Answer:
Yes, the value of g affected by the radius.
Explanation:
The formula for the force of gravity of 2 objects is
, where m1 and m2 are the masses of the 2 objects, r is the radius, and G is the gravitational constant, which is approximately 
.
Therefore, as the radius if bigger, the force of gravity is going to be smaller exponentially.
If 50 identical light bulbs are connected in series across
a single power source, then the voltage across each bulb
is ( 1/50 ) of the voltage delivered by the power source.
<h2>The frequency of tuning fork B is 475 Hz</h2>
Explanation:
The number of beats produced is equal to the frequency difference between the two tuning forks .
The frequency of A = 480 Hz
Thus the frequency of B can be = 485 or 475 Hz
When B is filed , its frequency increases and it start producing 2 beats with A .
Thus it frequency must be 475 Hz , by filing , it increases to 478 Hz .
By which it gives 480 - 478 = 2 beats per second .
This cannot happen with frequency 485 , because by filing this difference increases in place of decreasing .
Well, first of all, you really shouldn't use ' W ' for the unit when you
talk about resistors.
You may have seen the resistors written as 6ω, 12ω, and 2ω in your
book or on the homework sheet. But that little symbol ' ω ' is not a ' w '.
It's the small Greek letter 'omega'. The CAPITAL omega is ' Ω '. It's used
to label resistors because it's short for "ohms". So the resistors in this
problem have resistances of 6Ω, 12Ω, and 2Ω, and we have to do some
manipulating of the individual resistors to find out what resistance the
battery actually sees.
The parallel combination of the first two resistors looks like a single
resistor, whose value is
1 / (1/6 + 1/12)
= 1 / (2/12 + 1/12)
= 1 / (3/12)
= 12/3 = 4Ω .
Now, that parallel combination is connected in series with 2Ω .
All three resistors together look like a single resistor of
4Ω + 2Ω = 6Ω .
So the battery thinks there's a single resistor connected to it,
with 6Ω of resistance. The current out of the battery is
I = V / R = (24v) / (6Ω) = 4 Amperes.
That 4 Amperes of current will split between the parallel resistors,
but it will ALL flow through the series 2Ω resistor because there's
no other path through that part of the circuit.
So the current through the 2Ω resistor is 4 Amperes. (B).
Note:
The POWER dissipated by the 2Ω resistor is
P = I² R = (4A)² · (2Ω) = 32 watts .
This is a fair amount of heat, so you'll need to provide some way
to remove the heat from the resistor, otherwise it'll burn or crack.
