Answer:
C = 0.0125 m/s⁴. The calculation procedure can be found in the attachment below. The concept of motion along a straight line with constant acceleration has been applied to solve the problem.
Explanation:
The sign convention chosen in this problem solution is upwards as positive and downwards negative. The equation of motion v = u + at has been used to calculate the constant C as only one unknown is contained in this equation. This is so because we have been given the initial velocity, the acceleration and the time taken. To solve future problems of this kind, first thing to check for is an equation of motion with the least number of unknown. This helps to reduce the complexity of the problem solution.
Answer:
The charges under study are of the same sign
The calculation of the electric field for each charge separately, there is no relationship between the charges
Explanation:
Let's start by writing the equation for the electric field
E = k q / r²
where q is the charge under analysis and r the distance from this charge to a positive test charge.
When analyzing the statement the student has some problems.
* The charges under study are of the same sign, it does not matter if positive or negative.
* The calculation of the electric field for each charge separately, there is no relationship between the charges for the calculation of the electric field.
* What is added is the interaction of the electric field with the positive test charge, in this case each field has the opposite direction to the other, so the vector sum gives zero
Answer:
300 N/m
Explanation:
given,
Load attached to the spring, W = 54 N
length of stretch of the spring, x = 0.15 m
spring constant= ?
Force applied on the spring is calculated by the equation
F = k x
where k is the spring constant
x is the displacement of the spring due to applied load
now,
54 = k × 0.15
hence, the spring constant is equal to 300 N/m
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.
