Answer:
The dependent variable is academic performance
The independent variable is the presence/absence of tutorial support
The control group are students who did not get the tutorial support.
The experimental group were students that got the tutorial support
Explanation:
In every experiment, there is a dependent and independent variable as well as an experimental and a control group.
The experimental group receive the treatment while the control group do not receive the treatment. The independent variable is manipulated and its impact on the dependent variable is evaluated.
The control group are students who did not receive the tutorial support while the experimental group are students that received the tutorial support.
The dependent variable in this case is academic performance. Its outcome depends on the presence or absence of tutorial support (independent variable).
Answer:
yes ( true)
Explanation:
positive effects on all the body systems.
Answer:
empty space
Explanation:
Our solar system comprises of the sun as the star, the planets, the dwarf planets, various moons, and plenty of asteroids, comets, and meteoroids. However, the majority part of the solar system consists of a void or empty space. These empty spaces basically composed of planetary dust and gas.
Hence, it can be concluded that Most of our Solar system is composed of "Empty Spaces."
Answer:
impulse is the product of a force and the time during which the force acts
Explanation:
i hope this will help you :)
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
