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user100 [1]
3 years ago
12

How high would the level be in a gasoline barometer at normal atmospheric pressure?

Physics
1 answer:
Sergio [31]3 years ago
7 0

Answer:

h = 13.06 m

Explanation:

Given:

- Specific gravity of gasoline S.G = 0.739

- Density of water p_w = 997 kg/m^3

- The atmosphere pressure P_o = 101.325 KPa

- The change in height of the liquid is h m

Find:

How high would the level be in a gasoline barometer at normal atmospheric pressure?

Solution:

- When we consider a barometer setup. We dip the open mouth of an inverted test tube into a pool of fluid. Due to the pressure acting on the free surface of the pool, the fluid starts to rise into the test-tube to a height h.

- The relation with the pressure acting on the free surface and the height to which the fluid travels depends on the density of the fluid and gravitational acceleration as follows:

                                         P = S.G*p_w*g*h

Where,                              h = P / S.G*p_w*g

- Input the values given:

                                         h = 101.325 KPa / 0.739*9.81*997

                                         h = 13.06 m

- Hence, the gasoline will rise up to the height of 13.06 m under normal atmospheric conditions at sea level.

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A roller of radius 12.5 cm turns at 14 revolutions per second. What is the linear velocity of the roller in meters per second?
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Two asteroids identical to those above collide at right angles and stick together; i.e, their initial velocities were perpendicu
11111nata11111 [884]

Answer:

velocity = 62.89 m/s  in 58 degree measured from the x-axis

Explanation:

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Two asteroids moving with velocities collide at right angles and stick together. Asteroid A initially moving to right direction and asteroid B initially move in the upward direction.

Before collision Momentum of A = 1000 x 100 = $ 10^5$ kg - m/s in the right direction.

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Now applying the Momentum Conservation, we get

Initial momentum in right direction = final momentum in right direction = $ 10^5$

And, Initial momentum in upward direction = Final momentum in upward direction = 1.6 x $ 10^5$

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Therefore, velocity is = $ \sqrt{V_x^2 + V_y^2} $

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And direction is

tan θ = $ \frac{V_y}{V_x}$     = 1.6

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worty [1.4K]

Answer:

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