In mammals and amphibians? An enucleated egg, a donor nucleus (preferably from an early developmental stage such as a blastocyst), and a means to stimulate the egg to be activated as if it had just been fertilized (poking with a needle is sometimes enough)
Or cloning into a vector as in at the level of DNA?
A vector with positve and negative selection markers (like antibiotic resistance and drug susceptibility), an insert, DNA ligase and restriction enzymes, buffer for restriction and ligation. Or if you are doing cloning by PCR, then you need primers, template, nucleotides, RNA pol like Taq polymerase etc.
What is your question about it?
Answer:
C) It will accelerate.
Explanation:
According to Newton’s second law of motion, when an object is acted on by an unbalanced force, it will accelerate.
An unbalanced force will change the speed or direction (or both) of an object. A change in speed and/or direction is acceleration.
A) is wrong. The object will stop moving only if there is a balanced force in the opposite direction.
B) is wrong. The object will decrease speed only if the unbalanced force has a component opposite to the direction of motion.
d) is wrong. The object will increase speed only if the unbalanced force has a component in the direction of motion.
Answer:
D: It will increase because smaller particles provide more surface area to react.
Explanation:
When the large iron is broken up into smaller pieces, there are more places for the iron to react (meaning there's more surface area). Think of it like taking the surface area of a big cube compared to the surface area of a bunch of small cubes. The sum of the surface areas of the small cubes will be greater than that of the large cube. As a result, more places for the iron to react will cause for a greater reaction.
H = planks constant
<span>m = mass of the object </span>
<span>u = velocity of the object </span>
<span>h = 6.626 * 10^-34 J/s </span>
<span>the mass of an electron is 9.12*10^-31 kg </span>
<span>10% speed of light = 10% * 3*10^8 = 3*10^7 m/s, i dont have my graphing calc with me right now so i leave the technicalities up to you </span>
