A). 600,000 Hz or 600 KHz
Yes. Commercial broadcasters operate here.
This is the '600' on your AM radio dial.
B). 60 Hz
No. In principle, this frequency might be used for communication or
commercial broadcasting, but it suffers from two inconvenient truths:
-- An efficient antenna for 60 Hz ... either transmitting or receiving ...
needs to be almost 780 miles long.
-- This is the frequency of the electric power utility in the US and
Canada, so every outlet, wire, cable, lamp cord, and electric line
on a pole RADIATES a little bit of signal at this frequency. That's
an awful lot of interference.
C). 6,000,000 Hz or 6 MHz
There's a lot of broadcasting activity here, but it's not commercial
music, news, and sports into local homes and cars.
It's foreign short-wave broadcast, bringing news, propaganda, and
culture from one country to another. Pretty interesting to browse.
D). 6,000 Hz or 6 KHz.
No. Not used for communication, for an interesting reason:
This frequency is smack in the middle of the human hearing range.
So if it were used for communication ... with high-power transmitters
here and there ... then you wouldn't hear it in the air. But wherever
wires were being used to carry sound ... your stereo's speaker wires,
wires from your player to your ear-buds, wires to the telephones in
your house etc ... the wires would act as antennas, picking up
broadcasts at 6 KHz, and the broadcasts would get into everything.
Not a smart plan.
Answer:
A) continue to move to the right, with its speed increasing with time.
Explanation:
As long as force is positive , even when it is decreasing , it will create positive increase in velocity . Hence the body will keep moving with increasing velocity towards the right . The moment the force becomes zero on continuously decreasing , the increase in velocity stops and the body will be moving with the last velocity uniformly towards right . When the force acting on it becomes negative , even then the body will keep on going to the right till negative force makes its velocity zero . D uring this period , the body will keep moving towards right with decreasing velocity .
Hence in the present case A , is the right choice.
Answer:
P = 33.6 [N]
Explanation:
To solve this problem we must use Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.
∑F = m*a
where:
F = forces [N]
m = mass = 14 [kg]
a = acceleration = 6 [m/s²]
![F = 14*6\\F = 84 [N]](https://tex.z-dn.net/?f=F%20%3D%2014%2A6%5C%5CF%20%3D%2084%20%5BN%5D)
In the second part of this problem we must find the work done, where the work in physics is known as the product of force by distance, it is important to make it clear that force must be applied in the direction of movement.

where:
W = work [J]
F = force = 84 [N]
d = displaciment = 40 [m]
![W = 84*40\\W = 3360 [J]](https://tex.z-dn.net/?f=W%20%3D%2084%2A40%5C%5CW%20%3D%203360%20%5BJ%5D)
Finally, the power can be calculated by the relationship between the work performed in a given time interval.

where:
P = power [W]
W = work = 3360 [J]
t = time = 100 [s]
Now replacing:
![P=3360/100\\P=33.6[W]](https://tex.z-dn.net/?f=P%3D3360%2F100%5C%5CP%3D33.6%5BW%5D)
The power is given in watts
Answer:
Explanation:
We know the frequency and the velocity, both of which have good units. All we have to do is rearrange the equation and solve for
λ
:
λ
=
v
f
Let's plug in our given values and see what we get!
λ
=
340
m
s
440
s
−
1
λ
=
0.773
m
Wood isn't as tough as rock. Wood also breaks down in weather and cracks under pressure. Plus rocks were more accessible.