Answer:
Between 128dB and 149dB
Explanation:
When howling the Howler monkies has a volume between 128db and 149dB. This is as a result of them having a large hyoid bone in the neck.
Answer:
Space probes are made to conduct science experiments. They do not have people on them. Space probes have helped scientists get information about our solar system. Most probes are not designed to return to Earth. Some have landed on other planets! Others have flown past the planets and taken pictures of them for scientists to see. There are even some space probes that go into orbit around other planets and study them for a long time. The information they gather is used to help us understand the weather and other changes which happen on planets other than the Earth. This information is important in helping to plan other space missions such as ones to Mars and to Saturn.
Explanation:
To solve this problem we will apply the linear motion kinematic equations. With the data provided we will calculate the time of the first object to fall. Later we will get the time difference between the two. This difference will allow us to find the free fall distance. Through the distance we will find the initial velocity, that is,



The second object is thrown downward at one second later and it meets the first object at the water is


The distance of the object will travel due to free fall acceleration is



The distance of the object will travel due to its initial velocity is




Therefore the initial speed of the second object is 21.06m/s
Answer:
(a) f= 622.79 Hz
(b) f= 578.82 Hz
Explanation:
Given Data
Frequency= 600 Hz
Distance=1.0 m
n=120 rpm
Temperature =20 degree
Before solve this problem we need to find The sound generator moves on a circular with tangential velocity
So
Speed of sound is given by
c = √(γ·R·T/M)
............in an ideal gas
where γ heat capacity ratio
R universal gas constant
T absolute temperature
M molar mass
The speed of sound at 20°C is
c = √(1.40 ×8.314472J/molK ×293.15K / 0.0289645kg/mol)
c= 343.24m/s
The sound moves on a circular with tangential velocity
vt = ω·r.................where
ω=2·π·n
vt= 2·π·n·r
vt= 2·π · 120min⁻¹ · 1m
vt= 753.6 m/min
convert m/min to m/sec
vt= 12.56 m/s
Part A
For maximum frequency is observed
v = vt
f = f₀/(1 - vt/c )
f= 600Hz / (1 - (12.56m/s / 343.24m/s) )
f= 622.789 Hz
Part B
For minimum frequency is observed
v = -vt
f = f₀/(1 + vt/c )
f= 600Hz / (1 + (12.56m/s / 343.24m/s) )
f= 578.82 Hz