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3 years ago

a stone attached to 1m long string is moving with the speed of 5ms in a circle find the centripetal acceleration of the stone

Physics

1 answer:

Dafna1 [17]3 years ago

5 0

Answer:

The centripetal acceleration of the stone is 5 m/s²

Explanation:

The length of the string to which the stone is attached, r = 1 m

The speed with which the string is rotated, v = 5 m/s

The centripetal acceleration, $a_c$ , is given as follows;

$a_c = \dfrac{v^2}{r}$

Therefore, the centripetal acceleration of the stone found as follows;

$a_c = \dfrac{(5 \ m/s)^2}{1 \ m} = 5 \ m/s^2$

The centripetal acceleration of the stone, $a_c$ = 5 m/s².

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A glider with mass 0.24 kg sits on a frictionless horizontal air track, connected to a spring of negligible mass with force cons

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Answer:

$v=2.556m/s$

Explanation:

From the conservation of mechanical energy

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3 years ago

Two identical asteroids travel side by side while touching one another. If the asteroids are composed of homogeneous pure iron a

victus00 [196]

Answer:

diameter = 21.81 ft

Explanation:

The gravitational force equation is:

$F=\frac{GMm}{R^{2} }$

Where:

F => Gravitational force or force of attraction between two masses
M => Mass of asteroid 1
m => Mass of asteroid 2
R => Distance between asteroids 1 and 2 (from center of gravity)

We also know that the asteroids are identical so their masses are identical:

Since R is the distance between centers of the two asteroids and their diameters are identical (see attachment), we can conclude that:

R=d=2r

We don´t know the mass of the asteroids but we know they are composed of pure iron, so we can relate their masses to their density:

m=ρV

This is going to be helpful because the volume of a sphere is:

$\frac{4}{3}\pi r^{3}$

And know we can write our original force of gravity equation in terms of the radius of the asteroids:

$F=\frac{GMm}{R^{2} } =\frac{Gmm}{(2r)^{2} } =\frac{Gm^{2} }{4r^{2} }$
$F=\frac{G ( \frac{4}{3}\pi r^{3}ρ)^{2} }{4r^{2} }$
$F= \frac{G(16)\pi ^{2} r^{6} ρ^{2}}{(9)(4)r^{2} } =\frac{G(16)\pi ^{2} r^{4}ρ^{2} }{36}$

Now let´s plug in the values we know:

$F = 1 lb$ mutual gravitational attraction force
$G = 6.67(10)^{-11}$ gravitational constant
$ρ_{iron} =491.5 \frac{lb}{ft^{3} }$

$1= \frac{6.67(10)^{-11} \pi ^{2} r^{4} (491.5)^{2}}{36}$

Solve for r and multiply by 2 because 2r = diameter

$d=2\sqrt[4]{\frac{1}{7.07(10)^{-5} } }$

Result is d = 21.81 Feet

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sasho [114]

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3 years ago

a stone attached to 1m long string is moving with the speed of 5ms in a circle find the centripetal acceleration of the stone​

a stone attached to 1m long string is moving with the speed of 5ms in a circle find the centripetal acceleration of the stone