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2 years ago

The amount of energy carried by a wave and the waves maximum displacement from the rest position describe what?

Physics

1 answer:

Mademuasel [1]2 years ago

6 0

What you are describing is Amplitude.

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Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 5.00 s, it rotates 14.8 rad. Du

11Alexandr11 [23.1K]

Answer:

Explanation:

Given:

Initial θ = 0 rad (from rest)

Final θ = 14.3 rad

Time, t = 5 s

Angular velocity, w = dθ / dt

= (14.3 - 0)/5

= 2.86 rad/s

Acceleration, ao = dw/dt

Initial angular velocity, wi = 0 rad/s (from rest)

Final angular velocity, wf = 2.86 rad/s

a = (2.86 - 0)/5

= 0.572 rad/s^2

3 0

3 years ago

Read 2 more answers

The tension in cable da has a magnitude of tda=6.27 lb. find the cartesian components of tension tda, which is directed from d t

Oliga [24]

Complete Question

The Complete Question is attached below

We have that the Cartesian components of tension $T_{da}$ is

$T_{DA}=-4.433i-3.49j+2.735k$

From the Question we are told that

$M_{da}=6.27 lb\\\\w=9.50ft\\\\d=6.60ft\\\\h=4.50ft$

$\vec {DA}=-4.7i-3.7j+2.9k)ft$

$\vec {DB}=-1.9i-3.7j+1.9k)ft\\\\\vec {DC}=-1.9i+5.8j-1.6k)ft$

Generally the equation for $T_{DA}$ is mathematically given as

$T_{DA}=\phi_{DA}* M_{da}$

Where

$\phi_{DA}=\frac{-4.7i-3.7j+2.9k}{(-4.7)^2+(-3.7)^2+(2.9)^2}\\\\\phi_{DA}=\frac{-4.7i-3.7j+2.9k}{6.65}$

Therefore

$T_{DA}=\phi_{DA}* M_{da}$

$T_{DA}=\frac{-4.7i-3.7j+2.9k}{6.65}* 6.27$

$T_{DA}=-4.433i-3.49j+2.735k$

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2 years ago

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Ivanshal [37]

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Explanation:

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Answer:

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Explanation:

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3 years ago