Answer:
I = 1.23 A
Explanation:
In an RL circuit current passing is described by
I = E / R (1 - [tex]e^{-Rt/L}[/tex])
Let's reduce the magnitudes to the SI system
L = 35 mH = 35 10⁻³ H
t = 5.0 ms = 5.0 10⁻³ s
let's calculate
I = 18/12 (1 - [tex]e^{-12 .. 5 {10}^{-3}/35 .. {10}^{-3} }[/tex]e (- 5 10-3 12/35 10-3))
I = 1.5 (1- [tex]e^{-1.715}[/tex])
I = 1.23 A
how many stars are in our solar system?
Answer:
there are over 100 billion stars in our galaxy.
In a double‑slit interference experiment, the wavelength is lambda=487 nm , the slit separation is d=0.200 mm , and the screen is D=48.0 cm away from the slits. What is the linear distance Δx between the eighth order maximum and the fourth order maximum on the screen?
Answer:
Δx = 4.68 x 10⁻³ m = 4.68 mm
Explanation:
The distance between the consecutive maxima, in Young's Double Slit Experiment is given bu the following formula:
Δx = λD/d
So, the distance between the eighth order maximum and the fourth order maximum on the screen will be given as:
Δx = 4λD/d
where,
Δx = distance between eighth order maximum and fourth order maximum=?
λ = wavelength = 487 nm = 4.87 x 10⁻⁷ m
d = slit separation = 0.2 mm = 2 x 10⁻⁴ m
D = Distance between slits and screen = 48 cm = 0.48 m
Therefore,
Δx = (4)(4.87 x 10⁻⁷ m)(0.48 m)/(2 x 10⁻⁴ m)
Δx = 4.68 x 10⁻³ m = 4.68 mm
Which of
of
these
following material is
used as fuse material?
carbon,
silver
Copper
Aluminium
The provided question is not correct as, there is more than one options are correct, however the explaining every correct option -
Answer:
The correct answer are - silver, copper and aluminium all three used as fuse material.
Explanation:
A safety device in any electric circuit of that prevents the electric system in case of short circuit by breaking the connection of electric system or circuit termed as the Fuse or fuse element. Normally the fuse are made up of wire or element of material that are low in melting point and high in resistance.
Zinc, lead, tin, silver, copper, aluminium, and alloy of tin and alloy are used as fuse element or material for their low melting point and high resistance these are easily breaks the electric path in case of short circuit.
Lamar has been running sprints to prepare for his next football game.He has found that he can maintain his maximum speed for 45 yards.He’s thinking of running in a 5km race in a few months,but doesn’t know if he can maintain his maximum speed for the entire 5 km.Can you help him determine how far he can?
Answer:
Kindly check explanation
Explanation:
Length of race = 5km
Maximum speed = 45 yards
Converting from yards to kilometer :
1km = 1093.613 yards
x = 45 yards
(1093.613 * x) = 45
x = 45 / 1093.613
x = 0.0411480 km
Where x = maximum length for which he can maintain his maximum speed expressed in kilometers.
Therefore, with the available information, it can be concluded that Lamar cannot maintain his maximum speed for the entire 5km race and will only be able maintain his maximum speed for 0.0411 kilometers.
Lamar cannot maintain his maximum speed for the entire 5km race and will only be able maintain his maximum speed for 0.0411 kilometers.
The calculation is as follows;
Length of race = 5km
Maximum speed = 45 yards
Converting from yards to kilometer :
1km = 1093.613 yards
x = 45 yards
[tex](1093.613 \times x) = 45[/tex]
[tex]x = 45 \div 1093.613[/tex]
x = 0.0411480 km
here x represent maximum length for which he can maintain his maximum speed expressed in kilometers.
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Helium-neon laser light (λ = 6.33 × 10−7 m) is sent through a 0.30 mm-wide single slit. What is the width of the central maximum on a screen 1.0 m from the slit?
