Answer:
Both b and d can be correct
Explanation:
Generally, diffusion does not require energy (<em>making option a wrong</em>) because it is the movement of particles from a region of high concentration to a region of low concentration hence diffusion moves particles in the direction of a concentration gradient. An example of this is the passive transport (for instance, uptake of glucose by a liver cell).
However, in some cases, when diffusion is against the concentration gradient (i.e when particles move from a region of low concentration to a region of high concentration), diffusion will require energy in a case like this (<em>making option c wrong</em>). An example of this is active transport (transport of protein called sodium-potassium pump which involves pumping of potassium into the cell and sodium out of the cell).
The explanation above shows that diffusion can require energy to move particles (in or out) of the cell through the cell membrane.
How many inches are in 450 cm?
0.393701 in = 1 cm
0.393701 × 450 = <span>177.16545
450 cm = </span><span>177.16545 in</span>
The energy would increase
b) increase kinetic energy
If u disturbed equilibrium position then this principal comes into effect deciding how to counteract the disturbance.
Answer:
d. 103.3
Explanation:
In the given question, the National Weather Service routinely supplies atmospheric pressure data to help pilots set their altimeters. And the units of atmospheric pressure used for reporting the atmospheric pressure data are inches of mercury. For a barometric pressure of 30.51 inches of mercury, we can calculate the pressure in kPa as follow:
In principle, 3.386 kPa is equivalent to the atmospheric pressure of 1 inch of mercury. Thus, 30.51 inches of mercury is equivalent to 30.51 in *(3.386 kPa/1 in) = 103.307 kPa.
Therefore, a barometric pressure of 30.51 inches of mercury corresponds to _____103.3_____ kPa.
