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3 years ago

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A wheel of diameter 30.0 cm starts from rest and rotates with a constant angular acceleration of 3.50 rad/s2 . At the instant th

e wheel has completed its second revolution, compute the radial acceleration of a point on the rim in two ways.

1 answer:

____ [38]3 years ago

5 0

Answer: The radial acceleration of a point on the rim in two ways is 13.20 m/s^2

Explanation: Please see the attachments below

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A rabbit with a mass of 2.0 kg accelerates at 2.0 meters per second squared. Find the net force of the rabbit

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4N

Explanation:

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What potential difference would an electron have to fall through to acquire a speed of 3.00*10^6 m/sec?

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25.6 V

Explanation:

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If the block is subjected to the force of F = 500 N, determine its velocity at s = 0.5 m. When s = 0, the block is at rest and t

STatiana [176]

Answer:

The velocity is $4.6 m/s^2$

Explanation:

Given:

Force = 500N

Distance s= 0

To find :

Its velocity at s = 0.5 m

Solution:

$\sum F_{x}=m a$

$F\left(\frac{4}{5}\right)-F_{S}=13 a$

$500\left(\frac{4}{5}\right)-\left(k_{s}\right)=13 a$

$400-(500 s)=13 a$

$a = \frac{400 -(500s)}{13}$

$a = (30.77 -38.46s) m/s^2$

Using the relation,

$a=\frac{d v}{d t}=\frac{d v}{d s} \times \frac{d s}{d t}$

$a=v \frac{d v}{d s}$

$v d v=a d s$

Now integrating on both sides

$\int_{0}^{v} v d v=\int_{0}^{0.5} a d s$

$\int_{0}^{v} v d v=\int_{0}^{0.5}(30.77-38.46 s) d s$

$\left[\frac{v^{2}}{2}\right]_{0}^{2}=\left[\left(30.77 s-19.23 s^{2}\right)\right]_{0}^{0.5}$

$\left[\frac{v^{2}}{2}\right]=\left[\left(30.77(0.5)-19.23(0.5)^{2}\right)\right]$

$\left[\frac{v^{2}}{2}\right]=[15.385-4.807]$

$\left[\frac{v^{2}}{2}\right]=10.578$

$v^{2}=10.578 \times 2$

$v^{2}=21.15$

$v = \sqrt{21.15}$

$v = 4.6 m/s^2$

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3 years ago