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Olegator [25]
3 years ago
10

Which formula is used to find an object's acceleration?

Physics
1 answer:
swat323 years ago
5 0

Answer:

its the third one

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Which of the following frequencies falls in the range of RF waves used by commercial radio broadcasting stations?
erma4kov [3.2K]

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.

7 0
3 years ago
Read 2 more answers
Coherent light with wavelength 598 nm passes through two very narrow slits, and the interference pattern is observed on a screen
Stella [2.4K]

Answer:

1.196 μm

Explanation:

D = Screen distance = 3 m

\lambda = Wavelength = 598 m

y = Distance of first-order bright fringe from the center of the central bright fringe = 4.84 mm

d = Slit distance

tan\theta=\frac{y}{D}\\\Rightarrow \theta=tan^{-1}{\frac{y}{D}}\\\Rightarrow \theta=tan^{-1}{\frac{4.84\times 10^{-3}}{3}}\\\Rightarrow \theta=0.09243\ ^{\circ}

sin\theta=\frac{\lambda}{d}\\\Rightarrow d=\frac{\lambda}{sin\theta}\\\Rightarrow d=\frac{598\times 10^{-9}}{sin0.09243}\\\Rightarrow d=0.00037066\ m

For first dark fringe

dsin\theta=\frac{\lambda'}{2}\\\Rightarrow \lambda'=2dsin\theta\\\Rightarrow \lambda'=2\times 0.00037066\times sin0.09243\\\Rightarrow \lambda'=1.196\times 10^{-6}\\\Rightarrow \lambda'=1.196\ \mu m

Wavelength of first-order dark fringe observed at this same point on the screen is 1.196 μm

3 0
3 years ago
What determines the path that an object in projectile motion follows? (Ignore air resistance)
blondinia [14]
Gravity only determines the path in projectile motion
8 0
3 years ago
PLEASE HELP ME ON THIS ONE ANYONE
andreev551 [17]

Answer:

A & B

Explanation:

A & B Would be the right answer since Morse code cannot be represented through the height of the fire.

6 0
2 years ago
How many minutes are in 2 1<br>4 hours?​
amid [387]

Answer:

If you are meaning to say 214 the answer is 12840 minutes

Explanation:

5 0
3 years ago
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