homer's physics corner

[#1procrastinator]

I'm having trouble understanding the concepts of apparent weight and weightlessness.
Suppose the normal force was greater than the gravity force would that mean the the person would feel lighter or heavier? also what happens to the apparent weight when a person is traveling in a lift? Thanks

Would feel heavier.
When it goes up you feel heavier cause the normal force increases (required to accelerate you up), when it goes down you feel lighter.

[Chenpionn]

normal force = apparent weight. The greater the normal force, the heavier you feel. In the case of free fall, there is no normal force acting on you, only gravitational (9.8m/s^2 down) and so you feel weightless.

^correct me if I'm wrong

[pi]

normal force = apparent weight. The greater the normal force, the heavier you feel. In the case of free fall, there is no normal force acting on you, only gravitational (9.8m/s^2 down) and so you feel weightless.

^correct me if I'm wrong

Correcting you on something: in physics they use 10m/s^2 not 9.8m/s^2 unless specified for the latter in the question.



edit: typo said "on" twice

[Chenpionn]

right, my bad Guess I'm used to the Heinemann textbooks (they use 9.8m/s^2) 

[Homer]

having trouble with the electronics,

I can do the first but not the second row, also could you please explain the basics and how you got the answer. Thanks

[availn]

having trouble with the electronics,

I can do the first but not the second row, also could you please explain the basics and how you got the answer. Thanks

Pretty much, the higher the resistance of a resistor, the higher the voltage drop across it. When I say "high", I mean compared to the total resistance in the circuit.

Therefore, for voltage dividers, V_R = V_supply x (R/R_total)

R_total in this case is just R₁ + R₂

So for row 2, V_supply is 20V, V_R is 5V, and R is 1000ohms. The 1000ohm resistor drops a quarter of the voltage, so the other resistor must drop the remaining three-quarters; from this we can tell that this resistor is three times more resistive, and is 3000ohms.

[Homer]

i need help me with these questions. i always get confused with such question

[Phy124]

If you're struggling with this question one way of going about it is to designate arbitrary values to the batteries and resistors and go from there. As you know the resistors are all the same value and the same for the batteries, too.

For simplicity's sake, let's say and

From here you can work out the current being supplied by the batteries in each case, the current through the resistors and the power dissipated in them, too.

[Homer]

could someone help me with a) and b)

[Homer]

suppose you have a device which is designed to operate at optimum efficiency when it is connected to an AC supply that has a peak voltage of 170V and a frequency of 60hz. The lamp has an operating resistance of 100ohms.

How much power would this device be consuming when operating at optimum efficiency on
AC?
DC?

how do we calculate this? I know that we have to use p=v^2/r , but what values do i use for v in AC and DC. thanks

[SocialRhubarb]

When using DC, you can directly apply the formula for power, , where V is the voltage drop across your lamp.

When using AC, you use the RMS value, which is kind of like an average. RMS voltage is your peak voltage divided by , which you can then use in your formula.

[lzxnl]

Strictly speaking, you can have DC currents that vary with time as well; just think full-wave rectifiers that effectively take the absolute value of the AC signal, but as the current doesn't change direction, it's DC. Or, the electricity generated from a generator; varies with time, but it is DC. As far as I know, only currents from a battery are effectively constant.

As for which formula to use, the instantaneous power P = V^2/R where V is the voltage at that time.
We want averages though, so we need to average V^2/R. R is a constant, so we need the average of V^2. For an AC circuit, V = Vpeak * sin(2pi*t/T) where T is the period. You could probably prove, as an exercise, that the average value of V^2 = Vpeak^2 * sin^2(2pi*t/T) over an entire period is half of the peak voltage squared by integrating. Square-rooting this "average" yields the RMS value of Vpeak/sqrt 2.

[SocialRhubarb]

In the context of the question, I think that DC would just be constant voltage.

[Homer]

how do they get the left side as the north end and the right as south? I get the opposite. How do we find these ends in a solenoid? Thanks

[BasicAcid]

Always look at the inside of the solenoid.

Grab onto the wirewith your right hand with your thumb pointing in the arrow's direction (the current) and as you can see, you fingers curl to the left on the inside of the solenoid hence why that side is north.

It's called the right hand grip rule and the Heinemann textbook actually explains this pretty well.