The two sides of the DNA double helix are connected by pairs of bases (adenine, thymine, cytosine, and guanine). Because of the
geometric shape of these molecules, adenine bonds with thymine and cytosine bonds with guanine. The figure (Figure 1) shows the thymine-adenine bond. Each charge shown is ±e, and the H−N distance is 0.110 nm.
A) Calculate the net force that thymine exerts on adenine. To keep the calculations fairly simple, yet reasonable, consider only the forces due to the O−H−N and the N−H−N combinations, assuming that these two combinations are parallel to each other. Remember, however, that in the O−H−N set, the O− exerts a force on both the H+ and the N−, and likewise along the N−H−N set.
B) Is the net force attractive or repulsive?
C) Calculate the force on the electron in the hydrogen atom, which is 5.29×10−2 nm from the proton.
D) Compare the strength of the bonding force of the electron in hydrogen with the bonding force of the adenine-thymine molecules.
A) Calculate the net force that thymine exerts on adenine. To keep the calculations fairly simple, yet reasonable, consider only the forces due to the O−H−N and the N−H−N combinations, assuming that these two combinations are parallel to each other. Remember, however, that in the O−H−N set, the O− exerts a force on both the H+ and the N−, and likewise along the N−H−N set.
B) Is the net force attractive or repulsive?
C) Calculate the force on the electron in the hydrogen atom, which is 5.29×10−2 nm from the proton.
D) Compare the strength of the bonding force of the electron in hydrogen with the bonding force of the adenine-thymine molecules.
1 answer:
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:
In the case of the N-H-N combination:
The total force acting from this combination is:
The force that thymine applies on adenine is:
When rounded to three significant figures, the net force is
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:
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:
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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Explanation:
The scenario is as follows:
1. A ball drops from the edge of a cliff.
2. Upon reaching the ground, the energy held in its gravitational potential energy transforms entirely into kinetic energy.
This implies K.E = P.E.
3. The ball impacts the ground at a speed of 20 m/s.
4. The gravitational field strength noted is 10 N/kg.
<pOur goal is to ascertain the height from which the ball was dropped.<pSince the ball was dropped from a cliff, its initial velocity is 0.<p→ K.E =where v is the final velocity, is the initial velocity, and m is the mass.
→ K.E =
→ K.E = 200 m joules when the ball strikes the ground.
<p→ P.E = mg hwhere g is the gravitational field strength, m is mass, and h signifies height.
<p→ g = 10 N/kg.<p→ P.E = m(10)(h)→ P.E = 10m h joules.
<p→ P.E = K.E.→ 10m h = 200 m.
Dividing through by 10m yields:
→ h = 20 meters.
The ball was released from a height of 20 meters.
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To understand more about gravitational potential energy, visit
Answer:
The equivalent distance in kilometers is 4012 × km.
Explanation:
It's known that 1 millimeter converts to meters. Then, 1 meter converts to
kilometers. Therefore, the conversion for 1 millimeter to kilometers can be stated as
1 mm = m
1 m = km
Thus, 1 mm = ×
km =
km.
Given the distance of 4012 mm, the corresponding distance in kilometers will be
4012 mm = 4012 × km.
The distance therefore is 4012 × km.
Response:
Clarification:
Refer to the diagram indicating the charges on the specified sphere (see attachment).
The electric field at the stated positions is
E(r) = 0 for r≤a. Equation 1
E(r) = kq/r² for a<r<b. Equation 2
E(r) = 0 for b<r<c. Equation 3
E(r) = kq/r² for r>c. Equation 4.
We understand that electric potential correlates with the electric field through
V = Ed
A. To compute the potential at the outer surface of the hollow sphere (r=c), we determine that the electric field there is
E = kQ / r²
Then,
V = Ed,
At d = r = c
Thus,
Vc = (kQ / c²) × c
Vc = kQ / c
As a result, the total charge Q consists of +q, -q, and +q
Hence, Q = q - q + q = q
V = kq / c
B. To calculate the potential at the inner surface of the hollow sphere (r=b), we have
V = kQ/r
V = kQ / b, noting that r = b
So, Q = q
V = kq / b
C. At r = a
Following from equation 1:
E(r) = 0 for r≤a. Equation 1
The electric field at the surface of the solid sphere is 0, E = 0N/C
Thus,
V = Ed = 0 V
Consequently, the electric potential at the solid sphere's surface is 0.
D. At r = 0
The electric potential can be determined by
V = kq / r
As r approaches 0,
V = kq / 0
V approaches infinity.
