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>
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<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 />
<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:
<em> B.0</em>
Explanation:
Change in momentum: This is defined as the product of mass and change in velocity of a body. or it can be defined as the product of force and time of a body. The fundamental unit of change in momentum is kg.m/s
Change in momentum = M(V-U)......................... Equation 1
where M = mass of the ball, V = final velocity of the ball, U = initial velocity of the ball.
Let: M = m kg and V = U = v m/s
Substituting these values into equation 1
Change in momentum = m(v-v)
Change in momentum = m(0)
Change in momentum = 0 kg.m/s
<em>Therefore the momentum of the ball has not changed.</em>
<em>The right option is B.0</em>
mass of iron block given as

density of iron block is

now the volume of the iron piece is given as


Now when this iron block is complete submerged in oil inside the beaker the buoyancy force on the iron block will be given as

here we know that
= density of liquid = 916 kg/m^3


Now for the reading of spring balance we can say the spring force and buoyancy force on the block will counter balance the weight of the block at equilibrium



So reading of spring balance will be 16.45 N
Now for other scale which will read the normal force of the surface we can write that normal force on the container will balance weight of liquid + container and buoyancy force on block



So the other scale will read 36.47 N