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zepelin [54]
3 years ago
7

A person standing at the top of mountain aconcagua would be approximately 4.3 mi high. the radius of earth is 3959 mi. what is t

he distance to the horizon from this point?

Physics
2 answers:
Fed [463]3 years ago
6 0

Explanation :

It is given that,

The radius of the earth is, r = 3959 mi

From the attached figure, we have to find the distance to the horizon from this point.

So, using Pythagoras theorem:

AP^2=OA^2-OP^2

x^2=(R+h)^2-R^2

x^2=(3959+4.3)^2-(3959)^2

x^2=34065.89\ mi

x=184.57\ mi

The distance to the horizon from this point is 184.57 mi.

Hence, this is the required solution.

PilotLPTM [1.2K]3 years ago
5 0
Besides ambitious, it would be 184.6 mi
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A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.
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Answer:

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part (c) F = -228.6 N towards left.

Explanation:

Given,

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Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = v_2\ =\ 2.31\ m/s.

Let v_1 be the final velocity of first puck after the collision.

From the conservation of linear momentum,

m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = 25\times 10^{-3}\ sec

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

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