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 statement of the student is correct.
Since B attained a higher velocity in a short amount of time, that is it accelerated faster(having a larger slope).
Slope = dy/dx
That is, <u>Velocity</u>
Time
which is acceleration.
That's my guess.
Hope it's right.
If an object is on a frictionless surface, to keep it at a constant velocity you can’t apply any force because otherwise, the object will accelerate, and the velocity will change.
It would be D I believe! Depending on the angle of the mirror and distance positioned!
Answer:
Explanation:
Given that :
mass of the SUV is = 2140 kg
moment of inertia about G , i.e 
= 875 kg.m²
We know from the conservation of angular momentum that:
![mv_1 *0.765 = [I+m(0.765^2+0.895^2)] \omega_2](https://tex.z-dn.net/?f=mv_1%20%2A0.765%20%3D%20%5BI%2Bm%280.765%5E2%2B0.895%5E2%29%5D%20%5Comega_2)
![2140v_1*0.765 = [875+2140(0.765^2+0.895^2)] \omega_2](https://tex.z-dn.net/?f=2140v_1%2A0.765%20%3D%20%5B875%2B2140%280.765%5E2%2B0.895%5E2%29%5D%20%5Comega_2)
From the conservation of energy as well;we have :
^2 -2140(9.81)[\sqrt{0.76^2+0.895^2} -0.765]] =0](https://tex.z-dn.net/?f=%5B%5Cfrac%7B1%7D%7B2%7D%20%5B875%2B2140%280.765%5E2%2B0.895%5E2%29%5D%280.4262%20%5C%20v_1%29%5E2%20-2140%289.81%29%5B%5Csqrt%7B0.76%5E2%2B0.895%5E2%7D%20-0.765%5D%5D%20%3D0)
