Answer:
Yes, it is reasonable to neglect it.
Explanation:
Hello,
In this case, a single molecule of oxygen weights 32 g (diatomic oxygen) thus, the mass of kilograms is (consider Avogadro's number):

After that, we compute the potential energy 1.00 m above the reference point:

Then, we compute the average kinetic energy at the specified temperature:

Whereas
stands for the Avogadro's number for which we have:

In such a way, since the average kinetic energy energy is about 12000 times higher than the potential energy, it turns out reasonable to neglect the potential energy.
Regards.
Answer:
0.0360531138247 V/m
Explanation:
= Resistivity of gold =
(General value)
I = Current = 940 mA
d = Diameter = 0.9 mm
A = Area = 
E = Electric field
Resistivity is given by

The electric field in the wire is 0.0360531138247 V/m
<h3><u>Answer;</u></h3>
Higher temperatures
A wave will go faster through a liquid at <em><u>highe</u></em><u>r </u>temperatures
<h3><u>Explanation;</u></h3>
- <em><u>Mechanical waves are types of waves that require a material medium for transmission.</u></em> An example of mechanical wave is the sound wave whose transmission occurs in medium such as solids, liquids and gases.
- <em><u>The transmission of mechanical waves involves vibration of particles through the medium of transmission, thus transfer of energy from one point to another. </u></em>The vibration of particle may be in the form of a longitudinal wave or a transverse wave.
- <em><u>Increasing the temperature in a medium increases the kinetic energy of the particles in the medium and thus increasing the speed at which the particles vibrates and thus aiding a faster transmission of a wave.</u></em>
Answer:
m= 10 kg a = 52 m / s²
Explanation:
For this problem we must use Newton's second law, let's apply it to each axis
X axis
F - fr = ma
The equation for the force of friction is
-fr = miu N
Axis y
N- W = 0
N = mg
Let's replace and calculate laceration
F - miu (mg) = ma
a = F / m - mi g
a = 527.018 / m - 0.17 9.8
We must know the mass of the body suppose m = 10 kg
a = 527.018 / 10 - 1,666
a = 52 m / s²