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andrew-mc [135]
4 years ago
10

If a small-diameter open glass tube is partially immersed in a vessel containing mercury, the surface of the mercury inside the

tube will be
A. higher than the mercury in the vessel, and shaped like the inside of a bowl.
B. lower than the mercury in the vessel, and shaped like the outside of an umbrella.
C. lower than the mercury in the vessel, and shaped like the inside of a bowl.
D. higher than the mercury in the vessel, and shaped like the outside of an umbrella.
Physics
2 answers:
klemol [59]4 years ago
8 0
The pressure outside (atmosphere) is greater than that inside the tube. This means that the mercury will be pushed up in the glass tube. This phenomenon is called the capillary rise. It shape would be like that of the outside of the umbrella. The answer is therefore letter D. 
Sphinxa [80]4 years ago
5 0
<span>B. lower than the mercury in the vessel, and shaped like the outside of an umbrella. </span>
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A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.
galina1969 [7]

Answer:

44.64 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.2\times 1180+80.6^2}\\\Rightarrow v=128.01\ m/s

v=u+at\\\Rightarrow 128.01=80.6+4.2t\\\Rightarrow t=\frac{128.01-80.6}{4.2}=11.29\ s

<u>Time taken to reach 1180 m is 11.29 seconds</u>

v=u+at\\\Rightarrow 0=128.01-9.8t\\\Rightarrow t=\frac{128.01}{9.8}=13.06\ s

<u>Time the rocket will keep going up after the engines shut off is 13.06 seconds.</u>

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-128.01^2}{2\times -9.8}\\\Rightarrow s=836.05\ m

The distance the rocket will keep going up after the engines shut off is 836.05 m

Total distance traveled by the rocket in the upward direction is 1180+836.05 = 2016.05 m

The rocket will fall from this height

s=ut+\frac{1}{2}at^2\\\Rightarrow 2016.05=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{2016.05\times 2}{9.8}}\\\Rightarrow t=20.29\ s

<u>Time taken by the rocket to fall from maximum height is 20.29 seconds</u>

Time the rocket will stay in the air is 11.29+13.06+20.29 = 44.64 seconds

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