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sdas [7]
4 years ago
10

Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.55 times a second. A tack is stuck in the tire a

t a distance of 0.357 m from the rotation axis. Noting that for every rotation the tack travels one circumference, find the tack's tangential speed. tangential speed: m / s What is the tack's centripetal acceleration
Physics
2 answers:
earnstyle [38]4 years ago
7 0

Answer:

Tangential speed = 5.72 m/s

Centripetal acceleration = 91.6\text{ m/s}{}^2

Explanation:

The tangential speed, V, is given by

v=\omega r

where \omega is the angular speed and is given by 2\pi f (f is the angular frequency or frequency of rotation)

Thus,

v=2\pi f r = 2\times3.14\times2.55\times0.357 = 5.72\text{ m/s}

The centripetal acceleration,a, is given by

a=\dfrac{v^2}{r}

a=\dfrac{5.72^2}{0.357} = 91.6\text{ m/s}{}^2

balandron [24]4 years ago
7 0

Explanation:

Given:

Time for 1 rev = 1/2.55

= 0.392s

To rad/s,

= 2π/0.392

w = 16.03 rad/s

v = wr

= 16.03 * 0.357

= 5.72 m/s

B.

a = v²/r

= 5.72²/0.357

= 91.72 m/s²

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3 years ago
A train, traveling at a constant speed of 25.8 m/s, comes to an incline with a constant slope. While going up the incline, the t
zhannawk [14.2K]

The final velocity of the train after 8.3 s on the incline will be 12.022 m/s.

Answer:

Explanation:

So in this problem, the initial speed of the train is at 25.8 m/s before it comes to incline with constant slope. So the acceleration or the rate of change in velocity while moving on the incline is given as 1.66 m/s². So the final velocity need to be found after a time period of 8.3 s. According to the first equation of motion, v = u +at.

So we know the values for parameters u,a and t. Since, the train slows down on the slope, so the acceleration value will have negative sign with the magnitude of acceleration. Then

v = 25.8 + (-1.66×8.3)

v =12.022 m/s.

So the final velocity of the train after 8.3 s on the incline will be 12.022 m/s.

8 0
3 years ago
O'Malley is riding on a bus which is moving at 10 m/s, and he throws a ball which he observes to be moving at 10 m/s relative to
Vikki [24]

Answer:

<em>20 m/s in the same direction of the bus.</em>

Explanation:

<u>Relative Motion </u>

Objects movement is always related to some reference. If you are moving at a constant speed, all the objects moving with you seem to be at rest from your reference, but they are moving at the same speed as you by an external observer.

If we are riding on a bus at 10 m/s and throw a ball which we see moving at 10 m/s in our same direction, then an external observer (called Ophelia) will see the ball moving at our speed plus the relative speed with respect to us, that is, at 20 m/s in the same direction of the bus.

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Air resistance allow objects to develop full acceleration due to gravity true or false
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The drawing shows a hydraulic chamber with a spring (spring constant = 1600 N/m) attached to the input piston and a rock of mass
Triss [41]

Answer:

\Delta x=245\ mm

Explanation:

Given:

  • spring constant of the spring attached to the input piston, k=1600\ N.m^{-1}
  • mass subjected to the output plunger, m=40\ kg

<u>Now, the force due to the mass:</u>

F=m.g

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F=392\ N

<u>Compression in Spring:</u>

\Delta x=\frac{F}{k}

\Delta x=\frac{392}{1600}

\Delta x=0.245\ m

or

\Delta x=245\ mm

8 0
4 years ago
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