Answer:
2 seconds
Explanation:
The frequency of a wave is related to its wavelength and speed by the equation
where
f is the frequency
v is the speed of the wave
is the wavelength
For the wave in this problem,
v = 2 m/s
So the frequency is
The period of a wave is equal to the reciprocal of the frequency, so for this wave:
This means that the wave takes 4 seconds to complete one full cycle.
Therefore, the time taken for the wave to go from a point with displacement +A to a point with displacement -A is half the period, therefore for this wave:
<span>(20 cm)/(5 sec) = (0.20 meters)/(5 seconds)
</span>
Answer
given,
Length of the string, L = 2 m
speed of the wave , v = 50 m/s
string is stretched between two string
For the waves the nodes must be between the strings
the wavelength is given by
where n is the number of antinodes; n = 1,2,3,...
the frequency expression is given by
now, wavelength calculation
n = 1
λ₁ = 4 m
n = 2
λ₂ = 2 m
n =3
λ₃ = 1.333 m
now, frequency calculation
n = 1
f₁ = 12.5 Hz
n = 2
f₂= 25 Hz
n = 3
f₃ = 37.5 Hz
Complete question:
The coordinate of a particle in meters is given by x(t)=1 6t- 3.0t³ , where the time tis in seconds. The
particle is momentarily at rest at t is:
Select one:
a. 9.3s
b. 1.3s
C. 0.75s
d.5.3s
e. 7.3s
Answer:
b. 1.3 s
Explanation:
Given;
position of the particle, x(t)=1 6t- 3.0t³
when the particle is at rest, the velocity is zero.
velocity = dx/dt
dx /dt = 16 - 9t²
16 - 9t² = 0
9t² = 16
t² = 16 /9
t = √(16 / 9)
t = 4/3
t = 1.3 s
Therefore, the particle is momentarily at rest at t = 1.3 s
Explanation:
It is given that,
Mass of the tackler, m₁ = 120 kg
Velocity of tackler, u₁ = 3 m/s
Mass, m₂ = 91 kg
Velocity, u₂ = -7.5 m/s
We need to find the mutual velocity immediately the collision. It is the case of inelastic collision such that,
v = -1.5 m/s
Hence, their mutual velocity after the collision is 1.5 m/s and it is moving in the same direction as the halfback was moving initially. Hence, this is the required solution.
