Cells a and f shows an early and late stage of the same phase of mitosis . what phase is it?

A

The reason this is is because she the first sentence says that she believed that people would eat the mashed potatoes faster if they were brighter color.

6 0

3 years ago

Explain how the basic unit are combined to give the derived units of force, velocity, pressure and work

LuckyWell [14K]

Velocity:

Velocity is change in displacement with respect to time:

$\frac{\Delta x}{\Delta t}$

Analysing the units, meters (displacement) and seconds (time) are basic units:

$\frac{m}{s}$

Therefore the unit of velocity is m/s

Force:

Newton's second law of motion:

$F = ma$

Kilogram (mass) is a basic unit, and accelerations unit can be found using the equation:

$a=\frac{\Delta v}{\Delta t}$

Analysing the units:

$\frac{\frac{m}{s}}{s}=\frac{m}{s^2}$

Therefore, the unit of force is:

$kg\frac{m}{s^2}$

Pressure:

Pressure is given by the equation:

$P=\frac{F}{S}$ where S is area of effect, F is force

Area for a basic rectangle (geometric shape is arbitrary for dimensional analysis) is found by multiplying two lengths:

$[l^2]=m^2$ , the unit of area

Dividing the aforementioned unit of force by the unit of area:

$\frac{kg\frac{m}{s^2}}{m^2}=\frac{kg}{ms^2}$ , the unit of pressure

Work:

Work is given by the equation:

$W=\vec{F}\cdot \vec{x}$ , (dot product may be assumed as normal multiplication for the purposes of unit analysis)

Knowing displacement's (x) unit is m:

$[W]=\frac{kgm}{s^2}m=\frac{kgm^2}{s^2}$ , the unit of work.

3 0

3 years ago

At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet

expeople1 [14]

As we know that KE and PE is same at a given position

so we will have as a function of position given as

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

also the PE is given as function of position as

$PE = \frac{1}{2}m\omega^2x^2$

now it is given that

KE = PE

now we will have

$\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2$

$A^2 - x^2 = x^2$

$2x^2 = A^2$

$x = \frac{A}{\sqrt2}$

so the position is 0.707 times of amplitude when KE and PE will be same

Part b)

KE of SHO at x = A/3

we can use the formula

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

now to find the fraction of kinetic energy

$f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}$

$f = \frac{A^2 - (\frac{A}{3})^2}{A^2}$

$f_k = \frac{8}{9}$

now since total energy is sum of KE and PE

so fraction of PE at the same position will be

$f_{PE} = 1 - f_k$

$f_{PE} = 1 - (8/9) = 1/9$

7 0

3 years ago