If we have less power, we most likely have (2 points)

Physics

1 answer:

iren2701 [21]3 years ago

4 0

<span>. the same amount of work being done over a shorter period of time.</span>

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How would the terminal velocity of a piece of tissue paper compare to the terminal velocity of a rock?

matrenka [14]

Answer: Rock require larger drag force and to achieve it rock need to move at a very high terminal velocity.

Explanation: Terminal velocity is defined as the final velocity attained by an object falling under the gravity. At this moment weight is balanced by the air resistance or drag force and body falls with zero acceleration i.e. with a constant velocity.

Case 1: Terminal velocity of a piece of tissue paper.

The weight of tissue paper is very less and it experiences an air resistance while falling downward under the effect of gravity.

Downward gravitational force, F = mg

Upward air resistance or friction or drag force will be $f_{1}$

So, paper will attain terminal velocity when mg = $f_{1}$

Case 2: Rock is very heavy and require larger air resistance to balance the weight of rock relative to the tissue paper case.

Downward force on rock, F = Mg

Drag force = $f_{2}$

Rock will attain terminal velocity when Mg = $f_{2}$

Mg > mg

so, $f_{2}$ > $f_{1}$

And rock require larger drag force and to achieve it rock need to move at a very high terminal velocity.

5 0

3 years ago

A tube is sealed at both ends and contains a 0.0100-m long portion of liquid. The length of the tube is large compared to 0.0100

Ahat [919]

Answer:

31.321 rad/s

Explanation:

L = Tube length

A = Area of tube

$\rho$ = Density of fluid

v = Fluid velocity

m = Mass = $\rho Al$

Centripetal force is given by

$F=\dfrac{mv^2}{L}\\ F=\dfrac{m(\omega L)^2}{L}\\ F=m\omega^2\\ F= 0.01A\rho\omega^2L$

Pressure is given by

$P=\dfrac{F}{A}=\rho gL\\\Rightarrow \dfrac{0.01A\rho\omega^2L}{A}=\rho gL\\\Rightarrow 0.01\omega^2=g\\\Rightarrow \omega^2=\dfrac{g}{0.01}\\\Rightarrow \omega=\sqrt{\dfrac{g}{0.01}}\\\Rightarrow \omega=\sqrt{\dfrac{9.81}{0.01}}\\\Rightarrow \omega=31.321\ rad/s$