All STEM subjects require ________________ skills. a. Same b. Different

Plz help ya' girl out! Label the parts of whatever that is Plz giving brainliest!

Burka [1]

Answer:

BRAINLIEST PLEASE!!!

A = Smooth Endoplasmic Reticulum

B = Vacuole

C = Mitochondria (Plural)

D = Nucleolus

E = Cell Wall

Explanation:

This is obviously a plant cell. Animal cells do not have cell walls(E), and most animal cells will not have a vacuole(B). Smooth Endoplasmic Reticulum: This produces protiens (contains ribosomes) but when looked through a microscope, will appear to be smooth. The vacuole(B) is like a storage for the plant cell, and it helps support the plant cell. This is the reason why plants “droop” when you don’t water them for a long time. Mitochondia (Plural) processes nutrients for the cell. The nucleolus(D) covers the nucleus. It’s main function is to process RNA and combine it with proteins. The cell wall(E) helps support and protect the cell.

4 0

2 years ago

This is very urgent please hurry!!?!!??!

yawa3891 [41]

Answer:

The greatest acceleration when the unbalanced force is applied will be experienced in :

A) The box with a mass of 2 kg

Explanation:

According to second law of motion the external unbalanced force is directly proportional to rate of change of momentum.

F = (Final momentum - initial momentum)/time

or

Force is equal to the product of mass and acceleration

F = m x a

Here a= acceleration

m = mass of the object

If Force is constant then acceleration is inversely proportional to mass

$a =\frac{F}{m}$

A) The box with a mass of 2kg

F = 8 N

$a =\frac{8}{2}$

a = 4 m/s2

B) The box with the mass of 4kg

$a =\frac{8}{4}$

a = 2 m/s2

C) The box with a mass of 6kg

$a =\frac{8}{6}$

a = 1.33 m/s2

D) The box with a mass of 8kg

$a =\frac{8}{8}$

a = 1 m/s2

3 0

3 years ago

What is the power of 10 when 75,00 is written in scientific notation

blagie [28]

Answer:

7.5X1000

scientific notation

6 0

3 years ago

ANSWER THIS QUICKLY Which of the following is an example of natural selection?

mojhsa [17]

B is a case of natural selection.

7 0

3 years ago

Read 2 more answers

Alice and Bob are experimenting with two moles of neon, a monatomic gas, that starts out at conditions of standard temperature a

Archy [21]

Answer:

29273.178 joules have been added to the gas for the entire process.

Explanation:

The specific heats of monoatomic gases, measured in joules per mol-Kelvin, are represented by the following expressions:

Isochoric (Constant volume)

$c_{v} = \frac{3}{2}\cdot R_{u}$ (1)

Isobaric (Constant pressure)

$c_{p} = \frac{5}{2}\cdot R_{u}$ (2)

Where $R_{u}$ is the ideal gas constant, measured in pascal-cubic meters per mol-Kelvin.

Under the assumption of ideal gas, we notice the following relationships:

1) Temperature is directly proportional to pressure.

2) Temperature is directly proportional to volume.

Now we proceed to find all required temperatures below:

(i) Alice heats the gas at constant volume until its pressure is doubled:

$\frac{T_{2}}{T_{1}} = \frac{P_{2}}{P_{1}}$ (3)

( $\frac{P_{2}}{P_{1}} = 2$ , $T_{1} = 273.15\,K$ )

$T_{2} = \frac{P_{2}}{P_{1}} \times T_{1}$

$T_{2} = 2\times 273.15\,K$

$T_{2} = 546.3\,K$

(ii) Bob further heats the gas at constant pressure until its volume is doubled:

$\frac{T_{3}}{T_{2}} =\frac{V_{3}}{V_{2}}$ (4)

( $\frac{V_{3}}{V_{2}} = 2$ , $T_{2} = 546.3\,K$ )

$T_{3} = \frac{V_{3}}{V_{2}}\times T_{2}$

$T_{3} = 2\times 546.3\,K$

$T_{3} = 1092.6\,K$

Finally, the heat added to the gas ( $Q$ ), measured in joules, for the entire process is:

$Q = n\cdot [c_{v}\cdot (T_{2}-T_{1})+c_{p}\cdot (T_{3}-T_{2})]$ (5)

If we know that $R_{u} = 8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K}$ , $n = 2\,mol$ , $T_{1} = 273.15\,K$ , $T_{2} = 546.3\,K$ and $T_{3} = 1092.6\,K$ , the heat added to the gas for the entire process is:

$c_{v} = \frac{3}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} \right)$

$c_{v} = 12.471\,\frac{J}{mol\cdot K}$

$c_{p} = \frac{5}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} \right)$

$c_{p} = 20.785\,\frac{J}{mol\cdot K}$

$Q = (2\,mol)\cdot \left[\left(12.471\,\frac{J}{mol\cdot K} \right)\cdot (536.3\,K-273.15\,K)+\left(20.785\,\frac{J}{mol\cdot K} \right)\cdot (1092.6\,K-546.3\,K )\right]$

$Q = 29273.178\,J$

29273.178 joules have been added to the gas for the entire process.

6 0

2 years ago