Answer: v = 3.684 m/s
Explanation: The angular frequency (ω) of a loaded spring is given as
ω = √k/m
Where ω = angular frequency, k =spring constant = 11.9 N/m, m = mass of object = 40.1 g = 0.0401 kg.
The velocity of a simple harmonic motion is defied as
v = ω√A² - x²
Where A = Amplitude = 24.7cm = 0.247m and x = displacement.
For our question, we where asked to find velocity at half way, at half way, x = A/2
Hence at half way, x = 0.247/2 = 0.1235 m.
We need to get the value of angular frequency first.
ω = √(11.9/0.0401)
ω = √296.758
ω = 17.22 rad/s.
Then the velocity is
v = 17.22 √0.247² - 0.1235²
v = 17.22 √0.061009 - 0.01525225
v = 17.22 √0.04575675
v = 17.22 × 0.2139
v = 3.684 m/s
Answer:
L = 2.61 10³ kg m² / s
Explanation:
Let's approximate the drone to a particle the angular momentum is
L = r x p
L = m r v sin θ
in this case v = 9.5 m / s the mass of the drone is m = 11 kg.
The distance can therefore be found using the Pythagorean theorem, but we can see that the relation
r_perpendicular = r sin θ
r_perpendicular = 25 m
as the flight is horizontal this height does not change
L = m v r_perpendicular
let's calculate
L = 11 9.5 25
L = 2.61 10³ kg m² / s
Power = work / time --> time = work / power = 3600 J / 275 watts = 13.1 seconds.
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Answer:
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Explanation:
Answer: 1.77kg
Explanation:
Using the concept or hooke's law, the force (F) applied to a string is directly proportional its extension (e)
F=ke
K is the elastic constant
If the mass applied initially is 0.59kg i.e 5.9N force, we have
5.9=ke... (1)
If a second block is attached let's say 'm', its force will be 10m, and the extension will become 3e
10m = 3ke ... (2)
Dividing eqn 1 by 2,
5.9/10m = ke/3ke
5.9/10m = 1/3
10m = 5.9×3
m = 5.9×3/10
m = 1.77kg