Answer:
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e
Explanation:
For this problem let's use Newton's second law applied to each body
Body A
X axis
T = m_A a
Axis y
N- W_A = 0
Body B
Vertical axis
W_B - T = m_B a
In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension
We write the equations
T = m_A a
W_B –T = M_B a
We solve this system of equations
m_B g = (m_A + m_B) a
a = m_B / (m_A + m_B) g
In this initial case
m_A = M
m_B = M
a = M / (1 + 1) M g
a = ½ g
Let's find the tension
T = m_A a
T = M ½ g
T = ½ M g
Now we change the mass of the second block
m_B = 2M
a = 2M / (1 + 2) M g
a = 2/3 g
We seek tension for this case
T’= m_A a
T’= M 2/3 g
Let's look for the relationship between the tensions of the two cases
T’/ T = 2/3 M g / (½ M g)
T’/ T = 4/3
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e
Answer:
A. The sum of all the forces acting on an object.
Answer:
d = 10 inch
Explanation:
The farthest distance between the centers, is along the diagonal of the rectangle. Therefore, we need to calculate the diagonal of the rectangle, but counting the fact that we have both circles.
So if, one side is 12 inch, and the other is 14 inch, we can use the Pitagoras theorem which is:
d = √(a²) + (b)²
Where a and b, are the lenght of the rectangle, but without the lenght of the diameter of both circles.
With this, the expression is this:
d = √(14 - 6)² + (12 - 6)²
d = √64+36
d = √100
d = 10 inches
Answer:
The magnitude of Force is 8.58×10⁵N and direction is upwards
Explanation:
The work beam does on the pile driver is given by
W=(FCos180°)Δx= -F(0.088m)
From work energy theorem
Choosing y=0 at the the level where the driver first contacts the beam and vi=0 at yi=+3.40m and comes to rest again vf=0 at yf= -0.088m
So
The magnitude of Force is 8.58×10⁵N and direction is upwards
Maybe this can help:
https://blogs.bu.edu/ggarber/interlace/pendulum/energy-in-a-pendulum/
