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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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Power is work done over a what?
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3 years ago
At what location in a circuit is the electrical potential energy the greatest
Eduardwww [97]

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4 0
3 years ago
How much power does it take to lift 70.0 N to 5.0 m high in 5.00 s?
lesantik [10]

Answer:

Power = 70 W

Explanation:

Given that,

Force, F = 70 N

Height, h = 5 m

Time, t = 5 s

We need to find the power of the object. We know that,

Power = work done/time

Put all the values,

P=\dfrac{Fd}{t}\\\\P=\dfrac{70\times 5}{5}\\\\P=70\ W

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3 0
3 years ago
The outer layer of cable on a cable reel is 16.2 cm from the center of the reel. The reel is initially stationary and can rotate
ahrayia [7]

Answer:

B. w=12.68rad/s

C. α=3.52rad/s^2

Explanation:

B)

We can solve this problem by taking into account that (as in the uniformly accelerated motion)

\theta=\omega_{0}t+\frac{1}{2}\alpha t^{2}\\\theta = \frac{s}{r}      ( 1 )

where w0 is the initial angular speed, α is the angular acceleration, s is the arc length and r is the radius.

In this case s=3.7m, r=16.2cm=0.162m, t=3.6s and w0=0. Hence, by using the equations (1) we have

\theta=\frac{3.7m}{0.162m}=22.83rad

22.83rad=\frac{1}{2}\alpha (3.6s)^2\\\\\alpha=2\frac{(22.83rad)}{3.6^2s}=3.52\frac{rad}{s^2}

to calculate the angular speed w we can use\alpha=\frac{\omega _{f}-\omega _{i}}{t _{f}-t _{i}}\\\\\omega_{f}=\alpha t_{f}=(3.52\frac{rad}{s^2})(3.6)=12.68\frac{rad}{s}

Thus, wf=12.68rad/s

C) We can use our result in B)

\alpha=3.52\frac{rad}{s^2}

I hope this is useful for you

regards

3 0
3 years ago
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Which of the following items best embodies the physical property of conductivity?
Ostrovityanka [42]

Copper penny is the answer

8 0
3 years ago
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