It is given that the surface area of sphere is 4 π r² and its volume is (4/3 π r³)
With a diameter of 1.2 mm you have a radius of 0.6 mm so the surface area about 4.5 mm² and the volume is about 0.9 mm³
The total surface energy of the original droplet is (4.5 x 10⁻⁶ m x 72) = 3.24 x 10⁻⁴mJ
The five smaller droplets need to have the same volume as the original so:
5 V = 0.9 mm³ so the volume of smaller sphere will equal 0.18 mm³
Since this smaller volume still have volume (4/3 π r³) so r = 0.35 mm
Each of the smaller droplets has a surface are = 1.54 mm²
The surface energy of the 5 smaller droplet is then (5 x 1.54 x 10⁻⁶ m x 72) = 5.54 x 10⁻⁴ mJ
From this radius the surface energy of all smaller droplets is 5.54 x 10⁻⁴ and the difference in energy is (5.54 x 10⁻⁴) - (3.24 x 10⁻⁴) = 2.3 x 10⁻⁴ mJ
Therefore we need about 2.3 x 10⁻⁴ mJ of energy to change a spherical droplet of water of diameter 1.2 mm into 5 identical smaller droplets
Answer:
C. 75%
Explanation:
From a heterozygous cross (Tt x Tt) the produced offspring would consist of:
- 25% TT -it would be tall-.
- 50% Tt -it would be tall as the gene T is dominant-.
- 25% tt -it would not be tall-.
Thus the produced offspring that would be tall is (50% + 25%) 75%. The answer is option C.
It speeds up chemical reactions
Answer:
The wavelength the student should use is 700 nm.
Explanation:
Attached below you can find the diagram I found for this question elsewhere.
Because the idea is to minimize the interference of the Co⁺²(aq) species, we should <u>choose a wavelength in which its absorbance is minimum</u>.
At 400 nm Co⁺²(aq) shows no absorbance, however neither does Cu⁺²(aq). While at 700 nm Co⁺²(aq) shows no absorbance and Cu⁺²(aq) does.
Answer:
a mass is constant during -free-weight training,a greater impulse will result in greater velocity.therefore as a generation of force greater than weight of the resistance increases.
