I am thinking that maybe the problem is not with the calibration. It might be that the buffered solution is already expired since at this point the solution is already not stable and will give a different pH reading than what is expected.
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:
![\sum F_y=m_1*a_m_i_n = T-m_1*g](https://tex.z-dn.net/?f=%5Csum%20F_y%3Dm_1%2Aa_m_i_n%20%3D%20T-m_1%2Ag)
If the package is barely lifted, that means that T=m_2*g; then:
![\sum F_y =m_1*a_m_i_n=m_2*g-m_1*g](https://tex.z-dn.net/?f=%5Csum%20F_y%20%3Dm_1%2Aa_m_i_n%3Dm_2%2Ag-m_1%2Ag)
Solving the equation for a_mín, we have:
![a_m_i_n=((m_2-m_1)/m_1)*g = ((15kg-10kg)/10kg)*9,8 m/s^2 =4,9 m/s^2](https://tex.z-dn.net/?f=a_m_i_n%3D%28%28m_2-m_1%29%2Fm_1%29%2Ag%20%3D%20%28%2815kg-10kg%29%2F10kg%29%2A9%2C8%20m%2Fs%5E2%20%3D4%2C9%20m%2Fs%5E2)
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: ![\sum F_y = m_1*a \rightarrow T-m_1*g=m_1*a](https://tex.z-dn.net/?f=%5Csum%20F_y%20%3D%20m_1%2Aa%20%5Crightarrow%20T-m_1%2Ag%3Dm_1%2Aa)
For the package: ![\sum F_y = m_2*a \rightarrow m_2*g - T = m_2*a](https://tex.z-dn.net/?f=%5Csum%20F_y%20%3D%20m_2%2Aa%20%5Crightarrow%20m_2%2Ag%20-%20T%20%3D%20m_2%2Aa)
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: ![\sum F_y = -m_2*a \rightarrow T-m2*g=-m_2*a \rightarrow m_2*g -T=m_2 *a](https://tex.z-dn.net/?f=%5Csum%20F_y%20%3D%20-m_2%2Aa%20%5Crightarrow%20T-m2%2Ag%3D-m_2%2Aa%20%5Crightarrow%20m_2%2Ag%20-T%3Dm_2%20%2Aa)
We have two unknowns and two equations, so we can proceed. We can match both tensions and have:
![m_1*a+m_1*g=m_2*g-m_2*a](https://tex.z-dn.net/?f=m_1%2Aa%2Bm_1%2Ag%3Dm_2%2Ag-m_2%2Aa)
Solving a, we have
![(m_1+m_2)*a =(m_2 - m1)*g\\\\a=((m_2-m_1)/(m_1+m_2))*g \rightarrow a=((15kg-10kg)/(10kg+15kg))*9,8 m/s^2\\\\a= 2,0 m/s^2](https://tex.z-dn.net/?f=%28m_1%2Bm_2%29%2Aa%20%3D%28m_2%20-%20m1%29%2Ag%5C%5C%5C%5Ca%3D%28%28m_2-m_1%29%2F%28m_1%2Bm_2%29%29%2Ag%20%5Crightarrow%20a%3D%28%2815kg-10kg%29%2F%2810kg%2B15kg%29%29%2A9%2C8%20m%2Fs%5E2%5C%5C%5C%5Ca%3D%202%2C0%20m%2Fs%5E2)
We can then replace this value of a in one for the sums of force and find the tension T:
![T = m_1*a+m_1*g \rightarrow T=m_1*(a+g)\\\\T = 10kg*(2,0 m/s^2+9,8 m/s^2) \\\\T = 120 N](https://tex.z-dn.net/?f=T%20%3D%20m_1%2Aa%2Bm_1%2Ag%20%5Crightarrow%20T%3Dm_1%2A%28a%2Bg%29%5C%5C%5C%5CT%20%3D%2010kg%2A%282%2C0%20m%2Fs%5E2%2B9%2C8%20m%2Fs%5E2%29%20%5C%5C%5C%5CT%20%3D%20120%20N)
Answer:
Y'now water vapor isn't half bad. It's like a vape but it can't affect your lungs. Anyway, Water vapor is water in gaseous instead of liquid form. It can be formed either through a process of evaporation or sublimation.
Answer:
I may not have the answer so i'll just give up some hints.
Multiply the time by the acceleration due to gravity to find the velocity when the object hits the ground. If it takes 9.9 seconds for the object to hit the ground, its velocity is (1.01 s)*(9.8 m/s^2), or 9.9 m/s. Choose how long the object is falling. In this example, we will use the time of 8 seconds. Calculate the final free fall speed (just before hitting the ground) with the formula v = v₀ + gt = 0 + 9.80665 * 8 = 78.45 m/s . Find the free fall distance using the equation s = (1/2)gt² = 0.5 * 9.80665 * 8² = 313.8 m .h = 0.5 * 9.8 * (1.5)^2 = 11m. b. V = gt = 9.8 * 1.5 = 14.7m/s. A feather and brick dropped together. Air resistance causes the feather to fall more slowly. If a feather and a brick were dropped together in a vacuum—that is, an area from which all air has been removed—they would fall at the same rate, and hit the ground at the same time.When an object's point is taller the thing that is going down it will go faster than when the point is lower. EXAMPLE: The object is the tennis ball if you drop it down the higher hill it will be faster than if you drop it down a shorter hill. In other words, if two objects are the same size but one is heavier, the heavier one has greater density than the lighter object. Therefore, when both objects are dropped from the same height and at the same time, the heavier object should hit the ground before the lighter one.
I hope my little bit (big you may say) hint help you with your question.
If the rod is in rotational equilibrium, then the net torques acting on it is zero:
∑ τ = 0
Let's give the system a counterclockwise orientation, so that forces that would cause the rod to rotate counterclockwise act in the positive direction. Compute the magnitudes of each torque:
• at the left end,
τ = + (50 N) (2.0 m) = 100 N•m
• at the right end,
τ = - (200 N) (5.0 m) = - 1000 N•m
• at a point a distance d to the right of the pivot point,
τ = + (300 N) d
Then
∑ τ = 100 N•m - 1000 N•m + (300 N) d = 0
⇒ (300 N) d = 1100 N•m
⇒ d ≈ 3.7 m