How strong is the electric field between two parallel plates 4.2 mm apart if the potential difference between them is 220 V?

Under what conditions the reaction rate of an enzymolysis that follows Michaelis-Menten kinetics is a quarter of its maximum val

ExtremeBDS [4]

Solution :

Michaelis-Menten kinetics in the field of biochemistry is considered as one of the well known models for enzyme kinetics. The model represents an equation that describes the enzymatic reactions's rate by relating the reaction rate to the substrate's concentration. The equation is named after the two famous scientists, Leonor Michaelis and Maud Menten.

The formula is :

$v=\frac{V_{max}[S]}{K_M + [S]}$

where v = velocity of reaction

$V_{max}$ = maximum rate achieved

$K_M$ = Michaelis constant

[S] = concentration of the substrate, S

According to the question, by putting the velocity of reaction, v as $\frac{V_{max}}{4}$ , we get the above equation as

$[S]= \frac{K_M}{3}$

Therefore the answer is $[S]= \frac{K_M}{3}$

3 0

2 years ago

True or False? Energy is matter. 1. False 2. True

andriy [413]

Answer:

False because it has no mass.

Explanation:

7 0

2 years ago

Pls help me I don’t get it :(

Ksju [112]

Answer:

2nd and 4th

Explanation:

4 0

2 years ago

A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o

Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field $dE$ at the center of curvature of the arc is

$dE = k\dfrac{dQ}{r^2}cos(\theta )$ (we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )

where $r$ is the radius of curvature.

Now

$dQ = \lambda rd\theta$ ,

where $\lambda$ is the charge per unit length, and it has the value

$\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.$

Thus, the electric field at the center of the curvature of the arc is:

$E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta }{r^2} cos(\theta)$

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.$

Now, we find $\theta_1$ and $\theta_2$ . To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:

$fraction = \dfrac{3.0m}{2\pi (2.3m)} = 0.2076$

and this is

$0.2076*2\pi =1.304$ radians.

Therefore,

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.$

evaluating the integral, and putting in the numerical values we get:

$E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\$

$\boxed{ E = 31.329N/C.}$

4 0

3 years ago

Astronaut X of mass 50kg floats next to Astronaut Y of mass 100kg while in space, as shown in the figure. The positive direction

jonny [76]

Answer:

C

Explanation:

The change in momentum of x has to be the opposite of the change in momentum of Y because the momentum is just transferred from one to another. But I'm still trying to figure it out how to calculate.

5 0

2 years ago