Which will have greater impact on the electric force of attraction between two charged particles, doubling the distance between the two particles, or doubling the charge in one particle?
Because electric force is calculated as (K*q1*q2)/(r^2), doubling the charge in one particle will simply double the force, but increasing the distance to double will drop the force to a quarter of the original, having a greater impact.
Calculate the definite integral of the function 3x^2 + 1 from x=1 to x=5
First, to find the anti-derivative we can just undo the chain rule so that 3x^2 + 1 becomes x^3 + x + C. Using the first fundamental theorem of calculus, we can then calculate that the definite integral = (5^3)+5+C - ((1^3)+1+C). This simplifies to 5^3+3 or 128.
Consider a 400 N physics student riding a Ferris wheel. She decides to conduct an experiment to see what the difference in her weight is at the top of the wheel versus the bottom. If at the top of the wheel she her scale reads 300 N how much does her scale read at the bottom of the Ferris wheel?
At the top of the wheel the forces acting on the student in the y-direction can be limited to the force of her scale pushing her up and gravity pushing her down. Because of the motion of the Ferris wheel, part of the force of gravity maintains her circular motion. You can discount this when calculating her weight at the height of the wheel. This situation is mirrored at the bottom of the wheel, but with gravity instead adding to her weight. Therefore, the difference between the reading at the top of the wheel, 300 N, and her normal weight, 400 N is the negative of the difference between the reading at the bottom and her weight. Using this, we can understand that the reading at the bottom of the wheel will be 500 N.