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2 years ago

14

Find the radius of the circle formed by the intersection of a sphere of diameter 26 units and a plane that is 5 units away from

the center of the sphere.

1 answer:

horrorfan [7]2 years ago

8 0

Answer:

12 units

Explanation:

This problem can be solved if we take into account the equation for a sphere

$x^{2}+y^{2}+z^{2}=r^{2}\\x^{2}+y^{2}+z^{2}=(\frac{26}{2})^{2}=13^{2}$

where we took that the radius is 13 units. If we take z=5 and we replace this value in the equation of the sphere we have

$x^{2}+y^{2}+(5)^{5}=13^{2}\\x^{2}+y^{2}=144=(12)^{2}$

where we have taken x2 +y2 because if the equation of a circunference.

In this case the intersection is made when we take z=5, for this value the sphere and the plane coincides in values.

Hence, the radius is 12 units

I hope this is useful for you

regards

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3 years ago

: A 70 kg man and a 12 kg sled are on the frictionless ice of a frozen lake, 25 m apart but connected by a rope of negligible ma

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Answer:

$x_1 = 3.74m$

Explanation:

given,

mass of man = 70 kg

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F = m a_s

$a_s = \dfrac{F}{m}$

$a_s = \dfrac{8.2}{12}$

$a_s = 0.68\ m/s^2$

$F = m a_m$

$a_m = \dfrac{F}{m}$

$a_m = \dfrac{8.2}{70}$

$a_m = 0.12\ m/s^2$

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$\dfrac{1}{2}a_ct^2+ \dfrac{1}{2}a_mt^2 = 25$

$(a_c+a_m)t^2=50$

$(0.12+0.68)t^2=50$

$t = \sqrt{\dfrac{50}{0.8}}$

t = 7.90 s

$x_1 = \dfrac{1}{2}a_ct^2$

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3 years ago

Find its moment of inertia about an axis perpendicular to its plane and passing through the midpoint of the line connecting its

antoniya [11.8K]

A) Moment of inertia about an axis passing through the point where the two segments meet : $I_A=\frac{1}{12} M L^2$

B) Moment of inertia passing through the point where the midpoint of the line connects to its two ends: $I x=\frac{1}{3} M L^2$

What is Moment of inertia?

The term "moment of inertia" refers to a physical quantity that quantifies a body's resistance to having its speed of rotation along an axis changed by the application of a torque (turning force). The axis might be internal or exterior, fixed or not.

A) The moment of inertia about an axis passing through the point where the two segments meet is $I_A=\frac{1}{12} M L^2$ given that the rod is bent at the center and distance from all the points to the axis remains the same, the moment of inertia about the center will remain the same.

B) Determine the moment of inertia about an axis passing through the point midpoint of the line which connects the two ends

First step: determine the distance between the ends ( d )

After applying Pythagoras theorem $\mathrm{d}=\frac{\sqrt{2}}{2} L$

Next step : determine distance between the two axis $(\mathrm{x})$

After applying Pythagoras theorem

$\mathrm{x}=\frac{\sqrt{2}}{4} L$$$

Final step : Calculate the value of $\mathrm{I}_{\mathrm{x}}$

applying Parallel Axis Theorem

$I_x=I_8+M x^2$

$\begin{aligned}& =\frac{1}{12} M L^2+\frac{1}{4} M L^2 \\& \therefore \quad I x=\frac{1}{3} M L^2 \\&\end{aligned}$

Hence we can conclude that Moment of inertia about an axis passing through the point where the two segments meet: $I_A=\frac{1}{12} M L^2$ , Moment of inertia passing through the point where the midpoint of the line connects its two ends: $I x=\frac{1}{3} M L^2$

To learn more about moment of inertia visit:brainly.com/question/15246709

#SPJ4

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