Answer: 3.41 s
Explanation:
Assuming the question is to find the time
the ball is in air, we can use the following equation:

Where:
is the final height of the ball
is the initial height of the ball
is the initial velocity of the ball
is the time the ball is in air
is the acceleration due to gravity

Then:


Multiplying both sides of the equation by -1 and rearranging:

At this point we have a quadratic equation of the form
, which can be solved with the following formula:
Where:
Substituting the known values:
Solving the equation and choosing the positive result we have:
This is the time the ball is in air
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:
There are nine PCNs in the Edmonton area. They work alongside more than 1,100 family doctors in over 330 clinics to provide care for 1.2 million patients. PCN teams include more than 370 nurses, mental health clinicians and other health professionals.
Explanation:
Answer:
∑Fy = 0, because there is no movement, N = m*g*cos (omega)
Explanation:
We can solve this problem with the help of a free body diagram where we show the respective forces in each one of the axes, y & x. The free-body diagram and the equations are in the image attached.
If the product of mass by acceleration is zero, we must clear the normal force of the equation obtained. The acceleration is equal to zero because there is no movement on the Y-axis.
Answer:
Tangential
Explanation: This is a kind of force which act on a moving body in such a way that it is curved in the direction of the path of the body. This implies that when the velocity of the object is positive, the acceleration will be negative.