What type of structure is the cytoplasm? I Will mark as brainiest if right

Porfavor ayuden que es pa mañana :v

Tpy6a [65]

Answer:

what language is

thattranslate i will.answer then

3 0

2 years ago

HELP!!

Diano4ka-milaya [45]

The question in choice-C is the correct answer to your question.

(Is this confusing ?)

6 0

3 years ago

Read 2 more answers

Gold, which has a density of 19.32 g/cm3, is the most ductile metal and can be pressed into a thin leaf or drawn out into a long

NNADVOKAT [17]

Answer:

a) 578.0 cm²

b) 25.18 km

Explanation:

We're given the density and mass, so first calculate the volume.

D = M / V

V = M / D

V = (6.740 g) / (19.32 g/cm³)

V = 0.3489 cm³

a) The volume of any uniform flat shape (prism) is the area of the base times the thickness.

V = Ah

A = V / h

A = (0.3489 cm³) / (6.036×10⁻⁴ cm)

A = 578.0 cm²

b) The volume of a cylinder is pi times the square of the radius times the length.

V = πr²h

h = V / (πr²)

h = (0.3489 cm³) / (π (2.100×10⁻⁴ cm)²)

h = 2.518×10⁶ cm

h = 25.18 km

3 0

2 years ago

Explain the differences in structure and use for life between oxygen gas in the atmosphere and ozone.

navik [9.2K]

Answer:

Most of the oxygen in the atmosphere is in the form of O2, which means it is made up of molecules containing two oxygen atoms. Ozone, however is O3, which means it is made up of molecules containing three oxygen atoms. O2 is what we breath, and what plants release from photosynthesis. Ozone occurs naturally high in the stratosphere, where it absorbs ultraviolet light, protecting us here on the surface from skin cancer. Ozone can also occur closer to the surface of the earth as a pollutant. It is formed from reactions with O2 and chemicals emitted from factories and cars. It comes in the form of smog.

So in general:

Oxygen (O2): Essential to human life

Ozone (O3) in the stratosphere: essential to protecting life on earth

Ozone (O3) on the surface of the earth: toxic to human life, caused by pollution

5 0

2 years ago

Two isolated, concentric, conducting spherical shells have radii R1 = 0.500 m and R2 = 1.00 m, uniform charges q1=+2.00 µC and q

scZoUnD [109]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

a E = $1.685*10^3 N/C$

b E = $36.69*10^3 N/C$

c E = 0 N/C

d $V = 6.7*10^3 V$

e $V = 26.79*10^3V$

f V = $34.67 *10^3 V$

g $V= 44.95*10^3 V$

h $V= 44.95*10^3 V$

i $V= 44.95*10^3 V$

Explanation:

From the question we are given that

The first charge $q_1 = 2.00 \mu C = 2.00*10^{-6} C$

The second charge $q_2 =1.00 \muC = 1.00*10^{-6}$

The first radius $R_1 = 0.500m$

The second radius $R_2 = 1.00m$

$Generally \ Electric \ field = \frac{1}{4\pi\epsilon_0}\frac{q_1+\ q_2}{r^2}$

And $Potential \ Difference = \frac{1}{4\pi \epsilon_0} [\frac{q_1 }{r}+\frac{q_2}{R_2} ]$

The objective is to obtain the the magnitude of electric for different cases

And the potential difference for other cases

Considering a

r = 4.00 m

$E = \frac{((2+1)*10^{-6})*8.99*10^9}{16}$

$= 1.685*10^3 N/C$

Considering b

$r = 0.700 m \ , R_2 > r > R_1$

This implies that the electric field would be

$E = \frac{1}{4\pi \epsilon_0}\frac{q_1}{r^2}$

This because it the electric filed of the charge which is below it in distance that it would feel

$E = 8*99*10^9 \frac{2*10^{-6}}{0.4900}$

= $36.69*10^3 N/C$

Considering c

r = 0.200 m

=> $r$

The electric field = 0

This is because the both charge are above it in terms of distance so it wont feel the effect of their electric field

Considering d

r = 4.00 m

=> $r > R_1 >r>R_2$

Now the potential difference is

$V =\frac{1}{4\pi \epsilon_0} \frac{q_1 + \ q_2}{r} = 8.99*10^9 * \frac{3*10^{-6}}{4} = 6.7*10^3 V$

This so because the distance between the charge we are considering is further than the two charges given

Considering e

r = 1.00 m $R_2 = r > R_1$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{1.00} \frac{1.00*10^{-6}}{1.00} ] = 26.79 *10^3 V$

Considering f

$r = 0.700 m \ , R_2 > r > R_1$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.700} \frac{1.0*10^{-6}}{1.00} ] = 34.67 *10^3 V$

Considering g

$r =0.500\m , R_1 >r =R_1$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$

Considering h

$r =0.200\m , R_1 >R_1>r$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$

Considering i

$r =0\ m \ , R_1 >R_1>r$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$

8 0

3 years ago