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:
Hoop.
Explanation:
The angular acceleration performed at a given torque:

The moments of inertia of each element are described below:
Hoop

Solid sphere

Flat disk

Hollow sphere

The greater the moment of inertia, the greater the torque to obtain the same angular acceleration. Therefore, the hoop requires the largest torque to receive the same angular acceleration.
Answer:
320N
Explanation:
The magnitude of the torque required is expressed using the formula;
T = Fr sin theta where;
F is the force
r is the radius = 9cm = 0.09m
theta is the angle of inclination = 8 degrees
Torque T = 4Nm
Substitute the given values and get F
4 = F(0.09)sin8
4 = 0.0125F
F = 4/0.0125
F = 320N
Hence the magnitude of the force required when the force is applied at 8 degrees to the wrench is 320N
Answer:
jwhgrewhuejqiwmkosjcdihwbfuqjiwdkmojcshidvwuf hiiii againnnn :)) good luck
(1 cal/g °C) x (4000 g) x (45 - 25)°C = 80000 cal = 80 kcal. So the answer is 80 kcal .