Air resistance is a special kind of frictional force that acts on objects as they travel through air. Identify the situation whe
re this frictional force is ABSENT, from the ones given below:
1 answer:
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The butterfly takes a vertical perpendicular path equivalent to 9m and travels a horizontal distance of 17m. The net path between the two is equivalent to that of the hypotenuse, so we will apply the Pythagorean theorem.
Therefore the magnitude of the butterfly's displacement is 9m
Answer:
The final translational seed at the bottom of the ramp is approximately 4.84 m/s
Explanation:
The given parameters are;
The radius of the sphere, R = 2.50 cm
The mass of the sphere, m = 4.75 kg
The translational speed at the top of the inclined plane, v = 3.00 m/s
The length of the inclined plane, l = 2.75 m
The angle at which the plane is tilted, θ = 22.0°
We have;
+
=
+
K = (1/2)×m×v²×(1 + I/(m·r²))
I = (2/5)·m·r²
K = (1/2)×m×v²×(1 + 2/5) = 7/10 × m×v²
U = m·g·h
h = l×sin(θ)
h = 2.75×sin(22.0°)
∴ 7/10×4.75×3.00² + 4.75×9.81×2.75×sin(22.0°) = 7/10 × 4.75ײ + 0
7/10×4.75×3.00² + 4.75×9.81×2.75×sin(22.0°) ≈ 77.93
∴ 77.93 ≈ 7/10 × 4.75ײ
² = 77.93/(7/10 × 4.75)
≈ √(77.93/(7/10 × 4.75)) ≈ 4.84
The final translational seed at the bottom of the ramp, ≈ 4.84 m/s.
a) The spring constant is 12,103 N/m
b) The mass of the trailer 2,678 kg
c) The frequency of oscillation is 0.478 Hz
d) The time taken for 10 oscillations is 20.9 s
Explanation:
a)
When the two children jumps on board of the trailer, the two springs compresses by a certain amount
Since the system is then in equilibrium, the restoring force of the two-spring system must be equal to the weight of the children, so we can write:
(1)
where
m = 76.2 kg is the mass of each children
is the acceleration of gravity
is the equivalent spring constant of the 2-spring system
For two springs in parallel each with constant k,
Substituting into (1) and solving for k, we find:
b)
The period of the oscillating system is given by
where
And for the system in the problem, we know that
T = 2.09 s is the period of oscillation
m is the mass of the trailer
is the equivalent spring constant of the system
Solving the equation for m, we find the mass of the trailer:
c)
The frequency of oscillation of a spring-mass system is equal to the reciprocal of the period, therefore:
where
f is the frequency
T is the period
In this problem, we have
T = 2.09 s is the period
Therefore, the frequency of oscillation is
d)
The period of the system is
T = 2.09 s
And this time is the time it takes for the trailer to complete one oscillation.
In this case, we want to find the time it takes for the trailer to complete 10 oscillations (bouncing up and down 10 times). Therefore, the time taken will be the period of oscillation multiplied by 10.
Therefore, the time needed for 10 oscillations is:
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