This is physics, electricity.
I'll break it down for you, very easy:
you have to use the formula V=IR
where V= voltage in volts (V)
I= current in Amperes (A)
R= resistance in ohms (Ω)
so what you require is I, current there you have to rearrange the formula to subject current;
I=V/R
=9/5= 1.8Ω (ans)
Answer:
<u>independent variable:</u> size of soccer ball
<u>dependent variable</u>: period of time the balls stay in the air
<u>control variable</u>: parts of the body used to juggle the ball
Explanation:
The independent variable is the manipulatable variable supplied into an experiment. It is a variable that directly affects the dependent variable.
The independent variable is the outcome variable. It is the variable that is primarily measured from experiments whose value depends on the value of the independent variable.
The control variables are those that are kept constant throughout the course of experiments. In other words, they do not directly affect experimental outcomes.
Hence, in the illustration;
<u>independent variable</u>: size of soccer ball
<u>dependent variable</u>: period of time the balls stay in the air
<u>control variable</u>: parts of the body used to juggle the ball
Answer:
The correct answer would be phenotypic variation.
Phenotype refers to observable traits or characteristics of an individual. For example, height, color, shape, et cetera.
Phenotypic variation refers to the sum total of variations in characteristics within populations of the same species.
In contrast, genotypic variation refers to the sum total of variations (such as allele frequency) present in the genome of populations of the same species.
1. Action potential reaches the axon terminal and depolarizes it.
2. Depolarization opens voltage-gated calcium channels, enabling influx of Ca into the neuron.
3. Calcium binds to specialized proteins on vesicles (containing pre-made acetylcholine) and triggers them to fuse with the neuron membrane at the synapse.
4. Exocytosis of acetylcholine into the synaptic cleft occurs.
5. Acetylcholine diffuses across the synapse and binds to nicotinic receptors on the end plate of the myocyte.
6. Activated nicotinic receptors, themselves ion channels, cause cation influx into the myocyte and generate an end plate potential. This eventually gives rise to the full depolarization within the myocyte that enables contraction.
I would think it would kill it completely
