What investigations allow for the control of variables
2 answers:
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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:
W = 290.7 dynes*cm
Explanation:
d = 1/5 cm = 0.2 cm
The force is in function of the depth x:
F(x) = 1000 * (1 + 2*x)^2
We can expand that as:
F(x) = 1000 * (1 + 4*x + 4x^2)
F(x) = 1000 + 4000*x + 4000*x^2
Work is defined as
W = F * d
Since we have non constant force we integrate
W = [1000*x + 2000*x^2 + 1333*X^3] evaluated between 0 and 0.2
W = 1000*0.2 + 2000*0.2^2 + 1333*0.2^3 - 1000*0 - 2000*0^2 - 1333*0^3
W = 200 + 80 + 10.7 = 290.7 dynes*cm