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AlexFokin [52]
3 years ago
11

What force, in newtons, must you exert on the balloon with your hands to create a gauge pressure of 62.5 cm H2O, if you squeeze

on an effective area of 51 cm2? Assume the pressure in the balloon is one atmosphere before you squeeze.
Physics
1 answer:
yulyashka [42]3 years ago
6 0

Answer:54.70 N

Explanation:

Given

Gauge Pressure of 62.5 cm of H_2O

i.e. h=62.5 cm =0.625 m

Effective area A=51 cm^2

initial Pressure= 1 atm=101.325 kPa

Gauge Pressure P=\rho gh

\rho =density\ of\ water =1000 kg/m^3

P_{gauge}=1000\times 9.8\times 0.625=5.937 kPa

Force creates a pressure of P_1 which will be equal to Gauge Pressure

P_1=\frac{F}{A}

P_1=P_{gauge}

\frac{F}{A}=5.937 kPa

F=5.937\times 51\times 10^{-4}\times 10^3

F=30.27 N

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We cannot see a new moon in our sky because ________. We cannot see a new moon in our sky because ________. a new moon is quite
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a new moon is quite near the Sun in the sky

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As you move away from a positive charge distribution, the electric field:
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Answer:

The electric field always decreases.

Explanation:

The electric field due to a point charge is given by :

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Where

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Synchronous communications satellites are placed in a circular orbit that is 3.59 107 m above the surface of the earth. What is
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Answer: 0.223 m/s^{2}

Explanation:

We can solve this with the Law of Universal Gravitation and knowing the acceleration due gravity g of an object above the surface of the planet decreases with the distance (height) of this object from the center of the planet.

Well, according to the law of universal gravitation:

F=G\frac{m_{E}m}{r^2}  (1)

Where:

F is the module of the force exerted between both bodies

G=6.67(10)^{-11}\frac{m^{3}}{kgs^{2}} is the gravitational constant

m_{E}=5.98(10)^{24} kg is the mass of the Earth

m are the mass of each communications satellite

r=R_{E}+h is the distance between the center of the Earth and the satellite

R_{E}=6.38(10)^{6} m is the radius of the Earth

h=3.59(10)^{7} m is the height of the satellite, measured from the Earth's surface

On the other hand, we know according to <u>Newton's 2nd law of motion:</u>

F=mg  (2)

Combining (1) and (2):

G\frac{m_{E}m}{r^2}=mg  (3)

Isolating g:

g=\frac{G M_{E}}{r^2}  (4)

Remembering r=R_{E}+h:

g=\frac{G M_{E}}{(R_{E}+h)^2}  (5)

g=\frac{(6.67(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24} kg)}{(6.38(10)^{6} m+3.59(10)^{7} m)^2}  

Finally:

g=0.223 m/s^{2}  

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