How do i do question 9 & 11?you!

The equation for volume of a sphere is

... V = (4/3)π·r³

Differentiating gives

... dV = 4π·r²·dr

Filling in the given numbers gives

... change in volume ≈ 4π·(15 cm)²·(0.4 cm)

... = 360π cm³ ≈ 1130.97 cm³ . . . . . . volume of layer 4mm thick

The area is that of two cubes of dimension x joined together. The area of each such cube is 6x², but the two joined faces don't count in the external surface area. Thus the area of the cuboid is 10x².

The volume of the cuboid is that of two cubes joined, so is 2x³. Then the rate of change of volume is

... dV/dt = (d/dt)(2x³) = 6x²·dx/dt

We know x² = A/10, where A is the area of the cuboid, so the rate of change of volume is ...

... dV/dt = (6/10)A·dx/dt = 0.6·(129.6 cm²)(0.01 cm/s)

... dV/dt = 0.7776 cm³/s

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v