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3 years ago

11

Suppose a flagellum is rotating at a constant rps (revolution per second). What is the angular displacement of the flagellum in

seconds?

1 answer:

IgorLugansk [536]3 years ago

5 0

Answer:

1794.47 radian.

Explanation:

Suppose a flagellum is rotating at a constant 952 rps (revolution per second). What is the angular displacement of the flagellum in 0.3seconds? (report your answer in SI units)

Let us suppose that a flagellum is rotating at a constant 952 revolution per second.

Firstly, we will convert 952 revolution per second to radian per second as follows :

952 revolution per second = 2π × 952 rad/s

=5981.592 rad/s

The angular displacement is given by the expression as follows :

$\theta=\omega t\\\\\theta=5981.592\times 0.3\\\\\theta=1794.47\ \text{rad}$

So, the angular displacement of the flagellum in 0.3 seconds is 1794.47 radian.

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Charge q is accelerated starting from rest up to speed v through the potential difference V. What speed will charge q have after

adoni [48]

Answer: v = $1.19 * 10^{6} m/s$

Explanation: q = magnitude of electronic charge = $1.609 * 10^{-19} c$

mass of an electronic charge = $9.10 * 10^{-31} kg$

V= potential difference = 4V

v = velocity of electron

by using the work- energy theorem which states that the kinetic energy of the the electron must equal the work done use in accelerating the electron.

kinetic energy = $\frac{mv^{2} }{2}$ , potential energy = qV

hence, $\frac{mv^{2} }{2} = qV$

$\frac{9.10 *10^{-31} * v^{2} }{2} = 1.609 * 10^{-16} * 4\\\\\\\\9.10*10^{-31} * v^{2} = 2 * 1.609 *10^{-16} * 4\\\\\\9.10 *10^{-31} * v^{2} = 1.287 *10^{-15} \\\\v^{2} = \frac{1.287 *10^{-15} }{9.10 *10^{31} } \\\\v^{2} = 1.414*10^{15} \\\\v = \sqrt{1.414*10^{15} } \\\\v = 1.19 * 10^{6} m/s$

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2 years ago

If a volume of air is warmed, it expands and tends to

Butoxors [25]

if a volume of air is warmed it expands due to increased translational kinetic energy as it expands it will start to cool.

<h3>When does temperature increase volume?</h3>

We can then conclude that at constant pressure, temperature and volume are directly proportional: temperature increases, volume increases; decrease temperature, decrease volume.

In this case, the higher the temperature, the greater the kinetic energy that acts on the molecules of this gas, so when the gas expands, these molecules find more space and collide less, which will cause the gas to cool.

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2 years ago

Please please help

dsp73

Answer:

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Explanation:

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3 years ago

please help! find magnitude and direction (the counterclockwise angle with the +x axis) of a vector that is equal to a + c

-BARSIC- [3]

Answer:

Option (2)

Explanation:

From the figure attached,

Horizontal component, $A_x=A\text{Sin}37$

$A_x=12[\text{Sin}(37)]$

= 7.22 m

Vertical component, $A_y=A[\text{Cos}(37)]$

= 9.58 m

Similarly, Horizontal component of vector C,

$C_x$ = C[Cos(60)]

= 6[Cos(60)]

= $\frac{6}{2}$

= 3 m

$C_y=6[\text{Sin}(60)]$

= 5.20 m

Resultant Horizontal component of the vectors A + C,

$R_x=7.22-3=4.22$ m

$R_y=9.58-5.20$ = 4.38 m

Now magnitude of the resultant will be,

From ΔOBC,

$R=\sqrt{(R_x)^{2}+(R_y)^2}$

= $\sqrt{(4.22)^2+(4.38)^2}$

= $\sqrt{17.81+19.18}$

= 6.1 m

Direction of the resultant will be towards vector A.

tan(∠COB) = $\frac{\text{CB}}{\text{OB}}$

= $\frac{R_y}{R_x}$

= $\frac{4.38}{4.22}$

m∠COB = $\text{tan}^{-1}(1.04)$

= 46°

Therefore, magnitude of the resultant vector will be 6.1 m and direction will be 46°.

Option (2) will be the answer.

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