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3 years ago

How are velocity and displacement related?

1 answer:

7 0

Velocity is the rate at which displacement changes with time. It is a vector, too. The average velocity over some interval is the total displacement during that interval, divided by the time. ... Speed is a scalar, which describes the total distance travelled by some object during some time interval. Plze mark me brainiest

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3 years ago

In this relationship, the barnacles derive benefit. There is not benefit or harm to the whale. This relationship is an example o

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4 years ago

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An accelerometer has a damping ratio of 0.5 and a natural frequency of 18,000 Hz. It is used to sense the relative displacement

SOVA2 [1]

Answer:

A) i) Dynamic error ≈ 3.1%

ii) phase shift ≈ -12°

B) 79971.89 rad/s

Explanation:

Given data :

Damping ratio = 0.5

natural frequency = 18,000 Hz

<u>a) Calculate the dynamic error and phase shift in accelerometer output at an impart vibration of 4500 Hz</u>

i) Dynamic error

This can be calculated using magnitude ratio formula attached below is the solution

dynamic error ≈ 3.1%

ii) phase shift

This phase shift can be calculated using frequency dependent phase shift formula

phase shift ≈ -12°

<u>B) Determine resonance frequency </u>

Wr = 2 $\pi$ ( 18000 $\sqrt{0.5}$ ) = 79971.89 rad/s

C) The maximum magnitude ratio that the system can achieve

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3 years ago

A flywheel having constant angular acceleration requires 4.00 ss to rotate through 164 radrad . Its angular velocity at the end

ExtremeBDS [4]

Answer:

$\omega '=-13.5rad/s$

Explanation:

From the question we are told that:

Time $t=4sec$

Angular displacement $\theta= 161 rad$

Final Angular velocity $\omega = 100 rad / s$

Let

Angular acceleration $\alpha$

Generally the equation for Initial Angular velocity $\omega '$ is mathematically given by

$-\omega '^2=2 \alpha \theta -\omega^2$

$\omega '^2= \alpha 328 +11236$

Also,Initial Angular velocity $\omega '$ is mathematically given by

$\omega '=\omega - \alpha t$

Therefore substitution

$-\omega '^2=2 \alpha \theta -\omega^2$

$\omega '=\omega - \alpha t$

$(\omega - \alpha t)^2=2 \alpha \theta -\omega$

$-16\alpha^2+848\alpha+11236= \alpha 328 -11236$

$16\alpha^2-520\alpha=0$

$\alpha=29.875rads/s^2$

Substitution in the 2nd equation for Initial Angular velocity $\omega '$

$\omega '=106-(29.875rads/s^2*4)$

$\omega '=-13.5rad/s$

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3 years ago

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