Which is located furthest from our sun?
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There are 3 forces acting on the stoplight:
• its weight <em>W</em>, with magnitude <em>W</em> = 100 N, pointing directly downward
• two tension forces <em>T</em>₁ and <em>T</em>₂ with equal magnitude <em>T</em>₁ = <em>T</em>₂ = <em>T</em> = 1000 N, both making an angle of <em>θ</em> with the horizontal, but one points left and the other points right
The stoplight is in equilibrium, so by Newton's second law, the net vertical force acting on it is 0, such that
∑ <em>F</em> = <em>T</em>₁ sin(<em>θ</em>) + <em>T</em>₂ sin(180° - <em>θ</em>) - <em>W</em> = 0
We have sin(180° - <em>θ</em>) = sin(<em>θ</em>) for all <em>θ</em>, so the above reduces to
2<em>T</em> sin(<em>θ</em>) = <em>W</em>
2 (1000 N) sin(<em>θ</em>) = 100 N
sin(<em>θ</em>) = 0.05
<em>θ</em> ≈ 2.87°
If <em>y</em> is the vertical distance between the stoplight and the ground, then
tan(<em>θ</em>) = (15 m - <em>y</em>) / (100 m)
Solve for <em>y</em> :
tan(2.87°) = (15 m - <em>y</em>) / (100 m)
<em>y</em> = 15 m - (100 m) tan(2.87°)
<em>y</em> ≈ 9.99 m
A billiard ball collides with a stationary identical billiard ball to make it move. If the collision is perfectly elastic, the first ball comes to rest after collision.
<h3>Why does the first ball comes to rest after collision ?</h3>Let m be the mass of the two identical balls.
u1 = velocity before the collision of ball 1
u2 = 0 = velocity of second ball that is at rest
v1 and v2 are the velocities of the balls after the collision.
From the conservation of momentum,
∴ mu1 + mu2 = mv1 + mv2
∴ mu1 = mv1 + mv2
∴ u1 = v1 + v2
In an elastic collision, the kinetic energy of the system before and after collision remains same.
∴
∴
∴ ₁
₂ = 0
- It is impossible for the mass to be zero.
- Because the second ball moves, velocity v2 cannot be zero.
- As a result, the velocity of the first ball, v1, is zero, indicating that it comes to rest after collision.
An elastic collision is a collision between two bodies in which the total kinetic energy of the two bodies remains constant. There is no net transfer of kinetic energy into other forms such as heat, noise, or potential energy in an ideal, fully elastic collision.
Can learn more about elastic collision from brainly.com/question/12644900
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