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kompoz [17]
4 years ago
6

A person throws a balloon straight down above level ground. The balloon bounces back up into the air. Drag force is significant

when the balloon is moving. Verbally describe a free-body diagram for the balloon for each of the following times (you must include the force type, source, receiver, and direction for each force; relative magnitudes, when known, must be clear):
The balloon is being thrown down and is speeding up.
The balloon is in the air on its way down and is moving at constant speed.
The balloon has hit the ground and is instantaneously at rest as it reverses direction.
The balloon is in the air on its way back up and slowing down.
The balloon is in the air and is instantaneously at rest at the peak of its trajectory.
Physics
1 answer:
love history [14]4 years ago
5 0

Explanation:

   a) Balloon is being thrown down and is speeding up;

           mg > F_drag

  b) Balloon is in the air on its way down and moving with constant speed.

           F_drag =mg

  c) The Balloon is on the ground and rest instantaneously

           mg = Normal

  d) Balloon is moving slowly downward;

   e) Because, at the peak of trajectory drag force  is 0.

             Drag force = 0

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Answer:

6.40 m/s

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V = root 2*10*20.5

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Consider a portion of a cell membrane that has a thickness of 7.50 nm and 1.4 µm ✕ 1.4 µm in area. A measurement of the potentia
Alla [95]

Answer:

Part a: The current is 1.49x10⁻¹² A

Part b: The current is decreased by a factor of 4.

Explanation:

Part a

The area is given as

A =(1.4*10⁻⁶)² =1.96*10⁻¹² m²

The resistance is given as where resistivity of the membrane material is 1.30 x 10⁷ ohms*m

R=pL/A =(1.3*10⁷)(7.5*10⁻⁹)/(1.96*10⁻¹²)

R=4.97x10¹⁰ ohms

So the resistance is 4.97x10¹⁰ ohms.

I=V/R =86.0x10⁻³/5.77x10¹⁰

I=1.49x10^⁻¹² A

So the current is 1.49x10⁻¹² A

b)

S=So/2=1.4/2 =0.7μm

A=(0.7*10^-6)^2=4.9*10⁻¹³ m²

R=pL/A =(1.3*10⁷)(7.5*10⁻⁹)/(4.9*10⁻¹³)

R=1.98x10¹¹ ohms

So the resistance is 1.98x10¹¹ ohms.

I=V/R =86.0x10⁻³/1.98x10¹¹

I2=4.52x10^⁻¹³ A

So the ratio of the new current vs the old current is as

I2/I=4.52x10^⁻¹³/1.49x10⁻¹²=0.25

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4 years ago
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4 years ago
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Answer:

B) 20N.s is the correct answer

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4 0
3 years ago
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