Answer:
1, 1583.33 V/m
2, 4.72*10^-12 C
3, 39.2*10^-11 C
Explanation:
1
E = V / d
E = 8.55 / 5.4*10^-3
E = 8.55 / 0.0054
E = 1583.33 V/m
2
Capacitance, C = (k * e0 * A) / d, where k = 1
A = area of capacitor, 3.37 cm² = 3.37*10^-4 m²
d = plate separation, 5.4 mm
e0 = Constant, 8.85*10^-12
Applying these, we have
C = (1 * 8.85*10^-12 * 3.37*10^-4) / 5.4*10^-3
C = 29.82*10^-16 / 0.0054
C = 5.52*10^-13 F
Since Q = CV, then
Q = 5.52*10^-13 * 8.55
Q = 4.72*10^-12 C
3
We are given that k = 83, so
Capacitance, C = (k * e0 * A) / d
C = (83 * 8.85*10^-12 * 3.37*10^-4) / 5.4*10^-3
C = 2.475*10^-13 / 0.0054
C = 4.58*10^-11 F
Q = CV
Q = 4.58*10^-11 * 8.55
Q = 39.2*10^-11 C
Answer:
a) {[1.25 1.5 1.75 2.5 2.75]
[35 30 25 20 15] }
b) {[1.5 2 40]
[1.75 3 35]
[2.25 2 25]
[2.75 4 15]}
Explanation:
Matrix H: {[1.25 1.5 1.75 2 2.25 2.5 2.75]
[1 2 3 1 2 3 4]
[45 40 35 30 25 20 15]}
Its always important to get the dimensions of your matrix right. "Roman Columns" is the mental heuristic I use since a matrix is defined by its rows first and then its column such that a 2 X 5 matrix has 2 rows and 5 columns.
Next, it helps in the beginning to think of a matrix as a grid, labeling your rows with letters (A, B, C, ...) and your columns with numbers (1, 2, 3, ...).
For question a, we just want to take the elements A1, A2, A3, A6 and A7 from matrix H and make that the first row of matrix G. And then we will take the elements B3, B4, B5, B6 and B7 from matrix H as our second row in matrix G.
For question b, we will be taking columns from matrix H and making them rows in our matrix K. The second column of H looks like this:
{[1.5]
[2]
[40]}
Transposing this column will make our first row of K look like this:
{[1.5 2 40]}
Repeating for columns 3, 5 and 7 will give us the final matrix K as seen above.
Answer:
Nine times
Explanation:
= Initial compression of the spring = 8 cm = 0.08 m
= Final compression of the spring = 24 cm = 0.24 m
= Spring constant of the spring
Initial spring energy is given as
Initial spring energy is given as
Answer:
Compared to the first student, the second student did twice as much work as the first student.
Explanation:
The work done by the first student will be equal to the Force exerted by the backpack on the student carrying it multiplied by one mile (Distance). The work done by the second student will be equal to the Force exerted by the backpack on the student carrying it multiplied by two miles (Distance).
When geophysicists measure the geothermal gradient in areas on the ridges where there is no activity hydrothermally, the gradient is far below than what is predicted theoretically, but when measured near hydrothermal vents it is more than what is predicted. This is because most of the heat is being carried through convection by hydrothermal systems so that the average gradient when measured far from the circulation would be depressed or lower.
