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Sergeeva-Olga [200]

3 years ago

11

Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.59 times a second. A tack is stuck in the tire a

t a distance of 0.329 m from the rotation axis. Noting that for every rotation the tack travels one circumference, find the tack's tangential speed.__________m/s
what is the tacks radial acceleration?
___________m/s^2

1 answer:

Vesna [10]3 years ago

6 0

Answer:

Tangential speed=5.4 m/s

Radial acceleration= $88.6m/s^2$

Explanation:

We are given that

Angular speed=2.59 rev/s

We know that

1 revolution= $2\pi rad$

2.59 rev= $2\pi\times 2.59=5.18\pi=5.18\times 3.14=16.27 rad/s$

By using $\pi=3.14$

Angular velocity= $\omega=16.27rad/s$

Distance from axis=r=0.329 m

Tangential speed= $r\omega=16.27\times 0.329=5.4m/s$

Radial acceleration= $\frac{v^2}{r}$

Radial acceleration= $\frac{(5.4)^2}{0.329}=88.6m/s^2$

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2 years ago

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3 years ago

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2 years ago

Read 2 more answers

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Answer:

$v_{o}=-14.60m/s$

Explanation:

<u>Kinematics equation for first Object:</u>

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

but:

$v_{o}=0m/s$ The initial velocity is zero

$y_{o}=20m$

it reach the water at in instant, t1, y(t)=0:

$0=y_{o}-1/2*g*t_{1}^{2}$

$t_{1}=\sqrt{2y_{o}/g}=\sqrt{2*20/9.81}=2.02s$

<u>Kinematics equation for the second Object:</u>

The initial velocity is zero

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

but:

$y_{o}=20m$

it reach the water at in instant, t2, y(t)=0. If the second object is thrown 1s later, t2=t1-1=1.02s

$0=y_{o}+v_{o}t_{2}-1/2*g*t_{2}^{2}$

$v_{o}=1/(t_{2})*(1/2*g*t_{2}^{2}-y_{o})=(1/1.02)*(1/2*9.81*1.02^{2}-20)=-14.60m/s$

The velocity is negative, because the object is thrown downwards.

6 0

3 years ago