Answer: its 15 i think i hope this helps
Explanation:
As an example, the distance from Earth to the Sun is about 150,000,000,000 meters – a very large distance indeed. In scientific notation, the distance is written as 1.5 × 1011 m.
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) <em>2.278 x 10^-5 volts</em>
b) <em>1.139 x 10^-6 Ampere</em>
c) <em>2.59 x 10^-11 W</em>
Explanation:
The radius of the wire r = 2 mm = 0.002 m
the number of turns N = 200 turns
direction of the magnetic field ∅ = 25°
magnetic field strength B = 0.02 T
varying time = 2 sec
The cross sectional area of the wire = 
==> A = 3.142 x 
= 1.257 x 10^-5 m^2
Field flux Φ = BA cos ∅ = 0.02 x 1.257 x 10^-5 x cos 25°
==> Φ = 2.278 x 10^-7 Wb
The induced EMF is given as
E = NdΦ/dt
where dΦ/dt = (2.278 x 10^-7)/2 = 1.139 x 10^-7
E = 200 x 1.139 x 10^-7 = <em>2.278 x 10^-5 volts</em>
<em></em>
<em></em>
b) If the two ends are connected to a resistor of 20 Ω, the current through the resistor is given as
= E/R
where R is the resistor
= (2.278 x 10^-5)/20 = <em>1.139 x 10^-6 Ampere</em>
<em></em>
<em></em>
<em> </em>c) power delivered to the resistor is given as
P = E
P = (1.139 x 10^-6) x (2.278 x 10^-5) = <em>2.59 x 10^-11 W</em>
Answer:
x = 1.6 + 1.7 t^2 omitting signs
a) at t = 0 x = 1.6 m
b) V = d x / d t = 3.4 t
at t = 0 V = 0
c) A = d^2 x / d t^2 = 3.4 (at t = 0 A = 3.4 m/s^2)
d) x = 1.6 + 1.7 * (4.4)^2 = 34.5 (position at 4.4 sec = 34.5 m)
The magnitude (in N) of the force she must exert on the wrench is 150.1 N.
<h3>
Force exerted by the wrench</h3>
The force exerted by the wrench is calculated using torque formula as follows;
torque, τ = F x r x sinθ
where;
- F is the applied force
- r is the perpendicular distance if force applied
F = τ /(r sinθ)
F = (39) / (0.3 sin 60)
F = 150.1 N
Thus, the magnitude (in N) of the force she must exert on the wrench is 150.1 N.
Learn more about torque here: brainly.com/question/14839816
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