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chubhunter [2.5K]
3 years ago
5

E. Describe in short the structure of a mercury barometer​

Physics
1 answer:
Readme [11.4K]3 years ago
6 0

Answer:

A mercury barometer has a glass tube that is closed at the top and open at the bottom.At the bottom of the tube is a pool of mercury.The mercury sits in a circular,shallow dish surrounding the tube.The mercury in the tube will adjust itself to match the atmospheric pressure above the dish.

hope this helps you

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At this radius, what is the magnitude of the net force that maintains circular motion exerted on the pilot by the seat belts, th
Ainat [17]

Answer:

Fc=5253 N

Explanation:

Answer:

Fc=5253 N

Explanation:

sequel to the question given, this question would have taken precedence:

"The 86.0 kg pilot does not want the centripetal acceleration to exceed 6.23 times free-fall acceleration. a) Find the minimum radius of the plane’s path. Answer in units of m."

so we derive centripetal acceleration first

ac (centripetal acceleration) = v^2/r

make r the subject of the equation

r= v^2/ac

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substituing the parameters into the equation, to get the radius

(101^2)/(6.23*9.81) = 167m

Now for part

( b) there are two forces namely, the centripetal and the weight of the pilot, but the seat is exerting the same force back due to newtons third law.

he net force that maintains circular motion exerted on the pilot by the seat belts, the friction against the seat, and so forth is the centripetal force.

Fc (Centripetal Force) = m*v^2/r  

So (86kg* 101^2)/(167) =

Fc=5253 N

4 0
4 years ago
A sample of argon has a volume of 205 mL when its temperature is -44.0 oC and its pressure is 712 mm of Hg. What would be the vo
anzhelika [568]

Answer:

231.98 mL

Explanation:

V₁ = Initial volume = 205 mL

V₂ = Final volume

P₁ = Initial pressure = 712 mmHg

P₂ = Final pressure = 750 mmhg (STP)

T₁ = Initial temperature = -44 °C = 229.15 K

T₂ = Final temperature = 273.15 K (STP)

From Combined gas law

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\\\Rightarrow V_2=\frac{P_1V_1T_2}{T_1P_2}\\\Rightarrow V_2=\frac{712\times 205\times 273.15}{229.15\times 750}=231.98\ mL

Volume of the argon at STP is 231.98 mL

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