Answer:
(a) T = 10 s
(b) f = 0.1 Hz
(c) λ = 32 m
(d) v = 3.2 m/s
(e) Insufficient data
Explanation:
(a)
Time period is defined as the time interval required for one wave to pass. Therefore, the time period can be given as:
T = Period = Time Taken/No. of Waves
T = 50 s/5
<u>T = 10 s</u>
<u></u>
(b)
Frequency is the reciprocal of time period:
f = frequency = 1/T
f = 1/10 s
<u>f = 0.1 Hz</u>
<u></u>
(c)
Wavelength is the distance between two consecutive crests or troughs:
<u>λ = Wavelength = 32 m</u>
<u></u>
(d)
Speed of wave is given by the following formula:
Speed = v = fλ
v = (0.1 Hz)(32 m)
v = 3.2 m/s
(e)
Amplitude cannot be found with given data.
A change in position with respect to a reference point is called motion
hope it helps...
a₀). You know ...
-- the object is dropped from 5 meters
above the pavement;
-- it falls for 0.83 second.
a₁). Without being told, you assume ...
-- there is no air anyplace where the marshmallow travels,
so it free-falls, with no air resistance;
-- the event is happening on Earth,
where the acceleration of gravity is 9.81 m/s² .
b). You need to find how much LESS than 5 meters
the marshmallow falls in 0.83 second.
c). You can use whatever equations you like.
I'm going to use the equation for the distance an object falls in
' T ' seconds, in a place where the acceleration of gravity is ' G '.
d). To see how this all goes together for the solution, keep reading:
The distance that an object falls in ' T ' seconds
when it's dropped from rest is
(1/2 G) x (T²) .
On Earth, ' G ' is roughly 9.81 m/s², so in 0.83 seconds,
such an object would fall
(9.81 / 2) x (0.83)² = 3.38 meters .
It dropped from 5 meters above the pavement, but it
only fell 3.38 meters before something stopped it.
So it must have hit something that was
(5.00 - 3.38) = 1.62 meters
above the pavement. That's where the head of the unsuspecting
person was as he innocently walked by and got clobbered.
Data !
hope this helped <3