Answer:
A. 4,9 m/s2
B. 2,0 m/s2
C. 120 N
Explanation:
In the image, 1 is going to represent the monkey and 2 is going to be the package. Let a_mín be the minimum acceleration that the monkey should have in the upward direction, so the package is barely lifted. Apply Newton’s second law of motion:

If the package is barely lifted, that means that T=m_2*g; then:

Solving the equation for a_mín, we have:

Once the monkey stops its climb and holds onto the rope, we set the equation of Newton’s second law as it follows:
For the monkey: 
For the package: 
The acceleration a is the same for both monkey and package, but have opposite directions, this means that when the monkey accelerates upwards, the package does it downwards and vice versa. Therefore, the acceleration a on the equation for the package is negative; however, if we invert the signs on the sum of forces, it has the same effect. To be clearer:
For the package: 
We have two unknowns and two equations, so we can proceed. We can match both tensions and have:

Solving a, we have

We can then replace this value of a in one for the sums of force and find the tension T:

Because they are caused by your exercise
Answer:
i believe its 26.7
Explanation:
if the runner goes 8.9 m/s each second while accelerating for 3 seconds to reach top speed, the top speed would be 26.7 m/s
Answer:
The dependent variable is academic performance
The independent variable is the presence/absence of tutorial support
The control group are students who did not get the tutorial support.
The experimental group were students that got the tutorial support
Explanation:
In every experiment, there is a dependent and independent variable as well as an experimental and a control group.
The experimental group receive the treatment while the control group do not receive the treatment. The independent variable is manipulated and its impact on the dependent variable is evaluated.
The control group are students who did not receive the tutorial support while the experimental group are students that received the tutorial support.
The dependent variable in this case is academic performance. Its outcome depends on the presence or absence of tutorial support (independent variable).
The energy absorbed by photon is 1.24 eV.
This is the perfect answer.