Welll ... if you read the question, you'll see what a jumbled mess you actually
posted, but I think I can pull enough out of it to give you some answers.
<span>One
light bulb in a string of lights goes out. This causes all of the
other
lights in the string to also go out. The entire string of lights
must be one series circuit.
2. When a switch is turned from the off to the on position,
an open circuit is changed to a closed circuit.
3.Early telephone poles had wires that connected inside bell-shaped
glass enclosures like the ones shown below. (nothing is shown below)
several bell-shaped glass enclosures
These glass enclosures helped to
keep the electric current from moving
outside the circuit of wires. The
glass enclosures were used as insulators.
4.Use Ohm’s Law
to determine the resistance in a circuit if the
voltage is 12.0 volts
and the current is 4.0 amps.
Ohms law: Resistance = (voltage) / (current)
= (12.0 v) / (4.0 amp)
= 3 ohms .
5. "In a closed circuit, the
current only flows from the power source
to an electrical device such
as a lamp."
This statement is false. The current eventually has to get back
to the power source. If it doesn't then there's no 'circuit', and
nothing works. That's a big part of the reason why the plug
has 2 prongs on it, and the cord from the plug to the lamp
has 2 wires in it. </span>
Answer:
the frequency of the oscillation is 1.5 Hz
Explanation:
Given;
mass of the spring, m = 1500 kg
extention of the spring, x = 5 mm = 5 x 10⁻³ m
mass of the driver = 68 kg
The weight of the driver is calculated as;
F = mg
F = 68 x 9.8 = 666.4 N
The spring constant, k, is calculated as;
k = F/m
k = (666.4 N) / (5 x 10⁻³ m)
k = 133,280 N/m
The angular speed of the spring is calculated;
The frequency of the oscillation is calculated as;
ω = 2πf
f = ω / 2π
f = (9.426) / (2π)
f = 1.5 Hz
Therefore, the frequency of the oscillation is 1.5 Hz
Nebular cloud should be the answer, but i believe nuclear fusion is the closest answer, because it fuses together to become a cloud
hope this helps
The initial velocity of the ball is 0. Applying:
v = u + at
v = 0 + 229 x 0.08
v = 18.3 m/s
a)
Vx = Vcos(∅)
Vx = 18.3cos(52.3)
Vx = 11.2 m/s
b)
Vy = Vsin(∅)
Vy = 18.3sin(52.3)
Vy = 14.5 m/s
Answer:
The distance between the 2 points is 4.6558 meters.
Explanation:
The distance between 2 points 
is given by
In our case the given 2 points are 
hence the distance is
