Answer: 3.41 s
Explanation:
Assuming the question is to find the time
the ball is in air, we can use the following equation:

Where:
is the final height of the ball
is the initial height of the ball
is the initial velocity of the ball
is the time the ball is in air
is the acceleration due to gravity

Then:


Multiplying both sides of the equation by -1 and rearranging:

At this point we have a quadratic equation of the form
, which can be solved with the following formula:
Where:
Substituting the known values:
Solving the equation and choosing the positive result we have:
This is the time the ball is in air
Answer: The molar heat capacity of aluminum is 
Explanation:
As we know that,
.................(1)
where,
q = heat absorbed or released
= mass of water = 130.0 g
= mass of aluminiunm = 23.5 g
= final temperature
=
= temperature of water =
= temperature of aluminium =
= specific heat of water= 
= specific heat of aluminium= ?
Now put all the given values in equation (1), we get
Molar mass of Aluminium = 27 g/mol
Thus molar heat capacity =
Between the top of the first and the top of the second loop, the coaster has lost potential energy = mgh, where h = 22.2 - 15 = 7.2m
This energy would have converted to Kinetic. Write out an equation and the masses will cancel out. Does that hint help you to find the solution? If not, I will give you another hint.
Answer:
biopsychlogy
Explanation:https://www.answers.com/Q/Which_perspective_is_most_directly_involved_with_the_study_of_how_brain_development_might_affect_behavior
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