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IceJOKER [234]
3 years ago
12

Why does a partially inflated weather balloon expand as it rises?

Physics
2 answers:
Tomtit [17]3 years ago
5 0

Answer: So in an Inflated balloon, the pressure in the inside surface (the pressure by the gas inside the ballon) must be equal to the pressure applied on the outside surface. As the balloon rises, the outside pressure starts to decrease, so the pressure inside must decrease, this allows the surface of the balloon to grow, and the pressures can equilibrate.

Alson you can see it this way, as the outside pressure starts to decrease, there are less force holding the ballon size, then it can get a little bigger.

marusya05 [52]3 years ago
3 0
Air pressure pushing in on the balloon decreases as the balloon rises.
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Digiron [165]

32 = 0 \times t +  \frac{1}{2} \times 2.5 \times t^{2}     \\  32 = 0 + 1.25 \times t {}^{2}  \\ 32 = 1.25t {}^{2}   \\   \frac{32}{1.25}  =  \frac{1.25t {}^{2} }{1.25}  \\ t {}^{2}  = 25.6 \\  \sqrt{t {}^{2} }  =  \sqrt{25.6}  \\ t = 5.1seconds \\

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3 years ago
An airplane propeller is 2.68 m in length (from tip to tip) and has a mass of 107 kg. When the airplane's engine is first starte
telo118 [61]

Answer:

a)30.14 rad/s2

b)43.5 rad/s

c)60633 J

d)42 kW

e)84 kW

Explanation:

If we treat the propeller is a slender rod, then its moments of inertia is

I =\frac{mL^2}{12} = \frac{107*2.68^2}{12} = 64.04 kgm^2

a. The angular acceleration is Torque divided by moments of inertia:

\alpha = \frac{T}{I} = \frac{1930}{64.04} = 30.14 rad/s^2

b. 5 revolution would be equals to 10\pi rad, or 31.4 rad. Since the engine just got started

\omega^2 = 2\alpha\theta = 2*30.14*31.4 = 1893.5

\omega = \sqrt{1893.5} = 43.5 rad/s

c. Work done during the first 5 revolution would be torque times angular displacement:

W = T*\theta = 1930 * 31.4 = 60633 J

d. The time it takes to spin the first 5 revolutions is

t = \frac{\omega}{\alpha} = \frac{43.5}{30.14} = 1.44 s

The average power output is work per unit time

P = \frac{W}{t} = \frac{60633}{1.44} = 41991 W or 42 kW

e.The instantaneous power at the instant of 5 rev would be Torque times angular speed at that time:

P_i = T*\omega = 1930*43.5=83983 W or 84 kW

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3 years ago
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Industrial Waste. ..

Fertilizer and Lawn Chemicals. ...

Silt and Soil. ...

Humans and Water Pollution Through Detergents. ...

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Explanation:

8 0
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Answer:

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