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3 years ago

What is the formula for finding the time it takes for an object to reach terminal velocity?

Physics

1 answer:

Ivan3 years ago

5 0

Answer:

$t=\tau(\tanh^{-1}(\frac{v}{v_t}))$

Explanation:

Terminal velocity is the maximum velocity attained by a body when the net upward force on it is equal to the net downward force as it falls through a fluid from a certain height.

The time required to reach terminal velocity is infinite as the object never reaches absolute terminal velocity and always approaches terminal velocity.

The velocity $v$ of a falling object expressed in terms of terminal velocity $v_t$ and time 't' is given as:

$v=v_t\tanh(\frac{t}{\tau})$

Where, $\tau$ is a constant and has units of time.

Now, expressing the above in terms of time 't', we get:

$t=\tau(\tanh^{-1}(\frac{v}{v_t}))$

Therefore, the formula to find time to approach terminal velocity is:

$t=\tau(\tanh^{-1}(\frac{v}{v_t}))$

When the velocity is 76% of terminal velocity, the time taken is $\tau$ .

When the velocity is 96% of terminal velocity, the time taken is $2\tau$ .

When the velocity is 99.5% of terminal velocity, the time taken is $3\tau$ and so on..

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A constant force of magnitude F acts on an object of mass 0.04kg initially at rest at a point O. If the speed of the object when

vampirchik [111]

Answer:

F = 100 Newtons

Explanation:

F = ?

m = 0.04kg

u = 0m/s ==> u is just an abbreviation for initial velocity, it is conventional.

s = 50m ==> s is just an abbreviation for distance, it is conventional.

v = 500m/s ==> v is just an abbreviation for final velocity, it is conventional.

$v^{2} = u^{2} + 2as\\\\=> a = \frac{v^{2} - u^{2}}{2s}\\a = \frac{500^{2} }{2*50}\\a = 2500ms^{-2}$

Then F = ma = 0.04 x 2500 = 100N

5 0

3 years ago

What does the slope of the curve on a velocity vs. time graph represent?

Mademuasel [1]

Answer:

the slope of velocity-time graph represent an object acceleration

5 0

3 years ago

The tape in a videotape cassette has a total

xxMikexx [17]

Answer:

$w=19.76 \ rad/s$

Explanation:

<u>Circular Motion </u>

Suppose there is an object describing a circle of radius r around a fixed point. If the relation between the angle of rotation by the time taken is constant, then the angular speed is also constant. If that relation increases or decreases at a constant rate, the angular speed is given by:

$w=w_o+\alpha \ t$

Where $\alpha$ is the angular acceleration and t is the time. If the object was instantly released from the circular path, it would have a tangent speed of:

$\displaystyle v_t=w.r$

We have two reels: one loaded with the tape to play and the other one empty and starting to fill with tape. They both rotate at different angular speeds, one is increasing and the other is decreasing as the tape goes from one to the other. We'll assume the tangent speed is constant for both (so the tape can play correctly). Let's call $w_1$ the angular speed of the loaded reel and $w_2$ that from the empty reel. We have

$w_1=w_{o1}+\alpha_1 \ t$

$w_2=w_{o2}+\alpha_2 \ t$

If $r_f=35\ mm=0.035\ m$ is the radius of the reel when it's full of tape, the angular speed for the loaded reel is computed by

$\displaystyle w_{01}=\frac{v_t}{r_f}$

The tangent speed is computed by knowing the length of the tape and the time needed to fully play it.

$t=1.8\ h=1.8*3600=6480\ sec$

$\displaystyle v_t=\frac{x}{t}=\frac{249\ m}{6480\ sec}=0.0384 \ m/s$

$\displaystyle w_{01}=\frac{0.0384}{0.035}=1,098 \ rad/s$

If $r_e=10\ mm=0.001\ m$ is the radius of the reel when it's empty, the angular speed for the empty reel is computed by

$\displaystyle w_{02}=\frac{0.0384}{0.001}=38.426 \ rad/s$

The full reel goes from $w_{01}$ to $w_{02}$ in 6480 seconds, so we can compute the angular acceleration:

$\displaystyle \alpha_1=\frac{w_{02}-w_{01}}{6480}$

$\displaystyle \alpha_1=\frac{38.426-1.098}{6480}=0.00576 \ rad/sec^2$

The empty reel goes from $w_{02}$ to $w_{01}$ in 6480 seconds, so we can compute the angular acceleration:

$\displaystyle \alpha_2=\frac{1.098-38.426}{6480}=-0.00576 \ rad/sec^2$

So the equations for both reels are

$w_1=1.098+0.00576 \ t$

$w_2=38.426-0.00576 \ t$

They will be the same when

$1.098+0.00576 \ t=38.426-0.00576 \ t$

Solving for t

$\displaystyle t=\frac{38.426-1.098}{0.0115}$

$t=3240 \ sec$

The common angular speed is

$w_1=1.098+0.00576 \ 3240=19.76 \ rad/s$

$w_2=38.426-0.00576 \ 3240=19.76 \ rad/s$

They both result in the same, as expected

$\boxed{w=19.76 \ rad/s}$

6 0

4 years ago

Define in own words.

pickupchik [31]

Complementary is essentially two things that are "combined" with one another (complement one another) to make a bigger piece or make it better.

5 0

4 years ago

The core of the Sun is Group of answer choices

kap26 [50]

Answer:

2. much hotter and much denser than its surface

Explanation:

We know that the temperature around the center of the Sun is about 1.57×10⁷ K and its density is about 162 g/cm³.

Now, the temperature and the density decrease as one moves outward from the center of the Sun, the temperature at the surface of the sun is about 5×10³ K and the density as an average in the surface is about 1.4 g/cm³.

Therefore the answer is:

2. much hotter and much denser than its surface.

I hope it helps you!

Have a nice day!

3 0

4 years ago