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Digiron [165]
3 years ago
11

This rock known as balanced rock sits on a thin spike of rock in a canyon in Idaho. Explain the that keep the rock balanced on i

ts tiny pedestal

Physics
2 answers:
Ivahew [28]3 years ago
6 0
Based on the given picture above, I can say that this kind of rock is what we call a PRECARIOUS BOULDER, or in simpler terms, it is called a BALANCING ROCK. It is a geological formation that occurs naturally. This can be found in several parts of the earth. How the rock is kept balanced on its tiny pedestal is that it is firmly attached to a base rock or a stem. 
gizmo_the_mogwai [7]3 years ago
6 0

Answer:

Now, suppose you have a more common figure, like a square, and you want to balance it over a pencil for example.

When you put the square over the pencil, you want that the point where the square touches the pencil is right in the middle of one of the faces of the square. This is because, in this situation, you know that the mass is evenly distributed over all sides now, and none side of the square is pulling down with more force than another side.

Now, this is the same with this rock, the mass is evenly distributed around the contact point, so no side has more force downwards than another side, then all the forces are canceled, and if there are no forces acting on the rock, it will remain in place.

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murzikaleks [220]

Answer:

160 Nm

Explanation:

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The fragment of an asteroid or any interplanetary material is known as a: A) limestone dignitary satellite. B) moon. C) shower m
ElenaW [278]
The fragment of an asteroid or any interplanetary material is known as a a : D. Meteroid

Human came in contact with this material mostly because it penetrate the atmosphere and fall within the earth surface

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What type of acceleration does an object moving in a circular path with constant speed experience?
Marysya12 [62]

An object moving in a circular path has centripetal acceleration. <em>(A)</em>

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3 years ago
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A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strike
Anvisha [2.4K]

Answer:

(a)Magnitude=28.81 m/s

Direction=33.3 degree below the horizontal

(b) No, it is not perfectly elastic collision

Explanation:

We are given that

Mass of stone, M=0.150 kg

Mass of bullet, m=9.50 g=9.50\times 10^{3} kg

Initial speed of bullet, u=380 m/s

Initial speed of stone, U=0

Final speed of bullet, v=250m/s

a. We have to find the magnitude and direction of the velocity of the stone after it is struck.

Using conservation of momentum

mu+ MU=mv+ MV

Substitute the values

9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V

3.61i=2.375j+0.150V

3.61 i-2.375j=0.150V

V=\frac{1}{0.150}(3.61 i-2.375j)

V=24.07i-15.83j

Magnitude of velocity of stone

=\sqrt{(24.07)^2+(-15.83)^2}

|V|=28.81 m/s

Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s

Direction

\theta=tan^{-1}(\frac{y}{x})

=tan^{-1}(\frac{-15.83}{24.07})

\theta=tan^{-1}(-0.657)

=33.3 degree below the horizontal

(b)

Initial kinetic energy

K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2

K_i=685.9 J

Final kinetic energy

K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2

=\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2

K_f=359.12 J

Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.

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I think it is A) but someone might need to double check that.

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