The (blank) of vibration of a wave is defined as that which has the lowest frequency
2 answers:
Answer:
the <u>Fundamental Frequency</u> of vibration of a wave is defined as that which has the lowest frequency
Explanation:
As we know that the frequency of vibrations on a string is given as
now we know that
N = number of harmonics
v = speed of the wave
L = length of the string
So here for the lowest frequency of the wave we know that
N = 1
so lowest frequency of the wave is known as fundamental frequency of the wave
the <u>Fundamental Frequency</u> of vibration of a wave is defined as that which has the lowest frequency
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Answer:
v₀ = 60.38 mi / h
With this stopping distance, the starting speed should have been 60.38 mi/h, which is much higher than the maximum speed allowed.
Explanation:
For this exercise let's start by using Newton's second law
Y axis
N-W = 0
N = W
X axis
fr = m a
the expression for the friction force is
fr = μ N
we substitute
μ mg = m a
μ g = a
calculate us
a = 0.620 9.8
a = 6.076 m / s²
now we can use the kinematics relations
v² = v₀² - 2 a x
suppose v = 0
v₀ = Ra 2ax
let's calculate
v₀ =
v₀ = 27.00 m / s
let's slow down to the english system
v₀ = 27.0 m / s (3.28 ft / 1m) (1 mile / 5280 ft) (3600s / 1h)
v₀ = 60.38 mi / h
With this stopping distance, the starting speed should have been 60.38 mi/h, which is much higher than the maximum speed allowed.
at the top most point if Rupert will not fall then normal force at the top point is almost zero for minimum speed
so here we can say
now if
so above will be the minimum speed