Many ways, but some of the most famous are kicks (side, back, front, snap) or a smash.
Hope it helped! :)
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:
No
Explanation:
Please let me know if my answer is correct
Their common speed is 1.53 m/s.
<h3>What is speed?</h3>
Speed can be defined as the ratio of the distance to the time of a body in motion.
To calculate their final speed, we use the formula below.
Formula:
- mu+m'u' = V(m+m')............. Equation 1
Where:
- m = mass of the Lee
- m' = mass of Mat
- u = initial speed of Lee
- u' = initial speed of Mat
- V = Their common speed.
make V the subject of the equation
- V = (mu+m'u')/(m+m')........... Equation 2
From the question,
Given:
- m = 40 kg
- m' = 80 kg
- u = 4.6 m/s
- u' = 0 m/s
Substitute these values into equation 2
- V = [(40×4.6)+(80×0)]/(40+80)
- V = 184/120
- V = 1.53 m/s.
Hence, Their common speed is 1.53 m/s.
Learn more about speed here: brainly.com/question/3004254
