A nuclear explosion may release tremendous amounts of energy in the form of noise, heat, visible light, radiation and an atmosph
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Newton's second law we can find that the correct answer is:
E) It cannot be determiner whick block has more masses from the information provided
Newton's second law establishes the relationship between force, mass, and acceleration of a body. Since force and acceleration are vector quantities, their components must be added on each axis
For this problem we have two bodies, let's write Newton's second law for the body B, we assume that the body B descends
W_b - T = m_b a
W_b = m_b g
m_b - T = m_b a
Where W_b is the weight of block B, T the tension of the string, mb the mass of block b and the acceleration
Now let's find the relation for block A
let's set a datum with the x axis parallel to the ramp
T - Wₓ = mₐ a
sin θ = Wₓ / W
Wₓ = Wₐ sin θ
Wₐ = mₐ g
Where Wₓ is the component of the weight, Wₐ the weight of the body A and θ the angle of the plane
Let's write our system of equations
m_b g - T = m_b a
T - mₐ g sin θ = mₐ a
let's add the equations
g (m_b - mₐ sin θ) = (m_b + mₐ) a
a =
Let's analyze this expression
- The numerator is positive the body B descends, this occurs when
m_b - mₐ sin θ > 0
- The numerator is negative, body B rises
m_b - mₐ sin θ <0
We can observe that the acceleration is positive or negative depending on the relation of the masses and the angle of the plane.
In conclusion using Newton's second law we find that the correct answer is
E ) It cannot be determiner whick block has more masses from the information provided
learn more about Newton's second law here:
Answer:
hello your question has some missing values attached below is the complete question with the missing values
answer :
a) 0.083 secs
b) 0.33 secs
c) 3e^-4/3
Explanation:
Given that
g = 32 ft/s^2 , spring constant ( k ) = 2 Ib/ft
initial displacement = 1 ft above equilibrium
mass = weight / g = 4/32 = 1/8
damping force = instanteous velocity hence β = 1
a<u>)Calculate the time at which the mass passes through the equilibrium position.</u>
time mass passes through equilibrium = 1/12 seconds = 0.083
<u>b) Calculate the time at which the mass attains its extreme displacement </u>
time when mass attains extreme displacement = 1/3 seconds = 0.33 secs
<u>c) What is the position of the mass at this instant</u>
position = 3e^-4/3
attached below is the detailed solution to the given problem
