An electromagnetic wave of a frequency between about 104<span> and 10</span>11<span> or 10</span>12<span> Hz, as used for long-distance communication</span>
Answer:
a)
b)
Explanation:
Given:
mass of bullet, 
compression of the spring, 
force required for the given compression, 
(a)
We know

where:
a= acceleration


we have:
initial velocity,
Using the eq. of motion:

where:
v= final velocity after the separation of spring with the bullet.


(b)
Now, in vertical direction we take the above velocity as the initial velocity "u"
so,

∵At maximum height the final velocity will be zero

Using the equation of motion:

where:
h= height
g= acceleration due to gravity


is the height from the release position of the spring.
So, the height from the latched position be:



Answer: 529.9 Hz
Explanation:
Here we need to use the Doppler equation, so we have:
f' = f*(v + v0)/(v - vs)
Here, f is the frequency = 500Hz
v is the velocity of the wave, = 334m/s
v0 is the velocity of the observer = 20m/s
vs is the velocity of the source = 0m/s
Then we have:
f' = 500Hz*(334m/s + 20m/s)/(334m/s) = 529.9 Hz
Answer:
Similarity: >>Time is independent variable and such is on the x-axis. ... >>Distance time graph tells you how much distance you have travelled, while velocity time graph tells you your acceleration. The difference between them is that the velocity-time graph reveals the speed of an object (and whether it is slowing down or speeding up), while the position-time graph describes the motion of an object over a period of time.
Explanation: