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3 years ago

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The large blade of a helicopter is rotating in a horizontal circle. The length of the blade is 7.26 m, measured from its tip to

the center of the circle. Find the ratio of the centripetal acceleration at the end of the blade to that which exists at a point located 4.35 m from the center of the circle.

1 answer:

Elan Coil [88]3 years ago

5 0

Answer:

$\frac{a_1}{a_2}=1.67$

Explanation:

The equation for centripetal acceleration is $a_{cp}=r\omega^2$ . If $r_1=7.26m$ is the distance from the center of the point at the tip of the blade and $r_2=4.35m$ is the distance from the center of another point in the blade, and since both points are rotating at the same angular velocity $\omega$ because both of them belong to the blade, the ratio between their centripetal accelerations will be:

$\frac{a_1}{a_2}=\frac{r_1 \omega^2}{r_2 \omega^2}=\frac{7.26m}{4.35m}=1.67$

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The resistance of a conductive wire is given by:
$R= \frac{\rho L}{A}$
where
$\rho$ is the material resistivity
$L$ is the wire length
$A$ is the cross-sectional area of the wire

The length of the wire is quadrupled, so if we call L the original length and L' the new length, we can write
$L'=4 L$

Similarly, the radius of the wire is doubled (r'=2r), so the new area is
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And if we substitute into the equation, we find that the new resistance of the wire is
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3 years ago

A thin walled spherical shell is rolling on a surface. What

ExtremeBDS [4]

Answer:

$=\frac{1/3}{5/6} = 0.4$

Explanation:

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where

M represent sphere mass

R -sphere radius

we know linear speed is given as $v = r\omega$

translational $K.E = \frac{1}{2} mv^2 = \frac{1}{2} m(r\omega)^2$

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total kinetic energy will be

$K.E = \frac{1}{2} m(r\omega)^2 + \frac{1}{2} \frac{2}{3} MR^2 \omega^2$

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Answer:

0.00903 rad

0.00926 rad

6.268\times 10^{-6}

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$\theta_m=\dfrac{3.48\times 10^6}{3.85\times 10^8}\\\Rightarrow \theta_m=0.00903\ rad$

The angle subtended by the Moon is 0.00903 rad

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The angle subtended by the Sun is 0.00926 rad

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Answer:

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x = -F / k

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