To develop this problem it is necessary to use the equations of description of the simple harmonic movement in which the acceleration and angular velocity are expressed as a function of the Amplitude.
Our values are given as
The angular velocity of a body can be described as a function of frequency as
PART A) The expression for the maximum angular velocity is given by the amplitude so that
PART B) The maximum acceleration on your part would be given by the expression
Answer:
5.03 m
Explanation:
The wavelength of a wave is given by
where
v is the speed of the wave
f is the frequency of the wave
For the sonar signal in this problem,
Substituting into the equation, we find the wavelength:
Answer:
3.14 × 10⁻⁴ m³ /s
Explanation:
The flow rate (Q) of a fluid is passing through different cross-sections remains of pipe always remains the same.
Q = Area x velocity
Given:
Diameters of 3 sections of the pipe are given as
d1 = 1.0 cm, d2 = 2.0 cm and d3 = 0.5 cm.
Speed in the first segment of the pipe is
v1 = 4 m/s.
From the equation of continuity the flow rate through different cross-sections remains the same.
Flow rate = Q = A1 v1 = A2 v2 = A3 v3.
Q = A1v1
=π/4 d²1 v1 = π/4 * 0.01² ×4.0 m³/s = 3.14 × 10⁻⁴ m³ /s
Answer:
a. 4v
Explanation:
Alf moves with speed v
Alf travel during the same amount of time that is Δt = (1/4)s
v = (1/4)s / Δt = s / 4 Δt
s / Δt = 4 v
Beth travels a distance s during time Δt,
speed of Beth = s / Δt = 4 v .
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The average speed of an object is defined as the distance traveled divided by the time elapsed. Velocity is a vector quantity, and average velocity can be defined as the displacement divided by the time.
