Answer:
Explanation:
Let the bigger crate be in touch with the ground which is friction less. In the first case both m₁ and m₂ will move with common acceleration because m₁ is not sliding over m₂.
1 ) Common acceleration a = force / total mass
= 234 / ( 25 +91 )
= 2.017 m s⁻².
2 ) Force on m₁ accelerating it , which is nothing but friction force on it by m₂
= mass x acceleration
= 25 x 2.017
= 50.425 N
The same force will be applied by m₁ on m₂ as friction force which will act in opposite direction.
3 ) Maximum friction force that is possible between m₁ and m₂
= μ_s m₁g
= .79 x 25 x 9.8
= 193.55 N
Acceleration of m₁
= 193 .55 / 25
= 7.742 m s⁻²
This is the common acceleration in case of maximum tension required
So tension in rope
= ( 25 +91 ) x 7.742
= 898 N
4 ) In case of upper crate sliding on m₂ , maximum friction force on m₁
= μ_k m₁g
= .62 x 25 x 9.8
= 151.9 N
Acceleration of m₁
= 151.9 / 25
= 6.076 m s⁻².
11,000 m = 11 km
11 km/s over 380,000km
380,000 / 11 = 34545.4 seconds
34545.4 / 60 = 575.7 minutes
Momentum = (mass) x (velocity)
Original momentum before the hit =
(0.16 kg) x (38 m/s) this way <==
= 6.08 kg-m/s this way <==
Momentum after the hit =
(0.16) x (44 m/s) that way ==>
= 7.04 kg-m/s that way ==>
Change in momentum = (6.08 + 7.04) = 13.12 kg-m/s that way ==> .
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Change in momentum = impulse.
Impulse = (force) x (time the force lasted)
13.12 kg-m/s = (force) x (0.002 sec)
(13.12 kg-m/s) / (0.002 sec) = Force
6,560 kg-m/s² = 6,560 Newtons = Force
( about 1,475 pounds ! ! ! )
Answer:
= 0.55 m
Explanation:
A standing wave is characterized by anti-nodes and nodes.
Antinodes are points on a standing wave at maximum amplitude, while nodes are points on the standing wave that are stationary and have zero amplitude.
The distance between two adjacent nodes or two adjacent anti-nodes is equivalent to half the wavelength.
Therefore, in this case the half wavelength is 27.5 cm.
Thus, wavelength = 27.5 × 2
= 55 cm
<u>= 0.55 m</u>
Most likely gravity, because the gravity would pull it off course or wobble.
