Answer:
C) 16.3 ml
Explanation:
Density is equal to the ratio between the mass of an object and its volume:

where
m is the mass
V is the volume
In our problem, we know:
- density of aluminium: 
- mass of the aluminium foil: 
So we can re-arrange the equation above and use these data to find the volume of the piece of aluminium foil:

The condition is a neuron in when the outside of the neuron has a net positive charge and the inside has a net negative charge (due to accumulation of more sodium ions) is C. resting potential. T<span>he </span>resting membrane<span> </span>potential<span> of a </span>neuron<span> is approximately -70 mV (mV=</span><span>millivolt)</span>
Answer:
A fair test.
Explanation:
Hi, a fair test is used to do scientifically valuable experiments, is a controlled investigation to answer a scientific question.
In a fair test two or more things are compared.
It consists in changing only one factor (the one bieng tested) and keeping all the other conditions the same during an experiment.
The factor is called a variable.
To me, that sounds like the "Law of Conservation of Energy".
Answer:
FB = 0.187 N
Explanation:
To find the magnetic force FB in the wire you use the following formula:

the angle between B and L is given by:

Due to B depends on "y" you take into account the contribution of each element dy of the wire to the magnitude of the magnetic force. Thus, you have to integrate the following expression:
![|\vec{F_B}|=Isin\theta\int_0^{0.25}B(y)dy=Isin\theta\int_0^{0.25}(0.5y)dy\\\\|\vec{F_B}|=(2.0*10^{-3}A)(sin36.86\°)(0.5T)[\frac{0.25^2}{2}m]=0.187\ N](https://tex.z-dn.net/?f=%7C%5Cvec%7BF_B%7D%7C%3DIsin%5Ctheta%5Cint_0%5E%7B0.25%7DB%28y%29dy%3DIsin%5Ctheta%5Cint_0%5E%7B0.25%7D%280.5y%29dy%5C%5C%5C%5C%7C%5Cvec%7BF_B%7D%7C%3D%282.0%2A10%5E%7B-3%7DA%29%28sin36.86%5C%C2%B0%29%280.5T%29%5B%5Cfrac%7B0.25%5E2%7D%7B2%7Dm%5D%3D0.187%5C%20N)
hence, the magnitude of the magnetic force is 0.187N