Consider a wave on a string with constant tension. If the frequency of the wave is doubled, by what multiplicative factor does a
1 answer:
Explanation:
It is given that, a wave on a string with constant tension. The frequency of a wave and its speed is related by the following relation as :
............(1)
Where
v is the speed of the wave
l is the length of the string
(a) It is clear from equation (1) that the frequency of a wave is directly proportional to its speed. If the frequency of the wave is doubled, the speed of the wave also becomes double i.e. the speed increases by a factor of 2.
(b) The relationship between the frequency and the wavelength is given by :
It is clear from the above formula that the frequency is inversely proportional to the wavelength of the wave. If the frequency of a wave is doubled, the wavelength becomes half.
Hence, this is the required solution.
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Answer:
a) a = 3.72 m / s², b) a = -18.75 m / s²
Explanation:
a) Let's use kinematics to find the acceleration before the collision
v = v₀ + at
as part of rest the v₀ = 0
a = v / t
Let's reduce the magnitudes to the SI system
v = 115 km / h (1000 m / 1km) (1h / 3600s)
v = 31.94 m / s
v₂ = 60 km / h = 16.66 m / s
l
et's calculate
a = 31.94 / 8.58
a = 3.72 m / s²
b) For the operational average during the collision let's use the relationship between momentum and momentum
I = Δp
F Δt = m v_f - m v₀
F =
F = m [16.66 - 31.94] / 0.815
F = m (-18.75)
Having the force let's use Newton's second law
F = m a
-18.75 m = m a
a = -18.75 m / s²
Answer:
Explanation:
The strength of the electric field produced by a charge Q is given by
where
Q is the charge
r is the distance from the charge
k is the Coulomb's constant
In this problem, the electric field that can be detected by the fish is
and the fish can detect the electric field at a distance of
Substituting these numbers into the equation and solving for Q, we find the amount of charge needed: