the expression for diffraction grating allows to find the results for the questions for the angular separation are:
i) The third order is Δθ = 0.203 rad.
ii) The first order with water is Δθ = 0.046 rad.
The diffraction grating is a system formed by a large number of equally spaced lines whose diffraction is given by the expression.
d sin θ = m λ
Where d is the distance between two lines, θ is the angle of diffraction, the order of diffraction and λ is the wavelength.
i) Let's start by looking for the separation between two lines
Let's use a rule of direct proportions. If there are 300 lines in 1 mm, what distance is there between two lines.
d = 1 lines (1 mm / 300 lines) = 3,333 10⁻³ mm
d = 3.333 10⁻⁶ m
Let's find the angle of diffraction for the third order (m = 3) for each wavelength.
λ₁ = 400 nm = 400 10⁻⁹ m
sin θ₁ = [tex]\frac{m \ \lambda }{d}[/tex]m λ/ d
sin θ₁ = [tex]\frac{3 \ 400 \ 10^{-9} }{3.333 \ 10^{-6} }[/tex]
θ₁ = sin⁻¹ 0.3600
θ₁ = 0.368 rad
λ₂ = 600 nm = 600 10⁻⁹ m
sin θ₂ = [tex]\frac{3 \ 600 \ 10^{-9} }{3.333 \ 10^{-6} }[/tex]
θ₂ = sin⁻¹ 0.5401
θ₂ = 0.571 rad
The angular separation is
Δθ = θ₂ - θ₁
Δθ = 0.571 - 0.368
Δθ = 0.203 rad
ii) In this case, the separation between the network and the observation screen is filled with water.
When the rays leave the network they undergo a refraction process, for which they must comply with the relationship.
[tex]n_i \ sin \theta_1 = n_r \ sin \theta_r[/tex]
The incident side is in the air, therefore its refractive index is n_i = 1 and when it passes into the water with refractive index n_r = 1.33.
Let's start looking for the incident angles for the first order of diffraction.
m = 1
λ₁ = 400 nm
θ₁ = sin⁻¹ [tex]\frac{1 \ 400 \ 10^{-9}}{3.33 \ 10^{-6}}[/tex]
θ₁ = 0.120 rad
λ₂ = 600 nm
θ₂ = sin⁻¹¹ [tex]\frac{1 \ 600 \ 10^{-9} }{3.33 \ 10^{-6}}[/tex]
θ₂ = 0.181 rad
we use the equation of refraction.
[tex]\theta_r[/tex] = sin⁻¹ ([tex]\frac{n_i}{n_r} \ sin \ \theta_i[/tex] )
λ₁ = 400 nm
θ₁ = sin¹ ([tex]\frac{1 sin 0.120}{1.33}[/tex]
θ₁ = 0.090 rad
λ₂ = 600 nm
θ₂ =sin⁻¹ [tex]\frac{1 sin 0.181}{1.33}[/tex]
θ₂ = 0.1358 rad
The angular separation is
Δθ = 0.1358 - 0.090
Δθ = 0.046 rad.
In conclusion using the relation for the diffraction grating we can find the results for the questions about angular separation are:
i) The third order is Δθ = 0.203 rad.
ii) The first order with water is Δθ = 0.046 rad.
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Am I correct? Plz answer ASAP! I will give brainilest:D (Science)
Answer:
I would agree with your selection.
Explanation:
In 1962 measurements of the magnetic field of a large tornado were made at the Geophysical Observatory in Tulsa, Oklahoma. If the magnitude of the tornado's field was B = 17.50 nT pointing north when the tornado was 9.10 km east of the observatory, what current was carried up or down the funnel of the tornado? Model the vortex as a long, straight wire carrying a current. A (conventional current) flowing ---Direction--- the tornado.
Answer:
796.25 A
Explanation:
B= (μI)/(2πr)
I= (B*2πr)/(μ)
μ= 4π*10^-7
I= ((17.50*10^-9)(2π)(9.10*10^3)) / (4π*10^-7)
= 796.25 A
A (conventional current) flowing "DOWN" the tornado.
A trouble-making youth is standing on a bridge, and wants to drop a water balloon on an unsuspecting passerby. A man is jogging on a path below the bridge with a constant speed of 4.2 m/s. The bridge is 11.6 m above the ground. If the balloon is to land right at the jogger's feet, at what horizontal distance x from the bridge should he be when the youth drops the balloon?
Answer:
Explanation:
Time needed for a balloon to drop from vertical rest a distance of 11.6 m
t = √(2h/g) = √(2(11.6)/9.8) = 1.538618
d = vt = 4.2(1.538618) = 6.462197...
d = 6.5 m
Horizontal distance x = 6.5 m from the bridge he should be when the youth drops the balloon with velocity 4.2 m/s.
What is velocity?When an item is moving, its velocity is the rate at which its direction is changing as seen from a certain point of view and as measured by a specific unit of time.
Uniform motion an object is said to have uniform motion when object cover equal distance in equal interval of time within exact fixed direction. For a body in uniform motion, the magnitude of its velocity remains constant over time.
Given in the question time needed for a balloon to drop from vertical rest a distance of 11.6 m
t = √(2h/g) = √(2(11.6)/9.8) = 1.538618
d = vt = 4.2(1.538618) = 6.462197...
d = 6.5 m
Horizontal distance x = 6.5 m from the bridge he should be when the youth drops the balloon with velocity 4.2 m/s.
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What was the initial speed of a car if its speed is 40 m/s after 5 seconds of
accelerating at 4 m/s27
A. 10 m/s
B. 60 m/s
C. 20 m/s
D. 25 m/s
Reset Selection
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Answer:
[tex]\boxed {\boxed {\sf C. \ 20 \ m/s }}[/tex]
Explanation:
We are asked to find the initial speed of a car.
We are given the final speed, the time, and the acceleration, so we will use the following kinematic equation:
[tex]v_f=v_i+at[/tex]
We know the final speed is 40 meters per second, the acceleration is 4 meters per second squared, and the time is 5 seconds.
[tex]v_f[/tex]= 40 m/s t= 5 s a= 4 m/s²Substitute the values into the formula.
[tex]40 \ m/s = v_i+(4 \ m/s^2 * 5 \ s)[/tex]
Multiply inside the parentheses.
[tex]40 \ m/s =v_i+ 20 \ m/s[/tex]
We are solving for the initial speed, so we must isolate the variable [tex]v_i[/tex].
20 meters per second is being added to [tex]v_i[/tex]. The inverse operation of addition is subtraction. Subtract 20 m/s from both sides of the equation.
[tex]40 \ m/s - 20 \ m/s = v_i + 20 \ m/s - 20 \ m/s[/tex]
[tex]40 \ m/s - 20 m/ s = v_i[/tex]
[tex]20 \ m/s=v_i[/tex]
The initial speed of the car is 20 meters per second and choice C is correct.
A massive launcher sends a projectile vertically upwards from the surface of a planet of mass M and radius R. You may assume that this planet has no atmosphere. Part A If the projectile is launched at the escape speed, how do the magnitudes of its initial KE and gravitational potential energy (GPE) compare
Answer:
It can be shown that the potential energy of an object at the surface of the planet would be -G M / R if the potential at infinity is chosen to be zero.
Kinetic energy of G M / R would be required for the escape speed of such an object. The total energy in all such cases is zero.
This can easily be seen by considering the speed of an object falling from infinity towards the planet - the total energy will remain zero if it was zero when the object started to fall.
The potential at infinity is set to zero while the kinetic energy will be [tex]\rm \frac{GM}{R}[/tex] and total energy will be zero.
What is escape speed?Escape speed is the minimum speed required for a free, non-propelled object to escape from the gravitational pull of the main body and reach an infinite distance from it in celestial physics.
It is proven that if the potential at infinity is set to zero, the potential energy of an item on the planet's surface is [tex]\rm \frac{-GM}{R}[/tex].
The escape speed of such an item would necessitate kinetic energy will be [tex]\rm \frac{GM}{R}[/tex]. In all of these circumstances, the total energy is zero.
Consider the speed of an item falling from infinity towards the planet: if the total energy was zero before the thing began to descend, the total energy will stay zero.
Hence the potential at infinity is set to zero while the kinetic energy will be [tex]\rm \frac{GM}{R}[/tex] and total energy will be zero.
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The gas state of water
Answer:
The gas state of water is water vapour. It is formed by boiling liquid water or from the sublimation of ice (sublimation is the process of converting a solid to a liquid). Hope this helps!
Ex 2) A cannon ball is shot straight up into the air with an initial velocity of 25 m/s[Up).
What is the maximum height of the cannonball?
Explanation:
The final velocity at the cannon ball's maximum height is zero ([tex]v_y = 0[/tex]). We can use the equation
[tex]v_y^2 = 0 = v_{0y}^2 - 2gy_{max}[/tex]
[tex]\Rightarrow y_{max} = \dfrac{v_{0y}^2}{2g}[/tex]
[tex]y_{max} = \dfrac{(25\:\text{m/s})^2}{2(9.8\:\text{m/s}^2)}[/tex]
[tex]\:\:\:\:\:\:\:=31.9\:\text{m}[/tex]
I NEED A 100% ACCURATE ANSWER FOR THIS QUESTION ASAP NO LINKS !!!
What describes the pressure in a contained fluid?
1: unknown as long as the fluid is contained
2: the same throughout the fluid
3: higher in some places and lower in other places
Answer:
3) higher in some places nd lower In other places
Explanation:
The pressure in a fluid is given by the law
p = p0 + pgh
where
p0 is the atmospheric pressure
p is the fluid's density
g is the acceleration of gravity
h is the depth at which the pressure is calculated
As we see, the pressure depends on the value of h (depth): therefore, points which are located at more depth experience a larger pressure than points located near the fluid's surface.
Two vectors P=40m/s and q=60mls are perpendicular to each other. Calculate the resultant force and the angle resultant make with either vectors
This question involves the concepts of the rectangular components of the vector and the resultant of the vectors.
The magnitude of the resultant force is "72.11 N", and it makes an angle of "33.7°" from one of the vectors.
The rectangular components of a vector are those components that have a right angle between them. The resultant of two rectangular components of a vector can be found using the following formula:
[tex]F = \sqrt{F_1^2+F_2^2}[/tex]
where,
F = resultant force = ?
F₁ = First component of force = 40 N
F₂ = Second component of force = 60 N
Correction: The unit in question is given as m/s, while the unit of force is N. Hence, N is taken here.
Therefore,
[tex]F = \sqrt{(40\ N)^2+(60\ N)^2}[/tex]
F = 72.11 N
The following formula can be used to find the angle of resultant from 60N vector:
[tex]\theta = tan^{-1}(\frac{40\ N}{60\ N})\\\\[/tex]
θ = 33.7°
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The attached picture shows the resultant calculation for the rectangular components of a vector.
If you traveled 50m/s for 60 seconds, how far did you travel? Remember speed=distance/time
Question options:
300 m/s
500 m/s
3,000 m/s
300 km/h
Answer:
3,000 m
Explanation:
speed = distance/time
distance = speed * time
= 50 * 60
= 3000 m
tính chỉ số ampe kế thay ampe kế bằng vô kế có điện trở rất lớn thì vô kế chỉ bao nhiêu
Answer:
I don't understand it.....
the electrical energy used to pump the water up to a mountain lake is 1.2 x 10 to the power 12 J only 6.2 x 10 to the power 11 J of electrical energy is generated when the water is released calculate the efficiency of this energy storage scheme
Answer:
Explanation:
ake is 1.2 x 10^12 J. Only 6.2 x 10^11 J of electrical energy is generated when the water is released. Calculate the efficiency of this energy
A car with a mass of 1500 kg is pulled by a rope that is horizontal to the ground. The tension in the rope is 2000 N and a friction force of 350 N opposes the car's motion. What is the magnitude of the car's acceleration?
Answer:
Explanation:
Assuming the ground is level as well.
F = ma
a = F/m
a = (2000 - 350) / 1500
a = 1.1 m/s²
The acceleration of the car is 1.1 m/s².
To calculate the magnitude of the acceleration of the car, we use the formula below.
