Answer :
Explanation :
It is given that
Mass of iron, m = 39.5 g
Volume of iron,
So, density is :
density,
As a rocket rises, its kinetic energy changes. At the time the rocket reaches its highest point, most of the kinetic energy of r
2 answers:
Answer:
D transformed into gravitational potential energy.
Explanation:
Kinetic energy is the energy associated with movement and potential energy is the energy associated with position in a system. Energy, in general, is the ability to do a job.
Both kinetic and potential energy represent the two fundamental types of existing energy. Any other energy is a different version of kinetic or potential energy or a combination of both. For example, mechanical energy is the combination of kinetic and potential energy.
Kinetic energy is the type of energy that is associated with movement. Anything that is moving has kinetic energy. It can be calculated using the formula:
Where:
Potential energy is the type of energy that is associated with the relative position within a system, that is, the position of an object with respect to another. It can be calculated using the formula:
Where:
So, at the highest point, the velocity of the rocket would be zero, and the distance respect to the launch point would be maximum. Hence according to the previous equation its kinetic energy would zero and its potential energy would be the highest. in another words the kinetic energy of rocket has been transformed into gravitational potential energy.
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Explanation:
Given that,
Mass of the object, m = 7.11 kg
Spring constant of the spring, k = 61.6 N/m
Speed of the observer,
We need to find the time period of oscillation observed by the observed. The time period of oscillation is given by :
Time period of oscillation measured by the observer is :
So, the time period of oscillation measured by the observer is 5.79 seconds.
Answer:
a) I= I₀ (cos²θ - cos⁴θ) b) 75.5º
Explanation:
a) For this exercise we must use Malus's law
I = I₀ cos² θ
where tea is the angle between the two polarizers.
We apply this expression to our case
* Polarizer 1 suppose that it is vertical and polarizer 2 (intermediate) is at an angle θ with respect to the vertical
I₁ = I₀ cos² θ
* We analyze for the polarity 2 and the last polarizer 3 which indicate that it must be at 90º from the first one, therefore it must be horizontal.
The angle of polarizers 2 and 3 is θ' measured from the horizontal, if we measure with respect to the vertical
θ₂ = 90- θ’ = θ
fiate that in the exercise we must take a reference system and measure everything with respect to this system.
I = I₁ cos² θ'
we substitute
I = (I₀ cos² tea) cos² (θ - 90)
cos (θ -90) = cos θ cos 90 + sin θ sin 90 = sin θ
I = Io cos² θ sin² θ
1= cos²θ+ sin²θ
sin²θ = 1 - cos²θ
I= I₀ (cos²θ - cos⁴θ)
b) to find when the intensity is maximum,
we can use that we have an extreme point when the drift is zero
= 0
\frac{dI}{d \theta}= Io (2 cos θ - 4 cos³θ) = 0
whereby
cos θ - 2 cos³ θ = 0
cos θ ( 1 - 2 cos² θ) = 0
The zeros of this function are in
θ = 90º
1-2cos²θ =0 cos θ = 0.25 θ = 75.5º
Let's analyze this two results for the angle of 90º the intnesidd is zero with respect to the first polarizer, so it is not an acceptable solution.
Consequently, the angle that allows the maximum intensity to pass is 75.5º