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3 years ago

What is the speed of terminal velocity?

1 answer:

ivann1987 [24]3 years ago

4 0

The so-called "terminal velocity" is the fastest that something can fall
through a fluid. Even though there's a constant force pulling it through,
the friction or resistance of plowing through the surrounding substance
gets bigger as the speed grows, so there's some speed where the resistance
is equal to the pulling force, and then the falling object can't go any faster.

A few examples:
-- the terminal velocity of a sky-diver falling through air,
-- the terminal velocity of a pecan falling through honey,
-- the terminal velocity of a stone falling through water.

It's not possible to say that "the terminal velocity is ----- miles per hour".
If any of these things changes, then the terminal velocity changes too:

-- weight of the falling object
-- shape of the object
-- surface texture (smoothness) of the object
-- density of the surrounding fluid
-- viscosity of the surrounding fluid .

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g In a certain underdamped RLC circuit, the voltage across the capacitor decreases in one cycle from 5.0 V to 3.8 V. The period

NeTakaya

The question is incomplete. The complete question is :

In a certain underdamped RLC circuit, the voltage across the capacitor decreases in one cycle from 5.0 V to 3.8 V. The period of the oscillations is 1.2 microseconds (1.2*10^-6). What is Q?

Solution :

The underdamped RLC circuit

$v_{t} = ve^{-\frac{R}{2L}t} \cos \omega t$

$\omega = \sqrt{\frac{1}{LC}-\frac{R^2}{4L^2}}= \frac{2 \pi}{T}$

We know in one time period, v = 2v, at t = T, $v_t = 3.8 v$

so, $9.8 = 5 e^{-\frac{R}{2L}T} \cos \frac{2 \pi}{T}T$

$e^{-\frac{R}{2L}T} = \frac{3.8}{5} \times 1$

$\frac{R}{2L}T= \ln \frac{5}{3.8}$

$\frac{R}{L}= \frac{2}{1.2 \times 10^{-6}} \ln \frac{5}{3.8}$

$\frac{R}{L} = 457.3 \times 10^3$

Now, Q value $= \frac{1}{R}\sqrt{\frac{L}{C}}$

$=\sqrt{\frac{L}{R^2C}\times \frac{L}{L}}$

$=\sqrt{(\frac{L}{R})^2 \times \frac{1}{LC}}$

$\frac{1}{LC}=27.43 \times 10^{12}$

∴ $Q=\sqrt{\left(\frac{1}{457.3 \times 10^3}\right)^2 \times 27.43 \times 10^{12}}$

$Q=\sqrt{131.166}$

= 11.45

4 0

2 years ago

A 20.0-kg block is initially at rest on a horizontal surface. A horizontal force of 77.0 N is required to set the block in motio

saveliy_v [14]

Answer: The coefficient of static friction is 3.85 and The coefficient of kinetic friction is 2.8

Explanation:

in the attachment

6 0

3 years ago

Bob has been Springfield's top meteorologist for ten years. He received data from satellite imaging and computer models that a h

djyliett [7]

The correct letter of the answer would be letter b, Bob is using his personal knowledge of the local area to make his forecast. It does not look like he's ignoring the computer for he's been the top meteorologist for ten years, meaning to say, he would use his knowledge about the local area and tell them his knowledge about the weather.

7 0

3 years ago

Read 2 more answers

Determine how many times per second it would move back and forth across a 6.0-m-long room on the average, assuming it made very

timofeeve [1]

Answer:

The right solution is "24.39 per sec".

Explanation:

According to the question,

⇒ $v=\frac{502.1}{\sqrt{3} }$