Answer:
0.25 m.
Explanation:
We'll begin by calculating the spring constant of the spring.
From the diagram, we shall used any of the weight with the corresponding extention to determine the spring constant. This is illustrated below:
Force (F) = 0.1 N
Extention (e) = 0.125 m
Spring constant (K) =?
F = Ke
0.1 = K x 0.125
Divide both side by 0.125
K = 0.1/0.125
K = 0.8 N/m
Therefore, the force constant, K of spring is 0.8 N/m
Now, we can obtain the number in gap 1 in the diagram above as follow:
Force (F) = 0.2 N
Spring constant (K) = 0.8 N/m
Extention (e) =..?
F = Ke
0.2 = 0.8 x e
Divide both side by 0.8
e = 0.2/0.8
e = 0.25 m
Therefore, the number that will complete gap 1is 0.25 m.
Answer:
the correct answer is False
Explanation:
i hope its right
Answer:
The correct answer is:
Transport occurs with the help of membrane proteins
Explanation:
Facilitated diffusion is the passive (no energy required) movement of molecules across a cell membrane, with the aid of a membrane protein. energy in form of ATP or GTP is not required since molecules are moving down their concentration gradient.
The need for facilitated diffusion occurs when substances cannot successfully cross the phospholipid bilayer of the cells membrane on their own either due to their polarity or large size. The proteins in facilitated diffusion are tthose that act as carriers and those the form the channels across the membrane, through which the diffusion occurs. Examples of where facilitated diffusion is seen is in the transport of glucose (relatively large molecule) across the cell membranes, with the help of the glucose transporter and the transportation of water molecules across the cell membrane through aquaporins.
The answer is C because they are made up of different sizes and mass.
Answer:
Explanation:
In this problem we can use Bohr's atomic model, to deal with the electronic transition, so we can have transitions between two given states.
ΔE = 
-E₀
Lower state final state energy
Fundamental first excited state ΔE₁ = -2.4 - (-4.7) = 2.3 eV
Fundamental second excited state ΔE₂ = -1.0 - (-4.7) = 3.7 eV
Fundamental third excited state ΔE₃ = -0.3 - (-4.7) = 4.4 eV
As they indicate that there are no electrons in the excited states these are the only possible transitions.
When a wide range of light strikes, the frequencies of these energies are absorbed and observed as black bands (absence of radiation) in the spectrum.
