In a football game, a receiver is standing still, having just caught a pass. Before he can move, a tackler, running at a velocit
1 answer:
Answer:
Explanation:
The football players collide in a completely inelastic collision, in other words they have the same velocity after the collision, this velocity has a magnitude V=1.6m/s.
We need to use the conservation of momentum Law, the total momentum is the same before and after the collision, at the initial point the receiver does not have any speed
(1)
We solve in order to find the receiver mass:
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The frequency of middle C on a string is
f = 261.6 Hz.
The given linear density is
ρ = 0.02 g/cm = (0.02 x 10⁻³ kg)/(10⁻² m)
= 0.002 kg/m
The length of the string is L = 1 m.
Let T = the tension in the string (N).
The velocity of the standing wave is
In the fundamental mode, the wavelength, λ, is equal to the length, L.
That is
Because v = fλ, therefore
From given information, obtain
T = (0.002 kg/m)*(261.6 1/s)²*(1 m)²
= 136.87 N
Answer: 136.9 N (nearest tenth)
f = 261.6 Hz.
The given linear density is
ρ = 0.02 g/cm = (0.02 x 10⁻³ kg)/(10⁻² m)
= 0.002 kg/m
The length of the string is L = 1 m.
Let T = the tension in the string (N).
The velocity of the standing wave is
In the fundamental mode, the wavelength, λ, is equal to the length, L.
That is
Because v = fλ, therefore
From given information, obtain
T = (0.002 kg/m)*(261.6 1/s)²*(1 m)²
= 136.87 N
Answer: 136.9 N (nearest tenth)
Answer:
0.22 m
Explanation:
m = 0.43 kg, K = 561 N/m
Vmax = 8 m/s
Let the amplitude of the oscillations be A.
The formula for the angular frequency of oscillation sis given by
ω = 36.1 rad/s
The formula for the maximum velocity is given by
Vmax = ω x A
A = Vmax / ω
A = 8 / 36.1 = 0.22 m