Answer:
b- how much work can be done in a given time
Hello,
Here is your answer:
The proper answer to this question is option A "puffin".
Your answer is A.
If you need anymore help feel free to ask me!
Hope this helps!
Answer:
This is a very straightforward problem, and not too difficult to do in your head. If we have two nearly equal resistances, the series resistance (Rs) is twice the average value (midpoint between the two) and the parallel resistance Rp) is approximately equal to half the midpoint value. So a rough initial guess might be 12 and 13 ohms, where Rs is 25 ohms but Rp is approximately 6.25 ohms. Since Rp needs to be lower, we can play around with resistor values. Here’s an easy shortcut:
Let’s start with the equation for two parallel resistors:
Rp = (R1 * R2)/(R1 + R2)
Since Rs = R1 + R2,
Rp = (R1 * R2)/Rs.
We can rewrite this as
Rp * Rs = R1 * R2
Plugging in our numbers (Rp = 6, Rs = 25), we see that R1 + R2 = 150. What two numbers in the vicinity of 12 and 13 have the product of 150 and a sum of 25? Immediately 10 and 15 come to mind! Easy enough!
Conclusions be adjusted as necessary to incorporate new knowledge.
Hope this helps =)
im taking an integrated physics and chemistry test rite now!
Answer:
The tension in the string is equal to Ct
And the time t0 when the rension in the string is 27N is 3.6s.
Explanation:
An approach to solving this problem jnvolves looking at the whole system as one body by drawing an imaginary box around both bodies and taking summation of the forces. This gives F2 - F1 = Ct. This is only possible assuming the string is massless and does not stretch, that way transmitting the force applied across it undiminished.
So T = Ct
When T = 27N then t = T/C = 27/7.5 = 3.6s
