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algol
1 month ago
9

A 75 kg skydiver can be modeled as a rectangular "box" with dimensions 20 cm * 40 cm * 180 cm. what is his terminal speed if he

falls feet first? use 0.8 for the drag coefficient.
Physics
1 answer:
kicyunya [3.2K]1 month ago
4 0

Terminal velocity=136.7 m/s

The formula for terminal velocity can be expressed as:

V= \sqrt{\frac{2 mg}{\rho a C}}

m= mass=75 kg

a= area=0.2 x 0.4=0.08 m²

ρ= density of air=1.23 kg/m³

C= drag coefficient =0.8

Thus, V= \sqrt{\frac{2 *75*9.8}{1.23*0.08*0.8}}

v= 136.7 m/s

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The right answer is (a).

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3 months ago
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The two sides of the DNA double helix are connected by pairs of bases (adenine, thymine, cytosine, and guanine). Because of the
kicyunya [3294]

Answer:

Explanation:

An image of the bond resulting from the search is attached.

Consider the force directed towards thymine as negative.

For the O-H-N combination:

The resulting force from this combination is:

F=-F_{OH}+F_{ON}\\\\=\frac{Ke^2}{r^2}+\frac{Ke^2}{r'^2}\\\\=Ke^2(\frac{-1}{r^2}+\frac{1}{r'^2})\\\\=(9.0\times 10^9Nm^2/kg^2)(1.6\times 10^{-19}C)^2[\frac{1}{[(0.280-0.110)\times 10^{-9}m]^2}+\frac{1}{0.280\times 10^{-9}m)^2}]\\\\=-5.03354\times 10^{-9}N

In the case of the N-H-N combination:

The total force acting from this combination is:

F'=F_{NN}-F_{HN}\\\\=\frac{Ke^2}{r^2}-\frac{Ke^2}{r'^2}\\\\=Ke^2(\frac{-1}{r^2}+\frac{1}{r'^2})\\\\=(9.0\times 10^9Nm^2/kg^2)(1.6\times 10^{-19}C)^2[\frac{1}{[0.300\times 10^{-9}m]^2}-\frac{1}{((0300-0..110)\times 10^{-9})m)^2}]\\\\=-3.822\times 10^{-9}N

The force that thymine applies on adenine is:

F_{net}=F+F'\\\\=-5.03354\times 10^{-9}N-3.822\times 10^{-9}N\\\\=-8.8558\times 10^{-9}N

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b)

The negative value indicates that the force is attractive, as it is aimed towards thymi.

c)

The force acting on the electron due to the proton is:

F=\frac{Ke^2}{r^2}\\\\=\frac{(9.0\times 10^9Nm^2/C^2)(1.6\times 10^{-9}C)^2}{(5.29\times 10^{-11}m)^2}\\\\=8.233\times 10^{-8}N

Since the electron and proton carry opposite charges, the force on the electron points towards the proton.

d)

The ratio of the above forces is:

\frac{F}{F_{net}}=\frac{8.233\times 10^{-8}N}{8.233\times 10^{-8}N}\\\\=9.3

Therefore, the bonding strength of the electron in the hydrogen atom is 9.3 times greater than the bonding force between adenine and thymine molecules.

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