Answer:
F = 0.78[N]
Explanation:
The given values correspond to forces, we must remember or take into account that the forces are vector quantities, that is, they have magnitude and direction. Since we have two X-Y coordinate axes (two-dimensional), we are going to decompose each of the forces into the X & y components.
<u>For F₁</u>
<u />![F_{y}=2[N]](https://tex.z-dn.net/?f=F_%7By%7D%3D2%5BN%5D)
<u />
<u>For F₂</u>
![F_{x}=2*cos(60)\\F_{x}=1[N]\\F_{y}=-2*sin(60)\\F_{y}=-1.73[N]](https://tex.z-dn.net/?f=F_%7Bx%7D%3D2%2Acos%2860%29%5C%5CF_%7Bx%7D%3D1%5BN%5D%5C%5CF_%7By%7D%3D-2%2Asin%2860%29%5C%5CF_%7By%7D%3D-1.73%5BN%5D)
<u>For F₃</u>
<u />![F_{x}=-1*sin(60)\\F_{x}=-0.866[N]\\F_{y}=1*cos(60)\\F_{y}=0.5 [N]](https://tex.z-dn.net/?f=F_%7Bx%7D%3D-1%2Asin%2860%29%5C%5CF_%7Bx%7D%3D-0.866%5BN%5D%5C%5CF_%7By%7D%3D1%2Acos%2860%29%5C%5CF_%7By%7D%3D0.5%20%5BN%5D)
<u />
Now we can sum each one of the forces in the given axes:
![F_{x}=1-0.866=0.134[N]\\F_{y}=2-1.73+0.5\\F_{y}=0.77[N]](https://tex.z-dn.net/?f=F_%7Bx%7D%3D1-0.866%3D0.134%5BN%5D%5C%5CF_%7By%7D%3D2-1.73%2B0.5%5C%5CF_%7By%7D%3D0.77%5BN%5D)
Now using the Pythagorean theorem we can find the total force.
![F=\sqrt{(0.134)^{2} +(0.77)^{2}}\\F= 0.78[N]](https://tex.z-dn.net/?f=F%3D%5Csqrt%7B%280.134%29%5E%7B2%7D%20%2B%280.77%29%5E%7B2%7D%7D%5C%5CF%3D%200.78%5BN%5D)
Answer:
The near point of an eye with power of +2 dopters, u' = - 50 cm
Given:
Power of a contact lens, P = +2.0 diopters
Solution:
To calculate the near point, we need to find the focal length of the lens which is given by:
Power, P = 
where
f = focal length
Thus
f = 
f = 
= + 0.5 m
The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.
Now, by using lens maker formula:
where
u = object distance = 25 cm = 0.25 m = near point of a normal eye
u' = image distance
Now,
Solving the above eqn, we get:
u' = - 0.5 m = - 50 cm
Answer:
liquid a particles slides past pother
Explanation:
mark brainliest :))
