A constant force of 15 N is being applied at a constant angle os 30 degree on an object at the same time that a constant force of 25 N at a constant angle of 60degree is acting on the object. What is the magnitude and direction of the resultant force?

Step-by-step explanation:

Solution:

(a) is the vertical direction, the force of gravity

on the object (or the wight force: w = Mg ) pulls the crate down while the surface reaction. N. pushes it 103N

rd, as shown:

w = Mg = (20.0kg 9.8m s3 - 196N

N- 196N as well

(b) In the horizontal direction, in the absence of

friction, there are two forces: (F) and the 53-N force. Applying the Newton's 2nd law helps the calculation of the unknown force. F. F - 53N (20.0 kg)(2.5 m 3

F = 103N plz mark me as brainliest

there are many solutions.

step-by-step explanation:

step 1: simplify both sides of the equation.

4x+6=4x^2+10x−4

step 2: subtract 4x^2+10x-4 from both sides.

4x+6−(4x^2+10x−4)=4x^2+10x−4−(4x^2+10x−4)

−4x^2−6x+10=0

step 3: factor left side of equation.

−2(2x+5)(x−1)=0

step 4: set factors equal to 0.

x=-5/2 or x=1