Answer:
Explanation:
Sample Response: Yes, the law of conservation of momentum is satisfied. The total momentum before the collision is 1.5 kg • m/s and the total momentum after the collision is 1.5 kg • m/s. The momentum before and after the collision is the same.
a) We use the formula:
T = [g sinθ – a] * m
where g is gravity, θ is the angle, a is acceleration, m is mass
T = [9.81 sin37 - 2.00] * 12 = 46.85 N <span>
b) We use the formula for moment of interia:</span>
I = T * r² / a
I = 46.85 * 0.10² / 2.00 = .2343 kg∙m² <span>
c) The formula we can use here is:</span>
w = α * t = (a/r) * t
w = (2/.10) * 2
<span>w = 40 rad/sec</span>
By power transmitted by string, the frequency must be 163.31 Hz.
We need to know about power transmitted to solve this problem. The power transmitted by a wave on string can be determined by this equation
P = 1/2 . μ . ω² . A² . v
where P is the power, μ is mass per unit length of string, ω is angular speed, A is amplitude and v is wave propagation speed.
the wave propagation can be determined as
v = √(F.l/m)
where F is the string tension, l is length and m is the mass.
From the question above, we know that:
l = 2.7 m
m = 260 g = 0.26 kg
F = 36 N
A = 7.7 mm = 0.0077 m
P = 58 W
Find the mass per unit length
μ = m / l
μ = 0.26 / 2.7
μ = 0.096 kg / m
Find the wave propagation speed
v = √(F.l/m)
v = √(36. 2.7 /0.26)
v = √(373.85)
v = 19.34 m/s
Find the angular speed
P = 1/2 . μ . ω² . A² . v
58 = 1/2 . 0.096 . ω² . 0.0077² . 19.34
ω² = 1053777.29
ω = √1053777.29
ω = 1026.54 rad/s
Find the frequency
ω = 2πf
1026.54 = 2 . 3.14 . f
f = 163.31 Hz
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Answer:
the same time
Explanation:
since the speed of light is very fast (3×10^8 ms^-1) the difference in time arrived from firecrackers A and B to Bob is negligible, and our brain can distinguish such small time difference,l.
