Answer:
A thin, taut string tied at both ends and oscillating in its third harmonic has its shape described by the equation y(x,t)=(5.60cm)sin[(0.0340rad/cm)x]sin[(50.0rad/s)t]y(x,t)=(5.60cm)sin[(0.0340rad/cm)x]sin[(50.0rad/s)t], where the origin is at the left end of the string, the x-axis is along the string, and the y-axis is perpendicular to the string. (a) Draw a sketch that shows the standing-wave pattern. (b) Find the amplitude of the two traveling waves that make up this standing wave. (c) What is the length of the string? (d) Find the wavelength, frequency, period, and speed of the traveling waves. (e) Find the maximum transverse speed of a point on the string. (f) What would be the equation y(x, t) for this string if it were vibrating in its eighth harmonic?
Answer: Relative motion
Explanation: If two objects are moving either towards or away from each other with both having their velocities in a reference frame and someone is outside this reference frame seeing the motion of the two objects.
The observer ( in his own frame of reference) will measure a different velocity as opposed to the velocities of the two object in their own reference frame. p
Both the velocity measured by the observer in his own reference frame and the velocity of both object in their reference is correct.
Velocities of this nature that have varying values based on motion referenced to another body is known as relative velocity.
Motion of this nature is known as relative motion.
<em>Note that the word reference frame is simply any where the motion is occurring and the specified laws of motion is valid</em>
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For this example of ours, the reference frame of the companion is the train and the telephone poles has their reference frame as the earth.
The companion will measure the velocity of the telephone poles relative to him and the velocity of the telephone pole relative to an observer outside the train will be of a different value.
A is the correct answer !!!
Answer:
a) 
b) 
Explanation:
Given:
- upward acceleration of the helicopter,
- time after the takeoff after which the engine is shut off,
a)
<u>Maximum height reached by the helicopter:</u>
using the equation of motion,
where:
u = initial velocity of the helicopter = 0 (took-off from ground)
t = time of observation
b)
- time after which Austin Powers deploys parachute(time of free fall),
- acceleration after deploying the parachute,
<u>height fallen freely by Austin:</u>
where:
initial velocity of fall at the top = 0 (begins from the max height where the system is momentarily at rest)
time of free fall
<u>Velocity just before opening the parachute:</u>
<u>Time taken by the helicopter to fall:</u>
where:
initial velocity of the helicopter just before it begins falling freely = 0
time taken by the helicopter to fall on ground
height from where it falls = 250 m
now,
From the above time 7 seconds are taken for free fall and the remaining time to fall with parachute.
<u>remaining time,</u>
<u>Now the height fallen in the remaining time using parachute:</u>
<u>Now the height of Austin above the ground when the helicopter crashed on the ground:</u>
Acceleration....................................... <span />
