The vector c has a magnitude of 24.6m and it is in the negative y direction. Therefore

The vector b is 41.4° up from the x-axis. Therefore
![\vec{b} = b[cos(41.4^{o}) \hat{i} + sin(41.4^{o}) \hat{j} ] =b(0.75\hat{i} + 0.6613 \hat{j})](https://tex.z-dn.net/?f=%5Cvec%7Bb%7D%20%3D%20b%5Bcos%2841.4%5E%7Bo%7D%29%20%5Chat%7Bi%7D%20%2B%20sin%2841.4%5E%7Bo%7D%29%20%5Chat%7Bj%7D%20%5D%20%3Db%280.75%5Chat%7Bi%7D%20%2B%200.6613%20%5Chat%7Bj%7D%29)
The vector a is 27.7° up from the x-axis. Therefore
![\vec{a} = a[cos(22.7^{o})\hat{i} + sin(27.7^{o})\hat{j}] = a(0.8854\hat{i} + 0.4648\hat{j})](https://tex.z-dn.net/?f=%5Cvec%7Ba%7D%20%3D%20a%5Bcos%2822.7%5E%7Bo%7D%29%5Chat%7Bi%7D%20%2B%20sin%2827.7%5E%7Bo%7D%29%5Chat%7Bj%7D%5D%20%3D%20%20a%280.8854%5Chat%7Bi%7D%20%2B%200.4648%5Chat%7Bj%7D%29)
Because

, the sum of the x and y components should be zero. Therefore,
For the x-component,
0.8854a + 0.75b = 0
or
a + 0.847b = 0 (1)
For the y-component,
0.4648a + 0.6613b - 24.6 = 0
or
a + 1.4228b = 52.926 (2)
Subtract (1) from (2).
0.5758b = 52.926
b = 91.917
a = -0.847b = -77.854
Answer:
The magnitude of vector a is -77.85 m
The magnitude of vector b is 91.92 m
Answer:
R1 + R2 = R = 12 for resistors in series - so R1 = R2 if they are identical
2 R1 = 12 and R1 = R2 = 6 ohms
1 / R = 1 / R1 + 1 / R2 for resistors in parallel
R = R1 * R2 / (R1 + R2) = 6 * 6 / (6 + 6) = 3
The equivalent resistance would be 3 ohms if connected in parallel
Answer:
-450 m/s
Explanation:
Momentum is conserved.
p₀ = p
0 = (1.5 kg) (1.5 m/s) + (0.005 kg) v
v = -450 m/s
Answer:

Explanation:
Let the length of the string is L.
Let T be the tension in the string.
Resolve the components of T.
As the charge q is in equilibrium.
T Sinθ = Fe ..... (1)
T Cosθ = mg .......(2)
Divide equation (1) by equation (2), we get
tan θ = Fe / mg




As θ is very small, so tanθ and Sinθ is equal to θ.


It depends on the type of interference.
For constructive interference, add the amplitudes to get |35 + 41| = 76 units.
For destructive, subtract them |35 - 41| = 6 units