One form of Ohm's Law says . . . . . Resistance = Voltage / Current .
R = V / I
R = (12 v) / (0.025 A)
R = (12 / 0.025) (V/I)
<em>R = 480 Ohms</em>
I don't know if the current in the bulb is steady, because I don't know what a car's "accumulator" is. (Floogle isn't sure either.)
If you're referring to the car's battery, then the current is quite steady, because the battery is a purely DC storage container.
If you're referring to the car's "alternator" ... the thing that generates electrical energy in a car to keep the battery charged ... then the current is pulsating DC, because that's the form of the alternator's output.
Answer:
Explanation:
Assuming no friction between the roller coaster car and the hill, and neglecting air resistance, the kinetic energy the roller coaster car would have at the bottom of the hill would be equal to its gravitational potential energy at the top of the hill, by conservation of energy.
Answer:
Explanation:
- The expression for acceleration of the rolling body on an inclined plane is given as a = gsinФ/1 + k²/R²
- where Ф is the angle of inclination, R is the radius, k is the radius of gyration.
- The potential energy of the system is given as ; PE = mgh
- The potential energy will be constant for ring, cylinder, solid sphere, and hollow sphere.
- The total kinetic energy of the rolling body is ; KE = mv²/2 + Iw²/2
- Hence, the total kinetic energy of the ring, cylinder, solid sphere and hollow sphere will be constant.
2. The moment of inertia of the ring is given as ;
I = mR²
The moment of inertia of the ring is maximum and therefore reaches the bottom last.
Answer: A
<u>Explanation:</u>
NOTES:
d = 650 meters
t = 10 seconds
**********************************
v = d/t
= 650 meters/10 seconds
= 65 meters/second
Answer:
Explanation:
Energy of signal being radiated per second on all sides = 71 x 10³ J .
At a distance of 220 m it is spread over an area of 4 π x (220)² because it is spreading uniformly on all sides.
So energy crossing per unit area
= 
= 11.67 x 10⁻² Wm⁻²s⁻¹.
This is the intensity of the signal.
At 2200 m this intensity will further reduce by 100 times
So there it becomes equal to
11.67 x 10⁻⁴ Wm⁻² s⁻¹.
