Answer:
a) C.M 
b) 
Explanation:
The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"
The center of mass on a two dimensional plane is defined with the following formulas:
Where M represent the sum of all the masses on the system.
And the center of mass C.M 
Part a
represent the masses.
represent the coordinates for the masses with the units on meters.
So we have everything in order to find the center of mass, if we begin with the x coordinate we have:
C.M
Part b
For this case we have an additional mass 
and we know that the resulting new center of mass it at the origin C.M 
and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)
If we solve for a we got:
And solving for b we got:
So the coordinates for this new particle are:
Answer: a) V = 9.81 m/a
b) S = 3.905m
c) V2 = 8.29m/s
d) Yes. The speed reduces.
Explanation:
Please find the attached files for the solution
Answer:
i think that it is b but I could be wrong
Answer:
a The velocity of the third particle is 
b The total kinetic energy increase 
Explanation:
To solve this question we are going to be using the concept of conservation of momentum which is mathematically represented as
From the question
Mass of unstable atomic nucleus 
Mass of x-axis particle 
Velocity of x-axis particle 
Mass of y-axis particle 
Velocity of y-axis particle 
Since the initial mass is known we can obtain the mass of the third particle
Using the conservation of mass mathematical expression
Using the mathematical expression above for conservation of momentum
The above is the vector solution now to obtain the value in numeric form we evaluate the magnitude of the vector
To obtain the total increase in kinetic energy which is mathematically represented as
![E =\frac{1}{2} [(5.20*10^{-27})(6.0*0^{6})^2 + (8.44*10^{-27})(4.00*10^6)^2+\\\\(2.16*10^{-27})(21.285*10^6)^2]](https://tex.z-dn.net/?f=E%20%3D%5Cfrac%7B1%7D%7B2%7D%20%5B%285.20%2A10%5E%7B-27%7D%29%286.0%2A0%5E%7B6%7D%29%5E2%20%2B%20%288.44%2A10%5E%7B-27%7D%29%284.00%2A10%5E6%29%5E2%2B%5C%5C%5C%5C%282.16%2A10%5E%7B-27%7D%29%2821.285%2A10%5E6%29%5E2%5D)
Answer:
W₃ = 3310.49 J
, W3 = 3310.49 J
Explanation:
We can solve this exercise in parts, the first with acceleration, the second with constant speed and the third with deceleration. Therefore it is work we calculate it in these three sections
We start with the part with acceleration, the distance traveled is y = 5.90 m and the final speed is v = 2.30 m / s. Let's calculate the acceleration with kinematics
v2 = v₀² + 2 a₁ y
as they rest part of the rest the ricial speed is zero
v² = 2 a₁ y
a₁ = v² / 2y
a₁ = 2.3² / (2 5.90)
a₁ = 0.448 m / s²
with this acceleration we can calculate the applied force, using Newton's second law
F -W = m a₁
F = m a₁ + m g
F = m (a₁ + g)
F = 69 (0.448 + 9.8)
F = 707.1 N
Work is defined by
W₁ = F.y = F and cos tea
As the force lifts the man, this and the displacement are parallel, therefore the angle is zero
W₁ = 707.1 5.9
W₁ = 4171.89 J W3 = 3310.49 J
Let's calculate for the second part
the speed is constant, therefore they relate it to zero
F - W = 0
F = W
F = m g
F = 60 9.8
F = 588 A
the job is
W² = 588 5.9
W2 = 3469.2 J
finally the third part
in this case the initial speed is 2.3 m / s and the final speed is zero
v² = v₀² + 2 a₂ y
0 = vo2₀² + 2 a₂ y
a₂ = -v₀² / 2 y
a₂ = - 2.3²/2 5.9
a2 = - 0.448 m / s²
we calculate the force
F - W = m a₂
F = m (g + a₂)
F = 60 (9.8 - 0.448)
F = 561.1 N
we calculate the work
W3 = F and
W3 = 561.1 5.9
W3 = 3310.49 J
total work
W_total = W1 + W2 + W3
W_total = 4171.89 +3469.2 + 3310.49
w_total = 10951.58 J
