Step-#1:
Ignore the wire on the right.
Find the strength and direction of the magnetic field at P,
caused by the wire on the left, 0.04m away, carrying 5.0A
of current upward.
Write it down.
Step #2:
Now, ignore the wire on the left.
Find the strength and direction of the magnetic field at P,
caused by the wire on the right, 0.04m away, carrying 8.0A
of current downward.
Write it down.
Step #3:
Take the two sets of magnitude and direction that you wrote down
and ADD them.
The total magnetic field at P is the SUM of (the field due to the left wire)
PLUS (the field due to the right wire).
So just calculate them separately, then addum up.
Answer:
Explanation:
Given that,
Mass of the heavier car m_1 = 1750 kg
Mass of the lighter car m_2 = 1350 kg
The speed of the lighter car just after collision can be represented as follows
![m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2}](https://tex.z-dn.net/?f=m_1u_1%2Bm_2u_2%3Dm_1v_1%2Bm_2v_2%5C%5C%5C%5Cv_2%3D%5Cfrac%7Bm_1u_1%2Bm_2u_2-m_1v_1%7D%7Bm_2%7D)
![v_2=\frac{(1850)(1.4)+(1450)(-1.10)-(1850)(0.250)}{1450} \\\\=\frac{2590+(-1595)-(462.5)}{1450} \\\\=\frac{2590-1595-462.5}{1450} \\\\=\frac{532.5}{1450}\\\\=0.367m/s](https://tex.z-dn.net/?f=v_2%3D%5Cfrac%7B%281850%29%281.4%29%2B%281450%29%28-1.10%29-%281850%29%280.250%29%7D%7B1450%7D%20%5C%5C%5C%5C%3D%5Cfrac%7B2590%2B%28-1595%29-%28462.5%29%7D%7B1450%7D%20%5C%5C%5C%5C%3D%5Cfrac%7B2590-1595-462.5%7D%7B1450%7D%20%5C%5C%5C%5C%3D%5Cfrac%7B532.5%7D%7B1450%7D%5C%5C%5C%5C%3D0.367m%2Fs)
b) the change in the combined kinetic energy of the two-car system during this collision
![\Delta K.E=(\frac{1}{2} m_1v_1^2+\frac{1}{2} m_2v_2^2)-(\frac{1}{2} m_1u_1^2+\frac{1}{2} m_2u_2^2)\\\\=\frac{1}{2} (m_1(v_1^2-u_1^2)+m_2(v_2^2-u_2^2))](https://tex.z-dn.net/?f=%5CDelta%20K.E%3D%28%5Cfrac%7B1%7D%7B2%7D%20m_1v_1%5E2%2B%5Cfrac%7B1%7D%7B2%7D%20m_2v_2%5E2%29-%28%5Cfrac%7B1%7D%7B2%7D%20m_1u_1%5E2%2B%5Cfrac%7B1%7D%7B2%7D%20m_2u_2%5E2%29%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B2%7D%20%28m_1%28v_1%5E2-u_1%5E2%29%2Bm_2%28v_2%5E2-u_2%5E2%29%29)
substitute the value in the equation above
![=\frac{1}{2} (1850((0.250)^2-(1.4)^2)+(1450((0.3670)^2-(-1.10)^2)\\\\=\frac{1}{2}(11850(0.0625-1.96)+(1450(0.1347)-(1.21))\\\\= \frac{1}{2}(11850(-1.8975))+(1450(-1.0753))\\\\=\frac{1}{2} (-3510.375+(-1559.185)\\\\=\frac{1}{2} (-5069.56)\\\\=-2534.78J](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B2%7D%20%281850%28%280.250%29%5E2-%281.4%29%5E2%29%2B%281450%28%280.3670%29%5E2-%28-1.10%29%5E2%29%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B2%7D%2811850%280.0625-1.96%29%2B%281450%280.1347%29-%281.21%29%29%5C%5C%5C%5C%3D%20%5Cfrac%7B1%7D%7B2%7D%2811850%28-1.8975%29%29%2B%281450%28-1.0753%29%29%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B2%7D%20%28-3510.375%2B%28-1559.185%29%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B2%7D%20%28-5069.56%29%5C%5C%5C%5C%3D-2534.78J)
Hence, the change in combine kinetic energy is -2534.78J
Answer:39.88 rad/s
Explanation:
Given
mass of cylinder m_1=18 kg
radius R=1.7 m
angular speed ![\omega =40rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D40rad%2Fs)
mass of
dropped at r=0.3 m from center
let
be the final angular velocity of cylinder
Conserving Angular momentum
![L_1=L_2](https://tex.z-dn.net/?f=L_1%3DL_2)
![\left ( \frac{m_1R^2}{2}\right )\omega =\left ( \frac{m_1R^2}{2}+m_2r^2\right )\omega _2](https://tex.z-dn.net/?f=%5Cleft%20%28%20%5Cfrac%7Bm_1R%5E2%7D%7B2%7D%5Cright%20%29%5Comega%20%3D%5Cleft%20%28%20%5Cfrac%7Bm_1R%5E2%7D%7B2%7D%2Bm_2r%5E2%5Cright%20%29%5Comega%20_2)
![\left ( \frac{18\cdot 1.7^2}{2}\right )\cdot 40=\left ( \frac{18\cdot 1.7^2}{2}+0.8\cdot 0.3^2\right )\omega _2](https://tex.z-dn.net/?f=%5Cleft%20%28%20%5Cfrac%7B18%5Ccdot%201.7%5E2%7D%7B2%7D%5Cright%20%29%5Ccdot%2040%3D%5Cleft%20%28%20%5Cfrac%7B18%5Ccdot%201.7%5E2%7D%7B2%7D%2B0.8%5Ccdot%200.3%5E2%5Cright%20%29%5Comega%20_2)
![26.01\times 40=26.082\times \omega _2](https://tex.z-dn.net/?f=26.01%5Ctimes%2040%3D26.082%5Ctimes%20%5Comega%20_2)
![\omega _2=39.88 rad/s](https://tex.z-dn.net/?f=%5Comega%20_2%3D39.88%20rad%2Fs)
C decreased the factor cuz the max is smaller
•Every action has an equal and opposite reaction (the object is putting force on the target, and the target is putting an equal amount of force back)
•Am object in motion (the object) will stay in motion unless an outside force acts upon it (the Target)
And as for the third one I really don’t know, those seem to be the only two, I’m sorry. I did what a could, Hope it kinda helps :)