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3 years ago

Ultraviolet light causes production of vitamin d3 in the cells of the __________.

Physics

1 answer:

dedylja [7]3 years ago

4 0

Answer:

Epidermis

Explanation:

The epidermis is an important layer forming the skin that is mainly responsible for providing protection from the incoming harmful solar radiations, that are comprised of ultraviolet particles. These UV rays can cause various diseases like skin cancer.

When the solar radiations containing UV radiations (mostly the UV-B type radiation having a wavelength 290-315 nanometer) strikes the cells of human beings, then the protein that is stored within the skin named '7- dehydrocholesterolit (DHC)' is transformed into the vitamin d3 inside the cells.

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Only one of three balls A, B, and C carries a net charge q. The balls are made from conducting material and are identical. One o

Zarrin [17]

Answer:

This is greater than the initial charge, which violates the principle that the charge cannot be created or destroyed, consequently this distribution is impossible to achieve

Explanation:

The metals distribute the charge on all surface when they touch the surface increases so that charge density decreases and when the charge is separated into smaller in each metal.

Let's apply this principle to our case.

One of the spheres is loaded with a charge q, when touching a ball its charge is reduced to 1 / 2q for each ball.

qA = ½ q

qB = ½ q

qC = 0

The total charge is q

we make a second contact

If we touch the ball A again with the other sphere not charged C, the chare is distributed and when separated it is reduced by half

qA = 1/2 (q / 2) = ¼ q

qC = ¼ q

qB = ½ q

At this point all spheres have a charge,

qA = ¼ q

qb = ½ q

qC = ¼ q

The total charge is q

Now let's contact spheres B and one of the other two

Q = ½ q + ¼ q = ¾ q

When splitting the charge

qB = ½ ¾ q = 3/8 q

qC = ½ ¾ q = 3/8 q

qA = ¼ q

The total charge is q

Note that the total load is always equal to q

Now let's analyze the given configuration

Let's look for the total load

Q = qA + QB + QC

Q = ½ q + 3/8 q + ¼ q

Q = 9/8 q

This is greater than the initial charge, which violates the principle that the charge cannot be created or destroyed, consequently this distribution is impossible to achieve

8 0

3 years ago

Violet light (λ = 400 nm) passing through a diffraction grating for which the slit spacing is 6.0 μm forms a pattern on a screen

sergey [27]

Answer:

3 order dark fringe

Explanation:

y = Distance from central bright fringe = 204 mm

λ = Wavelength = 400 nm

L = Distance between screen and source = 1 m

d = Slit distance = 6 μm

$tan\theta =\frac{y}{L}\\\Rightarrow tan\theta =\frac{204}{1000}=0.2012^{\circ}$

$dsin\theta=m\lambda\\\Rightarrow m=\frac{dsin\theta}{\lambda}\\\Rightarrow m=\frac{6\times 10^{-6}sin0.2012}{400\times 10^{-9}}=2.9982\approx 3$

Order of fringe is 3

So, it is a Dark order fringe

3 0

3 years ago

A uniform electric field exists in the region between two oppositely charged plane parallel plates. a proton is released from re

dedylja [7]

According to Newton's second law

E.e = a * mp ..... (1)

where

E is the magnitude of the electric field; e = 1.6 * 10^-19 is the elementary charge; mp = 1.67*10^-27 kg is the proton mass; a is the acceleration.

So, the distance

l = at^2/2 .......(2)

The proton accelerated

a = 2l / t^2 ...........(3)

From equations (1) and (3)

E= 32.51 V/m

Electric field

The physical field that surrounds electrically charged particles and exerts a force on all other charged particles in the field, either attracting or repelling them, is known as an electric field (also known as an E-field). It can also refer to a system of charged particles' physical field. Electric charges and time-varying electric currents are the building blocks of electric fields. The electromagnetic field, one of the four fundamental interactions (also known as forces) of nature, manifests itself in both electric and magnetic fields.

To learn more about an electric field refer here:

brainly.com/question/15800304

#SPJ4

5 0

2 years ago

Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces

Mandarinka [93]

Answer: $0.81\times 10^{16}$ beta particles

Explanation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass = 14.0 g

Molar mass = 137 g/mol

$\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles$

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

1 mole of cesium contains atoms = $6.023\times 10^{23}$

0.102 moles of cesium contains atoms = $\frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}$

The relation of atoms with time for radioactivbe decay is:

$N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}$

Where $N_t$ =atoms left undecayed

$N_0$ = initial atoms

t = time taken for decay = 3 minutes

${t_{\frac{1}{2}}}$ = half life = 30.0 years = $1.577\times 10^7$ minutes

The fraction that decays : $1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}$

Amount of particles that decay is = $0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}$

Thus $0.81\times 10^{16}$ beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

7 0

3 years ago

A 23 kg body is moving through space in the positive direction of an x axis with a speed of 130 m/s when, due to an internal exp

babymother [125]

Answer:

a) Vx = 1088m/s

b) Vy = -162.93m/s

c) 5246745J

Explanation:

Mass of unbroken body = 23kg

Its velocity along +ve X-axis = 130m/s

Mass of first broken body, m1= 9.4kg

Its velocity along +ve X-axis = 130m/s

Nass of 2nd broken body, m2 = 6.1kg

Its velocity long-lived X - axis = -550m/s

Mass of 3rd broken body = ?

m3 = (23 - 9.4 - 6.1)kg

m3 = 7.5kg

Let velocity along the x-axis = Vx

Let the velocity along the x-axis = Vy

Applying law of conservation of momentum along x-axis

a) m1×0 + m2×(-550) + m3×(Vx) =M × 130

9.4 × 0 + 6.1× (-550) + 7.5(Vx) = 23 ×130

0 + (-5170) + 7.5Vx = 2990

2990 + 5170 = 7.5Vx

8160 = 7.5Vx

Vx = 8160/7.5

Vx = 1088m/s

b) Aplying conservation of momentum along the x-axis

(m1×130) + (m2 × 0) + (m3× Vy) = 0

(9.4 × 130) + (6.1 ×550) + 7.5Vy = 0

1222 + 0 + 7.5Vy = 0

1222 = -7.5Vy

Vy = 1222/(-7.5)

Vy = -262.93m/s

c) The energy released or change in KE is given by:

1/2[(m1v1^2) + (m2v2^2) +(m3Vx^2) ]= MV^2

Change in KE = 1/2[ 9.4× 130^2 + 6.1 × 550^2 + 7.5 × 1088^2 ] - 1/2(23 × 130^2)

Change in KE = 1/2[158860 + 1845250 + 8878080] - 1/2[388700]

Change in KE = 5441095 - 194350

Change in KE = 5246745J

4 0

3 years ago