Answer:
[tex]\sf Proof \ below[/tex]
Explanation:
We know that acceleration is change in velocity over time.
[tex]\sf a=\frac{\triangle v}{t}[/tex]
[tex]\sf a=\frac{v-u}{t}[/tex]
v is the final velocity and u is the initial velocity.
Solve for v.
Multiply both sides by t.
[tex]\sf at=v-u[/tex]
Add u to both sides.
[tex]\sf at + u=v[/tex]
Answer:
Acceleration = v-u/t when we flip -u and t to right hand side
then -u changes to plus and denominator t changes to numerator
then then this equations becomes v=u+at
Explanation:
A ray of light is projected into a glass tube that is surrounded by air. The glass has an index of refraction of 1.50 and air has an index of refraction of 1.00. At what minimum angle will light in the glass tube be totally reflected at the glass/air interface?
Answer:
θ = 41.8º
Explanation:
This is an internal total reflection exercise, the equation that describes this process is
sin θ = n₂ / n₁
where n₂ is the index of the incident medium and n₁ the other medium must be met n₁> n₂
θ = sin⁻¹ n₂ / n₁
let's calculate
θ = sin⁻¹ (1.00 / 1.50)
θ = 41.8º
A very thin film of soap, of thickness 170 nm, in between air seems dark. On the other hand, when placed on top of glass some visible light is seen to shine from the film. How can this happen and what is the smallest visible light that creates constructive interference when we place the film on top of glass
Answer:
λ₀ = 2 d n
Explanation:
A soap film is a layer where the lus is reflected on the surface and on the inside of the film, these two reflected rays can interfere with each other either constructively or destructively.
Let's analyze the general conditions of this interference,
* When the ray of light reaches the surface of the film it is reflected, as the index of refraction of the air is less than the index of the film, the reflected ray has a phase change of 180º
* When the ray penetrates the film, its wavelength changes due to the refractive index of the film.
λ = λ₀ / n
where lick is the wavelength in the vacuum or air and n index of refraction of the film, in general this interference is observed perpendicular to the film, so the sine veils 1. the expression for constructive interference taking in what previous remains
2d = (m + ½) λ
the expression for destructive interference remains
2d = m λ
2d = m λ₀ / n
When the film is placed on a glass plate whose index of refraction is greater than the index of refraction of the film, in the reflection in the lower part of the film another phase difference of 180º is created, for which we have a difference of total phase of 180 +180 = 360º, which is equivalent to no phase difference, therefore the two previous equations are interchanged.
Therefore where we had destructive interference now a cosntructive interference happens we can see the reflected light.
Find us the wavelength that this constructive interference creates
2d n = m λ₀
λ₀ = 2 d n / m
To find the minimum wavelength, suppose we observe the first interference pattern m = 1
λ₀ = 2 d n
where d is the thickness of the film and n the index of refraction of the same
a 6 letter word a way of explaining an object or event using a set of facts
Explanation:
A theory is a way of explaining an object or event using a set of facts.
Increase in Space Suit Pressure 0.0/3.0 points (graded) If the pressure in a space suit increases, how will each of the following be affected? Flexilibity will: Increase Decrease Stay the same unanswered The required pre-breathe time will: Increase Decrease Stay the same unanswered The mass of the suit will: Increase Decrease Stay the same
Answer:
Flexibility Increases
Pre-breathe time decreases
Mass of suit decreases.
Explanation:
Spacesuits are designed for space shuttles when a person goes to explore the galaxy. The spacesuits shuttle era are pressurized at 4.3 pounds per inch. The gas in the suit is 100% of oxygen and there is more oxygen to breathe when the altitude of 10,000 is reached. This will decrease the breathing time and mass of suit.
Two kilograms of nitrogen (N2) at 25°C is contained in a 0.62 m3 rigid tank. This tank is connected by a valve to a 0.16 m3 rigid tank containing 0.8 kg of oxygen (O2) at 127°C. The valve is opened, and the gases are allowed to mix, achieving an equilibrium state at 87°C.
initial pressures of N2 is 5.7293 bar and O2 is 5.2 bar.
the final pressure is 6.44 bar.
the magnitude of the heat transfer for the process is 162.8 kJ, and the direction of energy flow is going in.
What is the entropy change for the mixing process, in kJ/K?
Answer:
Explanation:
For entropy change the formula is
ΔS = ΔQ / T
ΔQ = Δ H
ΔS = Δ H / T
Given
Δ H = + 162.8 kJ
We can take equilibrium temperature as average temperature of the whole process
So, T = 273 + 87 = 360 K
ΔS = Δ H / T
= 162.8 kJ / 360
= + 0.508 kJ / K .
When the magnitude of the heat transfer for the process is 162.8 kJ, Then the entropy change for the mixing process, in kJ/K is = + 0.508 kJ / K
What is Entropy change?
For The entropy change, the formula is
Then ΔS = ΔQ / T
After that ΔQ = Δ H
Then ΔS = Δ H / T
Given as per question are:
Then Δ H = + 162.8 kJ
Now We can take equilibrium temperature as average temperature of the whole process are:
So, T is = 273 + 87 = 360 K
Then ΔS = Δ H / T
After that = 162.8 kJ / 360
Therefore, = + 0.508 kJ / K.
Find more information about Entropy change here:
https://brainly.com/question/17241209
A uniform string of length 10.0 m and weight 0.32 N is attached to the ceiling. A weight of 1.00 kN hangs from its lower end. The lower end of the string is suddenly displaced horizontally. How long does it take the resulting wave pulse to travel to the upper end
Answer: 0.0180701 s
Explanation:
Given the following :
Length of string (L) = 10 m
Weight of string (W) = 0.32 N
Weight attached to lower end = 1kN = 1×10^3
Using the relation:
Time (t) = √ (weight of string * Length) / weight attached to lower end * acceleration due to gravity
g = acceleration due to gravity = 9.8m/s^2
Weight of string = 0.32N
Time(t) = √ (0.32 * 10) / [(1*10^3) * (9.8)]
Time = √3.2 / 9800
= √0.0003265
= 0.0180701s
A 24 cm radius aluminum ball is immersed in water. Calculate the thrust you suffer and the force. Knowing that the density of aluminum is 2698.4 kg / m3
Answer:
W =1562.53 N
Explanation:
It is given that,
Radius of the aluminium ball, r = 24 cm = 0.24 m
The density of Aluminium, [tex]d=2698.4\ kg/m^3[/tex]
We need to find the thrust and the force. The mass of the liquid displaced is given by :
[tex]m=dV[/tex]
V is volume
Weight of the displaced liquid
W = mg
[tex]W=dVg[/tex]
So,
[tex]W=dg\times \dfrac{4}{3}\pi r^3\\\\W=2698.4\times 10\times \dfrac{4}{3}\times \pi \times (0.24)^3\\\\W=1562.53\ N[/tex]
So, the thrust and the force is 1562.53 N.