F-F' = ma................ Equation 1Where:
F = Tension in the ropeF' = Friction forcem = mass of the cara = acceleration of the carMake a the subject of the equation
a = (F-F')/m................. Equation 2From the equation,
Given:
F = 2000 NF' = 350 Nm = 1500 kgSubstitute these values into equation 2
a = (2000-350)/1500a = 1650/1500a = 1.1 m/s²Hence, the acceleration of the car is 1.1 m/s².
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What will be the current through a resistance of 50Ω if the applied voltage across the resistance is 117V?
Formulas: V=IR, I=V/R, R=V/I
Note: No need to write the unit of your answer.
2.34 A
Explanation:
[tex]V = IR \Rightarrow I = \dfrac{V}{R} = \dfrac{117\:\text{V}}{50\:Ω} = 2.34\:\text{A}[/tex]
Which of the following formulas describes the change in momentum of an
object?
A. change in momentum = force x time over which force is applied
B. change in momentum = acceleration distance over which
acceleration is applied
C. change in momentum = force x distance over which force is
applied
O D. change in momentum = acceleration time over which
acceleration is applied
When converted to a household measurement, 9 kilograms is approximately equal to a
Answer:
D) 19.8 lbs
Explanation:
1kg in household measurement is equal to 35.274 ounces. 35.274*9=317.466 ounces.
1kg is also equal to 2.205 lbs. 9*2.205=19.8416
9 kg is also equal to 9000 grams, but grams are not a part of the household measurement system
a) 9000 grams. b) 9000 ounces. c) 19.8 ounces. d) 19.8 pounds.
This leaves us with 19.8 lbs
how long untill the semester is done
The rocket now has a thruster that malfunctions and is now pushing the rocket in the wrong direction. What is the new net force on the rocket if it is now accelerating at 12mls²??
From your previous question
Mass=30kgAcceleration=12m/s^2[tex]\\ \sf\longmapsto F=ma[/tex]
[tex]\\ \sf\longmapsto F=30(12)[/tex]
[tex]\\ \sf\longmapsto F=360N[/tex]
Describe what causes the planets to stay in orbit around the Sun
Answer:
Gravitional pull or gravity
Explanation:
Its that simple
A race car rounding a corner at a constant speed of 200 miles per hour.
Answer:
Yes since it is changing direction
Explanation:
A race car rounding a corner at a constant speed of 200 miles per hour
hola buen día por favor alguien me puede ayudar por favor1. Dos cargas puntuales q1=+4μC y q2=+ 6μC están separadas por 10 cm. Una carga
puntual
q3=+2μC se coloca a medio camino entre q1 y q2. Encuentra la magnitud y la dirección de la
fuerza resultante sobre q3. Asuma que q1 está a la izquierda de q3 y q2 a la derecha de q3.
Revisa el ejemplo de la clase del tema 6. Además, recuerda que las fuerzas no siempre actúan
en la misma dirección.
2. Se coloca una carga puntual de 4 μC en un punto P (x = 0.2 m, y = 0.4 m). ¿Cuál es el campo
eléctrico E debido a esta carga en el origen del plano cartesiano? Hint: asume un plano
cartesiano y dibuja el campo eléctrico que se busca. Además, recuerda el teorema de
Pitágoras.
3. Una carga puntual q = 25μC está ubicada en el centro de una esfera de radio R = 25 cm, el
campo eléctrico generado a esta distancia debido a esa carga es E=3.6 x 10^6 N/C. Se quita de
la superficie una sección circular con radio r = 5 cm. Encuentre el flujo eléctrico que pasa por
esta sección. Hint: solo quieres saber el flujo eléctrico sobre el área de la superficie que se
aisló, por lo que es importante recordar cómo se obtiene el área de un círculo.
4. En cierta región al norte del planeta tierra, existe un campo eléctrico uniforme, la dirección
del campo eléctrico es hacía el centro del planeta y según se sabe, tiene un valor de 1000 N/C.
Encuentra el cambio de la energía potencia eléctrica de una partícula que tiene una carga de
-1.6 x10 -19 C. Considera que la partícula se suelta en un punto A y llega hasta un punto B que
está a 50 m del punto A.
