} m = 1.44 \cdot 10^{8} km

Therefore, the order of magnitude of the distance from the sun to Earth is 10⁸ km.

I hope it helps you!

5 0
3 years ago
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