This is a question of resolving the forces in question into horizontal and vertical and then finding the action of the resultant force.
The first force acts at an angle less than 90° and thus its resolved forces are positive. The second force acts at angle larger than 90° and is incident and thus its horizontal value is positive while its vertical value is negative.
Therefore;
For force of 300 N at 30°;
Horizontal value = 300*Cos 30 = 259.81 N
Vertical value = 300*Sin 30 = 150 N
For 150 N at 135°;
Horizontal value = 150*Cos (180-135) = 106.07 N
Vertical value = -150*Sin (180-135) = -106.07 N
Then,
Resultant horizontal value = 259.81+106.07 = 365.88 N
Resultant vertical value = 150-106.07 = 43.93 N
Therefore,
Resultant force, v = Sqrt (365.88^2+43.93^2) = 368.51 N
Angle of action measured from horizontal = tan ^-1(43.93/365.88) 6.85°
Then,
v = 368.51 N at 6.85° from horizontal
Answer:
14n
Step-by-step explanation:
You multiply 14 by n, which is 14n.
Step-by-step answer:
Given:
Circle C1: x^2+y^2 = 45
Line L1: -3x+y=15
Need to find the points of intersection.
Solution:
basically we need to solve for the roots of equations C1 and L1.
Here, we can use substitution of L1 into C1.
Rewrite L1 as : y=3x+15
substitute into C1:
x^2+(3x+15)^2 = 45
Expand
x^2 + 9x^2+90x+225 = 45
Rearrange terms:
10x^2+90x+180 = 0
Simplify
x^2+9x+18 = 0
Factor
(x+6)(x+3) = 0
so
x=-6 or x=-3
Back-substitute x into L1 to calculate y:
x=-6, y=3*x+15 = 3(-6)+15 = -3 => (-6,-3)
x=-3, y=3*x+15 = 3(-3) + 15 = 6 => (-3, 6)
Therefore the intersection points are (-6,-3) and (-3,6)
Check using equation C1:
(-6)^2+(-3)^2 = 36+9 = 45 ok
(-3)^2+(6)^2 = 9 + 36 = 45 ok
Check using equation L1:
Point (-6,-3) : y = 3x+15 = 3(-6) +15 = -3 ok
Point (-3,6) : y = 3x+15 = 3(-3)+15 = 6 ok.
Answer:
Step-by-step explanation:
If you are trying to find the hypotenuse, then you want to use the equation a^2 + b^2= c^2
1. 7^2 + 9^2=c^2
2. 49+81=c^2
3. c^2=130
4. Square root both sides
Synthetic division yields
See attachment for work
