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:
The frictional force acting on the block is 14.8 N.
Explanation:
Given that,
Weight of block = 37 N
Coefficients of static = 0.8
Kinetic friction = 0.4
Tension = 24 N
We need to calculate the maximum friction force
Using formula of friction force

Put the value into the formula


So, the tension must exceeds 29.6 N for the block to move
We need to calculate the frictional force acting on the block
Using formula of frictional force

Put the value in to the formula


Hence, The frictional force acting on the block is 14.8 N.
D; solar system, because the planets are inside it.
Answer:
s = it+1/2 at²
s= 8×3+1/2 (10)(3)²
s = 24+45
s= 69
the object was thrown from a height of 69 meters
Answer:
A compound
Explanation:
A compound is a substance formed when two or more elements are chemically joined