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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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An underground cannon launches a cannonball from ground level at a 35 degree angle. the cannonball is shot with an initial veloc
user100 [1]

Answer:

Time = 1.75[s]; Distance traveled = 21.5 [m]; Max height = 15 [m]

Explanation:

First, we have to break down the velocity vector into the X & y components.

(v_{x})_{0} = 15 * cos( 35)= 12.28[m/s]\\(v_{y})_{0} = 15 * sin( 35)= 8.6[m/s]\\\\

To find the time t that lasts the ball of cannon in the air we must use the following equation of kinematics, in this equation the value of y is equal to zero because it will be proposed that the ball lands at the same level that was fired.

y=(v_{y} )_{0}-\frac{1}{2}*g*t^{2}   \\where:\\g=9.81[m/s^2]\\t = time[s]\\y=0[m]

0=8.6*t-\frac{1}{2}*9.81*t^{2}  \\4.905*t^{2}=8.6*t\\ t=1.75[s]

In order to find the distance traveled horizontally from the cannonball, we must use the speed kinematics equation in the X coordinate.

x = (v_{x})_{0}  *t\\x=12.28*1.75\\x=21.5 [m]

In order to find the last value, we must bear in mind that when the cannonball reaches the maximum height, the velocity in the component y is equal to zero, and we can find the value of and with the following kinematic equation

y = (v_{y})_{0} *t+\frac{1}{2} *g*(t)^{2} \\y = 0*t+\frac{1}{2} *9.81*(1.75)^{2}\\ y=15 [m]

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3 years ago
Can I get a direct answer please??
OleMash [197]
Get a direct answer of what???
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Which of the following explanations is correct about the wave that travels through the medium?
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The correct answer is letter A
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Answer:

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