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2 years ago

5

Question 5: A heavy rope is flicked upwards, creating a single pulse in the rope. Make a drawing of the rope and indicate the fo

llowing in your drawing
a. the direction of motion of the pulse
b. Amplitude
c. Pulse length
d. Position of the rest

1 answer:

Anastaziya [24]2 years ago

8 0

Answer:

the direction of motion of the pulse

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<span>This statement simply implies that the 50 million tons of coal will produce the same energy output in one year as the 0.6 million barrels per day of oil. Assuming coal is more plentiful than oil, this is significant as 50 million tons is much smaller than 219 million barrels of oil.</span>

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A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s2 starts from rest at t = 0. The wheel has

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Answer:

The total linear acceleration is approximately 0.246 meters per square second.

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The total linear acceleration ( $a$ ) consist in two components, <em>radial</em> ( $a_{r}$ ) and <em>tangential</em> ( $a_{t}$ ), in meters per square second:

$a_{r} = \omega^{2}\cdot r$ (1)

$a_{t} = \alpha \cdot r$ (2)

Since both components are orthogonal to each other, the total linear acceleration is determined by Pythagorean Theorem:

$a = \sqrt{a_{r}^{2}+a_{t}^{2}}$ (3)

Where:

$r$ - Radius of the wheel, in meters.

$\omega$ - Angular speed, in radians per second.

$\alpha$ - Angular acceleration, in radians per square second.

Given that wheel accelerates uniformly, we use the following kinematic equation:

$\omega = \omega_{o}+ \alpha\cdot t$ (4)

Where:

$\omega_{o}$ - Initial angular speed, in radians per second.

$t$ - Time, in seconds.

If we know that $r = 0.1\,m$ , $\alpha = 2\,\frac{rad}{s^{2}}$ , $\omega_{o} = 0\,\frac{rad}{s}$ and $t = 0.60\,s$ , then the total linear acceleration is:

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$\omega = 1.2\,\frac{rad}{s}$

$a_{r} = \omega^{2}\cdot r$

$a_{r} = 0.144\,\frac{m}{s^{2}}$

$a_{t} = \alpha \cdot r$

$a_{t} = 0.2\,\frac{m}{s^{2}}$

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$a \approx 0.246\,\frac{m}{s^{2}}$

The total linear acceleration is approximately 0.246 meters per square second.

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3 years ago

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sladkih [1.3K]

Answer:

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The part b and c of this question is done in the attachment

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