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Pachacha [2.7K]
3 years ago
9

A 1.00-kg mass is attatched to a string 1.0m long and completes a horizontal circle in 0.25s. What is the centripetal accelerati

on and force or the mass?
Physics
1 answer:
liraira [26]3 years ago
3 0

-- The string is 1 m long.  That's the radius of the circle that the mass is
traveling in.  The circumference of the circle is  (π) x (2R) = 2π meters .

-- The speed of the mass is (2π meters) / (0.25 sec) = 8π m/s .

-- Centripetal acceleration is  V²/R = (8π m/s)² / (1 m) = 64π^2 m/s²

-- Force = (mass) x (acceleration) = (1kg) x (64π^2 m/s²) =

                                                         64π^2 kg-m/s² = 64π^2 N = about <span>631.7 N .

</span>
That's it.  It takes roughly a 142-pound pull on the string to keep
1 kilogram revolving at a 1-meter radius 4 times a second !<span> 
</span>
If you eased up on the string, the kilogram could keep revolving
in the same circle, but not as fast.

You also need to be very careful with this experiment, and use a string
that can hold up to a couple hundred pounds of tension without snapping. 
If you've got that thing spinning at 4 times per second and the string breaks,
you've suddenly got a wild kilogram flying away from the circle in a straight
line, at 8π meters per second ... about 56 miles per hour !  This could definitely
be hazardous to the health of anybody who's been watching you and wondering
what you're doing.


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3 years ago
Hallar la distancia que recorre una movil al cabo
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Answer:

Distancia, S = 136 metros

Explanation:

Dados los siguientes datos;

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Velocidad inicial, u = 5 m/s

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Para encontrar la distancia recorrida, usaríamos la segunda ecuación de movimiento;

S = ut + ½at² Sustituyendo en la fórmula, tenemos;

S = 5 × 8 + ½ × 3 × 8²

S = 40 + 1,5 × 64

S = 40 + 96

Distancia, S = 136 metros

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Answer:

Explanation:

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Hope this helped and have a good day

3 0
2 years ago
Read 2 more answers
Jake drops a book out of a window from a height of 10 meters. At what velocity does the book hit the ground?
user100 [1]
G = 9.81 m/sec^2)     g = 9.81\frac{meters}{ second^{2} }

<span>Solving for velocity : </span>

velocity^{2}<span> = 2gh </span>
<span>v = </span>2gh^{ \frac{1}{2} }
<span>v = (2 x 9.81 x 10)^1/2 </span>
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