Answer:
Circle Q
Explanation:
The flux of a field through a surface is given by:
where
F is the field strength
A is the area of the surface
is the angle between the direction of the field and the normal to the surface
In the figure, we see that the 4 surfaces have different orientations, so different angle . In the formula, we see that the flux is maximum when 
, which occurs when the field is perpendicular to the surface, and the flux is zero when 
, (because 
) which occurs when the field is parallel to the surface.
In the figure, we see that circle Q is parallel to the direction of the field, so in this case the field is zero, while in all the other cases is not.
So, the correct answer is
Circle Q
Answer:
4E
Explanation:
From the question given above, the following data were obtained:
Initial elongation (e₁) = 4 cm = 4/100 = 0.04 m
Initial energy (E₁) = E
Final elongation (e₂) = 0.04 + 0.04 = 0.08 m
Final energy (E₂) =?
The energy stored in a s spring is given by:
E = ½Ke²
Where
E => is the energy
K => is the spring constant
e => is the elongation
From:
E = ½Ke²
Energy is directly proportional to the elongation. Thus,
E₁/e₁² = E₂/e₂²
With the above formula, we can obtain the final energy as follow:
Initial elongation (e₁) = 0.04 m
Initial energy (E₁) = E
Final elongation (e₂) = 0.08 m
Final energy (E₂) =?
E₁/e₁² = E₂/e₂²
E / 0.04² = E₂ / 0.08²
E / 0.0016 = E₂ / 0.0064
Cross multiply
0.0016 × E₂ = 0.0064E
Divide both side by 0.0016
E₂ = 0.0064E / 0.0016
E₂ = 4E
Therefore, the final energy is 4 times the initial energy i.e 4E
Answer:
the linear speed of the car is 28.83 m/s
Explanation:
Given;
radius of the car, r = 0.33 m
angular speed of each tire, ω = 13.9 rev/s = 13.9 x 2π = 87.35 rad/s
The linear speed of the car is calculated as;
V = ωr
V = 87.35 rad/s x 0.33 m
V = 28.83 m/s
Therefore, the linear speed of the car is 28.83 m/s
The final velocity of object 2 is: 3 m/sec
<h3>What is the conservation of mass?</h3>
According to the rule of conservation of mass, no atoms are created or destroyed during a chemical process. Instead, the atoms combine in a variety of ways to create goods. This explains why each element has the equal amount of atoms on both sides of an equation with balanced symbols.
According to conservation of mass,
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Given that,
m₁ = 50 kg
m₂ = 75 kg
u₁ = 4 m/s
u₂ = -1 m/s
v₁ = -2 m/s
v₂ = ?
Now, putting the values,
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(50 × 4) + [75 × (-1)] = [50 × (-2)] + (75 × v₂)
or, 125 = -100 + 75 v₂
or, 75 v₂ = 225
or, v₂ = 3 m/sec.
To know more about velocity refer to:
brainly.com/question/80295
#SPJ1
Answer:
Explanation:
The original has hybrid 15N/14N DNA, and the second generation has both hybrid 15N/14N DNA and 14N/14N DNA. No 15N/15N DNA was observed. In this experiment:
Nitrogen is a significant component of DNA. 14N is the most bounteous isotope of nitrogen, however, DNA with the heavier yet non-radioactive and 15N isotope is likewise practical.
E. coli was developed for several generations in a medium containing NH4Cl with 15N. When DNA is extracted from these cells and centrifuged on a salt density gradient, the DNA separates at which its density equals to the salt arrangement. The DNA of the cells developed in 15N medium had a higher density than cells developed in typical 14N medium. After that, E. coli cells with just 15N in their DNA were transferred to a 14N medium.
DNA was removed and compared to pure 14N DNA and 15N DNA. Immediately after only one replication, the DNA was found to have an intermediate density. Since conservative replication would result in equal measures of DNA of the higher and lower densities yet no DNA of an intermediate density, conservative replication was eliminated. Moreso, this result was consistent with both semi-conservative and dispersive replication. Semi conservative replication would result in double-stranded DNA with one strand of 15N DNA, and one of 14N DNA, while dispersive replication would result in double-stranded DNA with the two strands having mixtures of 15N and 14N DNA, either of which would have appeared as DNA of an intermediate density.
The DNA from cells after two replications had been completed and found to comprise of equal measures of DNA with two different densities, one corresponding to the intermediate density of DNA of cells developed for just a single division in 14N medium, the other corresponding to DNA from cells developed completely in 14N medium. This was inconsistent with dispersive replication, which would have resulted in a single density, lower than the intermediate density of the one-generation cells, yet at the same time higher than cells become distinctly in 14N DNA medium, as the first 15N DNA would have been part evenly among all DNA strands. The result was steady with the semi-conservative replication hypothesis. The semi conservative hypothesis calculates that each molecule after replication will contain one old and one new strand. The dispersive model suggests that each strand of each new molecule will possess a mixture of old and new DNA.
