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
Answer:
x = 0.9 m
Explanation:
For this exercise we must use the rotational equilibrium relation, we will assume that the counterclockwise rotations are positive
∑ τ = 0
60 1.5 - 78 1.5 + 30 x = 0
where x is measured from the left side of the fulcrum
90 - 117 + 30 x = 0
x = 27/30
x = 0.9 m
In summary the center of mass is on the side of the lightest weight x = 0.9 m
Answer:
Explanation:
The formula method is used to calculate termination payments on a prematurely ended swap, where the terminating party compensates the losses borne by the non-terminating party due to the early termination.
Answer: Option (b) is the correct answer.
Explanation:
The given data is as follows.
= 30 J, l = 10 cm = 0.1 m (as 1 m = 100 cm)
F = 10 N
The formula to calculate the work done is as follows.
W = 
Hence, putting the given values into the above formula as follows.
W = 
= 
= 1 J
As the dart loses 1 joule of energy when it goes through the barrel. Therefore, the kinetic energy it has when it comes out is calculated as follows.
K.E = 30 - 1
= 29 J
Thus, we can conclude that the kinetic energy of the fired dart is 29 J.