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zalisa [80]
3 years ago
5

A 29.0-kg child on a 3.00-m-long swing is released from rest when the ropes of the swing make an angle of 25.0° with the vertica

l. (a) Neglecting friction, find the child's speed at the lowest position.
Physics
1 answer:
KiRa [710]3 years ago
6 0

Answer:

v = 2.348 m/s

Explanation:

Let:

h be the height of the child above the swing's low point,

v be the speed at the low point.

When the ropes make angle 3

25 deg. with the vertical:

h = 3.00(1 - cos(25)).

Equating potential energy at the high point to kinetic energy at the low point:

29.0 * 9.81 * 3.00(1 - cos(25)) = 29.0v^2 / 2

79.96 =29.0v^2 / 2

v = 2.348 m/s

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Answer:

A

Explanation:

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An event occurs in system K' at x' = 2 m, y' = 3.7 m, z' = 3.7 m, and t' = 0. System K' and K have their axes coincident at t =
fiasKO [112]

Answer:

Coordinates of event in system K are (x,y,z,t)=(5.103m , 3.7m , 3.7m , 1.57×10⁻⁸s)

Explanation:

To find the coordinates of event in system K ,we have to use inverse Lorentz transformation

So

x=\frac{x^{|}+vt^{|} }{\sqrt{1-\frac{v^{2} }{c{2} } } } \\x=\frac{2m+0.92c(0) }{\sqrt{1-\frac{(0.92c)^{2} }{c{2} } } }\\x=5.103m\\y=y^{|}\\ y=3.7m\\z=z^{|}\\ z=3.7m

for t

r=\frac{1}{\sqrt{1-v^{2} } } \\r=\frac{1}{\sqrt{1-(0.92)^{2} } } \\r=2.551\\t=r(t^{|}+vx^{|}/c^{2}   )\\t=2.551(0s+(0.92c)(2)/c^{2} )\\t=1.57*10^{-8}s

Coordinates of event in system K are (x,y,z,t)=(5.103m , 3.7m , 3.7m , 1.57×10⁻⁸s)

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3 years ago
Which of the following questions best highlights a shortcoming of the big bang theory?
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<span>A. How could energy become the matter present today? </span>
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A block of 250-mm length and 48 × 40-mm cross section is to support a centric compressive load P. The material to be used is a b
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Answer:

153.6 kN

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The compressive load will generate a stress of

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F = 80*10^6 * 0.048 * 0.04 = 153.6 kN

The smallest admisible load is 153.6 kN

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