(6) Wagon B is at rest so it has no momentum at the start. If <em>v</em> is the velocity of the wagons locked together, then
(140 kg) (15 m/s) = (140 kg + 200 kg) <em>v</em>
==> <em>v</em> ≈ 6.2 m/s
(7) False. If you double the time it takes to perform the same amount of work, then you <u>halve</u> the power output:
<em>E</em> <em>/</em> (2<em>t </em>) = 1/2 × <em>E/t</em> = 1/2 <em>P</em>
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Explanation:
Given that,
Mass, m = 0.08 kg
Radius of the path, r = 2.7 cm = 0.027 m
The linear acceleration of a yo-yo, a = 5.7 m/s²
We need to find the tension magnitude in the string and the angular acceleration magnitude of the yo‑yo.
(a) Tension :
The net force acting on the string is :
ma=mg-T
T=m(g-a)
Putting all the values,
T = 0.08(9.8-5.7)
= 0.328 N
(b) Angular acceleration,
The relation between the angular and linear acceleration is given by :
(c) Moment of inertia :
The net torque acting on it is, 
, I is the moment of inertia
Also, 
So,
Hence, this is the required solution.
Answer:
Explanation:
Let h be the height .
initial velocity in first case u = 0
final velocity v = 6 m /s
acceleration due to gravity g = 9.8 m /s²
v² = u² + 2 g h
6² = 0 + 2 x 9.8 x h
h = 1.837 m .
For second case u = 3 m /s
v² = u² + 2 gh
= 3² + 2 x 1.837 x 9.8
= 9 + 36
= 45 m
v = 6.7 m /s
Answer:
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Explanation: North Korea and Cuba are the only places you can't buy Coca-Cola
