A circuit contains two light bulbs connected in parallel. What would happen to the brightness of each light bulb if two more lig
ht bulbs were added in parallel to the first ones? A. The brightness of each bulb would decrease because the total resistance of the circuit would decrease.
B. The brightness of each bulb would increase because the total resistance of the circuit would increase.
C. The brightness of each bulb would remain the same even though the total resistance of the circuit would decrease.
D. The brightness of each bulb would remain the same even though the total resistance of the circuit would increase.
In order to say definitely what would happen, we need to make sure that the light bulbs are the only items connected to the battery or power supply. What I mean is: You can't start out with, say, a string of Christmas tree lights, and at some point in the string, there's an extra little bulb in parallel with one of the lights in the string. If that's the situation ... if there's anything else in the circuit besides the two light bulbs in parallel ... then we'd need more information in order to predict what happens when we make changes to that part of the circuit.
So the whole circuit consists of: Two light bulbs in parallel, connected to a battery. That's all.
Now, we connect two MORE light bulbs in parallel with the first ones. Now we have FOUR light bulbs, all in parallel, connected to the battery.
-- The battery would see the total resistance of the circuit DECREASE. (An electron that leaves one terminal of the battery now has 4 ways to reach the other side of the battery, instead of just two. It's easier for electrons to get there, so more of them make the trip every second.)
-- Did you hear that ? "More electrons make the trip every second." That means there's more current leaving the battery.
-- IF the battery or power supply is able to deliver the increased current, then the original devices ... the original two light bulbs ... don't even know that anything has happened. Their brightness doesn't change.
I just told you that connecting more stuff in parallel makes no difference to the devices that are already there. If that sounds weird to you, then let me tell you this:
EVERYTHING in your house is in parallel with everything else ! Every light, kitchen appliance, motor, heater, electric furnace, radio, air conditioner, phone charger, TV, computer, mixer, microwave ... everything that plugs into a socket in the wall ... is in parallel with everything else that you plug in.
When you plug in something new, or turn on something that was off, it should have no effect on the things that were already running. (If it does ... if, say, the lights in kitchen go dim when you turn on the microwave ... then the wiring in your house is a little bit skimpy for running all the things that we run in our houses now.)
C. The brightness of each bulb would remain the same even though the total resistance of the circuit would decrease.
Explanation;
-If light bulbs are connected in parallel to a voltage source, the brightness of the individual bulbs remains more-or-less constant as more and more bulbs are added to the circuit.
-The current increases as more bulbs are added to the circuit and the overall resistance decreases. In addition, if one bulb is removed from the circuit the other bulbs do not go out. Each bulb is independently linked to the voltage source
C. The brightness of each bulb would remain the same even though the total resistance of the circuit would decrease.
Explanation;
-If light bulbs are connected in parallel to a voltage source, the brightness of the individual bulbs remains more-or-less constant as more and more bulbs are added to the circuit.
-The current increases as more bulbs are added to the circuit and the overall resistance decreases. In addition, if one bulb is removed from the circuit the other bulbs do not go out. Each bulb is independently linked to the voltage source
Let X (t) represent the equation of the position, then you have to d2x / dt2 = 5.
Applying the fundamental theorem of the calculation dx/dt = 5t + vo. The speed equation is V (t) = 5t + vo. Since the initial velocity is 30m/s, V (0) = 5 (0) + vo = 30. Therefore, V (t) = dx/dt = 5t + 30. Applying again the fundamental theorem of the calculation X (t) = 5t^2 / 2 + 30t + xo.
Displacement in 5 seconds is given by X (5) - X (0).
X (5) - X (0) = 5 (5)^2/2 +3 (5) + Xo - 5 (0)^2/2 -3 (0) -Xo = 155/2
It would need to be over 20 because if the load of the Pulley E is 20 and the effort is 20, then they will be equal and the Pulley would not move, so your answer is at least 20
In order to understand the results of another scientist's work, any scientist who reads the report MUST understand the measurement system in which the result is described. Otherwise the result is meaningless to him.
We can consider only the horizontal component since the horizontal component of the velocity is constant (), and we want to know how much time it takes for the balloon to travel a horizontal distance at that speed. The definition of (horizontal) velocity is , so we have:
