Answer:
its a sure for not but i think its a 192583
Answer:
1) R1 + ((R2 × R3)/(R2 + R3))
2) 0.5 A
3) 3.6 V
Explanation:
1) We can see that resistors R2 and R3 are in parallel.
Formula for sum of parallel resistors; 1/Rt = 1/R2 + 1/R3
Making Rt the subject gives;
Rt = (R2 × R3)/(R2 + R3)
Now, Resistor R1 is in series with this sum of R2 and R3. Thus;
Total resistance of circuit = R1 + ((R2 × R3)/(R2 + R3))
2) R_total = R1 + ((R2 × R3)/(R2 + R3))
We are given;
R1 = 7.2 Ω
R2 = 8 Ω
R3 = 12 Ω
R_total = 7.2 + ((8 × 12)/(8 + 12))
R_total = 7.2 + 4.8
R_total = 12 Ω
Formula for current is;
I = V/R
I = 6/12
I = 0.5 A
3) since current through the circuit is 0.5 and R1 is 7.2 Ω.
Thus, potential difference through R1 is;
V = IR = 0.5 × 7.2 = 3.6 V
I'm gonna have to assume the girl is on the right side and boy on left.
The net force is the sum of all forces on an object (includes negatives).
Let's say the force of the boy is variable <em>b</em>. Use the formula F = ma.
<em>b </em>+ 3.5 = 0.2(2.5)
This is now simple algebra. Solve to get that <em />the boty is exerting a force of -3N to the left.
Answer:
v = ![\left[\begin{array}{c}0.66&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0.66%260%5Cend%7Barray%7D%5Cright%5D)
m/s
Explanation:
The position vector r of the bug with linear velocity v and angular velocity ω in the laboratory frame is given by:
The velocity vector v is the first derivative of the position vector r with respect to time:
![\overrightarrow{v}=[vcos(\omega t)-\omega vtsin(\omega t)]\hat{x}+[vsin(\omega t)+\omega vtcos(\omega t)]\hat{y}](https://tex.z-dn.net/?f=%5Coverrightarrow%7Bv%7D%3D%5Bvcos%28%5Comega%20t%29-%5Comega%20vtsin%28%5Comega%20t%29%5D%5Chat%7Bx%7D%2B%5Bvsin%28%5Comega%20t%29%2B%5Comega%20vtcos%28%5Comega%20t%29%5D%5Chat%7By%7D)
The given values are:
