The equilibrium conditions allow to find the results for the balance forces are:
When the acceleration is zero we have the equilibrium conditions for both linear and rotational motion.
∑ F = 0
∑ τ = 0
Where F are the forces and τ the torques.
The torque is the product of the force and the perpendicular distance to the point of support,
The free-body diagrams are diagrams of the forces without the details of the bodies, see attached for the free-body diagram of the system.
We write the translational equilibrium condition.
F₁ - W₁ - W₂ + F₂ = 0
We write the equation for the rotational motion, set our point of origin at scale 1, and the counterclockwise turns are positive.
F₂ 2 - W₁ 1 - W₂ 1.5 = 0
Let's calculate F₂
F₂ = 
F₂ = (m g + M g 1.5)/ 2
F₂ = 
F₂ = 558.6 N
We substitute in the translational equilibrium equation.
F₁ = W₁ + W₂ - F₂
F₁ = (m + M) g - F₂
F₁ = (12 +68) 9.8 - 558.6
F₁ = 225.4 N
In conclusion using the equilibrium conditions we can find the forces of the balance are:
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Answer: Atoms are electrically neutral because they have equal numbers of protons (positively charged) and electrons (negatively charged). If an atom gains or loses one or more electrons, it becomes an ion.
Answer:
If the frequency is doubled the wavelength is only half as long.If you put your fingertip in a pool of water and repeatedly move it up and down,you will create circular water waves.
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Gravitational force is the result of the earth pulling down on the book, so the normal force is best described as the earth or more accurately the table pushing up on the book
Answer:
<em>Second option</em>
Explanation:
<u>Linear Momentum</u>
The linear momentum of an object of mass m and speed v is
P=mv
If two or more objects are interacting in the same axis, the total momentum is
Where the speeds must be signed according to a fixed reference
The images show a cart of mass 2m moves to the left with speed v since our reference is positive to the right
The second cart of mass m goes to the right at a speed v
The total momentum before the impact is
The total momentum after the collision is negative, both carts will join and go to the left side
The first option shows both carts with the same momentum before the collision and therefore, zero momentum after. It's not correct as we have already proven
The third option shows the 2m cart has a positive greater momentum than the other one. We have proven the 2m car has negative momentum. This option is not correct either
The fourth option shows the two carts keep separated after the collision, which contradicts the condition of the question regarding "they hook together".
The second option is the correct one because the mass 
has a negative momentum and then the sum of both masses keeps being negative
