15:) using more force in your muscle will increase the force used to bounce the basketball
16:) the pulling of gravity livitation does not allow the ball to go back up with the hieght it was dropped from on the scientifical drop point
14:) <span>a weight hung from a fixed point so that it can swing freely backward and forward, especially a rod with a weight at the end that regulates the mechanism of a clock that is the deffinition of to which of the word pendulum read it do not plagarize and i hope ii helped and have a great day bye.)::</span>
We have that the most stable nuclei are the ones with the highest average binding energy. We see that Nitrogen has a mass number of 15 and that in this region of the graph average binding energy is low. Silver and Gold are along a line where there is a constant decline in average binding energy; silver has more than gold. However, we see that at the start of this decline, there is Fe 56. This region has the elements with the highest average binding energy; Nickel with a mass number of 58 is right there and thus it is the most stable nucleus out of the listed ones.
B. It reflects light rays outward.
A convex mirror looks like this ⊂. So when light rays hit the outward curves, it bends the light rays outwards, and backwards, and it would appear to be coming from its center called the focus.
Answer:
L = μ₀ n r / 2I
Explanation:
This exercise we must relate several equations, let's start writing the voltage in a coil
= - L dI / dt
Let's use Faraday's law
E = - d Ф_B / dt
in the case of the coil this voltage is the same, so we can equal the two relationships
- d Ф_B / dt = - L dI / dt
The magnetic flux is the sum of the flux in each turn, if there are n turns in the coil
n d Ф_B = L dI
we can remove the differentials
n Ф_B = L I
magnetic flux is defined by
Ф_B = B . A
in this case the direction of the magnetic field is along the coil and the normal direction to the area as well, therefore the scalar product is reduced to the algebraic product
n B A = L I
the loop area is
A = π R²
we substitute
n B π R² = L I (1)
To find the magnetic field in the coil let's use Ampere's law
∫ B. ds = μ₀ I
where B is the magnetic field and s is the current circulation, in the coil the current circulates along the length of the coil
s = 2π R
we solve
B 2ππ R = μ₀ I
B = μ₀ I / 2πR
we substitute in
n ( μ₀ I / 2πR) π R² = L I
n μ₀ R / 2 = L I
L = μ₀ n r / 2I
