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3 years ago

The body needs small amounts of ________ to help enzymes break down proteins.

Physics

2 answers:

Olin [163]3 years ago

6 0

Answer:

The body needs small amount of vitamin B-6 to help enzymes break down protein and carry the dismantled amino acids to the blood stream.

Inessa [10]3 years ago

4 0

Answer:

I would say vitamin B-6

Explanation:

it is also known as pyridoxine. it helps ebzymes break down protein and carry the dismantled amino acids to the blood stream

Hope this helps ✌✌

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A photon in a laboratory experiment has an energy of 5 eV. What is the frequency of this photon? (using the idea of the electron

andreyandreev [35.5K]

Answer and working shown on photo

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3 years ago

A long metal cylinder with radius a is supported on an insulating stand on the axis of a long, hollow, metal tube with radius b.

bija089 [108]

i) Potential for r < a: $V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

ii) Potential for a < r < b: $V(r)=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}$

iii) Potential for r > b: $V(r)=0$

b) Potential difference between the two cylinders: $V_{ab}=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

c) Electric field between the two cylinders: $E=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}$

Explanation:

Here we want to calculate the potential for r < a.

Before calculating the potential, we have to keep in mind that the electric field outside an infinite wire or an infinite cylinder uniformly charged is

$E=\frac{\lambda}{2\pi \epsilon_0 r}$

where

$\lambda$ is the linear charge density

r is the distance from the wire/surface of the cylinder

By integration, we find an expression for the electric potential at a distance of r:

$V(r) =\int Edr = \frac{\lambda}{2\pi \epsilon_0} ln(r)$

Inside the cylinder, however, the electric field is zero, because the charge contained by the Gaussian surface is zero:

$E=0$

So the potential where the electric field is zero is constant:

$V=const.$

iii) We start by evaluating the potential in the region r > b. Here, the net electric field is zero, because the Gaussian surface of radius r here contains a positive charge density $+\lambda$ and an equal negative charge density $-\lambda$ . Therefore, the net charge is zero, so the electric field is zero.

This means that the electric potential is constant, so we can write:

$\Delta V= V(r) - V(b) = 0\\\rightarrow V(r)=V(b)$

However, we know that the potential at b is zero, so

$V(r)=V(b)=0$

ii) The electric field in the region a < r < b instead it is given only by the positive charge $+\lambda$ distributed over the surface of the inner cylinder of radius a, therefore it is

$E=\frac{\lambda}{2\pi r \epsilon_0}$

And so the potential in this region is given by:

$V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}$ (1)

i) Finally, the electric field in the region r < a is zero, because the charge contained in this region is zero (we are inside the surface of the inner cylinder of radius a):

E = 0

This means that the potential in this region remains constant, and it is equal to the potential at the surface of the inner cylinder, so calculated at r = a, which can be calculated by substituting r = a into expression (1):

$V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

And so, for r<a,

$V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

Here we want to calculate the potential difference between the surface of the inner cylinder and the surface of the outer cylinder.

We have:

- Potential at the surface of the inner cylinder:

$V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

- Potential at the surface of the outer cylinder:

$V(b)=0$

Therefore, the potential difference is simply equal to

$V_{ab}=V(a)-V(b)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

Here we want to find the magnitude of the electric field between the two cylinders.

The expression for the electric potential between the cylinders is

$V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}$

The electric field is just the derivative of the electric potential:

$E=-\frac{dV}{dr}$

so we can find it by integrating the expression for the electric potential. We find:

$E=-\frac{d}{dr}(\frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}$

So, this is the expression of the electric field between the two cylinders.

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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3 years ago

What is specific gravity

Mrac [35]

Answer:

It is the ratio of the density of a substance to the density of a given reference material.

Explanation:

<em>Specific gravity is also known as relative density.</em>

<u>To find the relative density of substance, you:</u>

Divide the density of substance measured

And divide that by the density of the reference

7 0

2 years ago

Hello I need help its due today!

telo118 [61]

Answer:This also means that Mercury's surface gravity is 3.7 m/s2, which is the equivalent of 38% of Earth's gravity (0.38 g). This means that if you weighed 100 kg (220 lbs) on Earth, you would weigh 38 kg (84 lbs) on Mercury.

Explanation:

5 0

3 years ago

At the end of the Shock Wave roller coaster ride, the 6000-kg train of cars (includes passengers) is allowed from a speed of 20

Angelina_Jolie [31]

Answer:

56250 N

Explanation:

mass, m = 6000 kg

initial speed, u = 20 m/s

final speed, v = 5 m/s

distance, s = 20 m

Use third equation of motion

$v^{2}=u^{2}+2as$

5 x 5 = 20 x 20 + 2 a x 20

25 = 400 + 40 a

a = - 9.375 m/s^2

Braking force, F = mass x acceleration

F = 6000 x 9.375

F = 56250 N

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3 years ago