5. Dos cargas q1 = −2 μC y q2 = + 2 μC se fijan en sus posiciones y se separan por una distancia d
= 10 cm. ¿Cuál es el potencial eléctrico en el origen debido a estas dos cargas?
Answer:
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Explanation:
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A closed, uninsulated system fitted with movable piston, so no matter is exchanged with the surroundings, was assembled. Introduction of 430 J of heat caused the system to expand, doing 238 J of work against a constant pressure of 101 kPa. What is the value of for this process
Answer: You do not specify what is being asked for. ∆E? ∆H?
∆E = (430 - 238) J = 192 J
∆H = 430 J
Explanation:
If asked for the value of ∆H the answer is simply the change in heat, and in the question, it states introduction of 430 J of heat is causing the system to expand.
Therefore ∆H = 430 J
If asked for ∆E, we know that ∆E = ±q (heat) + work (-P∆V) = ±q + w
The question states that 238 J of work are done AND the system expanded
(work is negative because expansion means work is done BY the system, releasing energy/heat... Conversely, if the system were compressed, work is done ON the system, absorbing heat/energy)
Therefore, ∆E = (430 - 238) J = 192 J
(a) Define the term ginning? (b) Name the simple device used for spinning?
A simple device used for spinning is a hand spindle also known as takli.
hope it helps
A student is asked to calculate the centripetal acceleration of a hummingbird that makes a complete circle in 0.43 seconds. If the circle has a radius of 0.25 m, what is
the correct setup for determining the centripetal acceleration of the hummingbird?
(3.65m/s)2/0.25m
(3.65m/s)2-0.43
O (0.435)2 / 25m
O (0.432.25m
The centripetal acceleration of the hummingbird is [tex]\frac{(3.65)^2}{0.25} = 53.29 \ m/s^2[/tex]
The given parameters;
time of motion of the hummingbird, t = 0.43 sradius of the circle, r = 0.25 mnumber of revolution of the hummingbird = 1 rev per 0.43 sThe angular speed of the hummingbird is calculated as follows;
[tex]\omega = \frac{1 \ rev}{0.43 \ s} \times \frac{2\pi \ rad}{1 \ rev} =14.61 \ rad/s[/tex]
The linear speed of the hummingbird is calculated as follows;
v = ωr
v = 14.61 x 0.25
v = 3.65 m/s
The centripetal acceleration of the hummingbird is calculated as follows;
[tex]a_c = \frac{v^2}{r} \\\\a_c = \frac{(3.65)^2}{0.25} \\\\a_c = 53.29\ m/s^2[/tex]
Thus, the centripetal acceleration of the hummingbird is [tex]\frac{(3.65)^2}{0.25} = 53.29 \ m/s^2[/tex]
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Which of the following statement describes a nonpolar molecule
Answer:
?
Explanation:
what are the statements
Una masa de aire de 20g absorbe 780 cal. Teniendo en cuenta que su temperatura inicial es de 30°C. Calcular la
temperatura que alcanzo al absorber el calor. Ce:0,24 cal/g °C
Answer:
sorry I don't know the answer
What is an involuntary and a voluntary muscle action?
Answer:
Involuntary muscle action are muscles that move WITHOUT conscious control, and voluntary muscle action are muscles that move WITH conscious control.
Explanation:
A example of a involuntary muscle action would be your heart. Your heart is a muscle beats without you thinking about it and is very vital to your body. A example of a voluntary muscle would be the muscles that are attached to your bones and allow you to move your body, and it's voluntary because your body won't move on it's own.
Hope this helps.
how can I become a good science student ?
Answer:
Study hard , focus on your studies and alyways ask questions .
Study, revise, write notes, listen in class, don't let yourself be distracted by others, and do the work in class...maybe join stem or science club if you wanna
Hope this helped you- have a good day bro cya)
I need help ASAP please.....
Which term describes energy stored in the bonds between atoms?
A. Nuclear energy
B. Chemical energy
O C. Thermal energy
D. Sound energy
Answer:
B. Chemical energy
Explanation:
chemical energy is the energy stored in the bonds between atoms