Answer:
The width is [tex]w_c = 0.00422 \ m[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 6.33*10^{-7} \ m[/tex]
The width of the slit is [tex]d = 0.3\ mm = 0.3 *10^{-3} \ m[/tex]
The distance of the screen is [tex]D = 1.0 \ m[/tex]
Generally the central maximum is mathematically represented as
[tex]w_c = 2 * y[/tex]
Here y is the width of the first order maxima which is mathematically represented as
[tex]y = \frac{\lambda * D}{d}[/tex]
substituting values
[tex]y = \frac{6.33*10^{-7} * 1.0}{ 0.30}[/tex]
[tex]y = 0.00211 \ m[/tex]
So
[tex]w_c = 2 *0.00211[/tex]
[tex]w_c = 0.00422 \ m[/tex]
Coherent light from a sodium-vapor lamp is passed through a filter that blocks everything except for light of a single wavelength. It then falls on two slits separated by 0.490 mm . In the resulting interference pattern on a screen 2.12 m away, adjacent bright fringes are separated by 2.86 mm . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Determining wavelength. Part A What is the wavelength of the light that falls on the slits
Answer:
λ = 6.61 x 10⁻⁷ m = 661 nm
Explanation:
From the Young's Double Slit experiment, the the spacing between adjacent bright or dark fringes is given by the following formula:
Δx = λL/d
where,
Δx = fringe spacing = 2.86 mm = 2.86 x ⁻³ m
L = Distance between slits and screen = 2.12 m
d = slit separation = 0.49 mm = 0.49 x 10⁻³ m
λ = wavelength of light = ?
Therefore,
2.86 x 10⁻³ m = λ(2.12 m)/(0.49 x 10⁻³ m)
(2.86 x 10⁻³ m)(0.49 x 10⁻³ m)/(2.12 m) = λ
λ = 6.61 x 10⁻⁷ m = 661 nm
light of wavelength 550 nm is incident on a diffraction grating that is 1 cm wide and has 1000 slits. What is the dispersion of the m = 2 line?
Answer:
The dispersion is [tex]D = 2.01220 *10^{5} \ rad/m[/tex]
Explanation:
From the question we are told that
The wavelength of the light is [tex]\lambda = 550 \ = 550 *10^{-9} \ n[/tex]
The width of the grating is[tex]k = 1\ cm = 0.01 \ m[/tex]
The number of slit is N = 1000 slits
The order of the maxima is m = 2
Generally the spacing between the slit is mathematically represented as
[tex]d = \frac{k}{N}[/tex]
substituting values
[tex]d = \frac{ 0.01}{1000}[/tex]
[tex]d = 1.0 *10^{-5} \ m[/tex]
Generally the condition for constructive interference is
[tex]d\ sin(\theta ) = m * \lambda[/tex]
substituting values
[tex]1.0 *10^{-5} sin (\theta) = 2 * 550 *10^{-9}[/tex]
[tex]\theta = sin^{-1} [\frac{ 2 * 550 *10^{-9}}{ 1.0 *10^{-5}} ][/tex]
[tex]\theta = 6.315^o[/tex]
Generally the dispersion is mathematically represented as
[tex]D = \frac{ m }{d cos(\theta )}[/tex]
substituting values
[tex]D = \frac{ 2 }{ 1.0 *10^{-5} cos(6.315 )}[/tex]
[tex]D = 2.01220 *10^{5} \ rad/m[/tex]
Two parallel slits are illuminated with monochromatic light of wavelength 567 nm. An interference pattern is formed on a screen some distance from the slits, and the fourth dark band is located 1.83 cm from the central bright band on the screen. (a) What is the path length difference corresponding to the fourth dark band? (b) What is the distance on the screen between the central bright band and the first bright band on either side of the central band? (Hint: The angle to the fourth dark band and the angle to the first bright band are small enough that tan θ ≈ sin θ.)
Answer:
a)1984.5nm
b)523mm
Explanation:
A)A destructive interference can be explained as when the phase shifting between the waves is analysed by the path lenght difference
θ=(m+0.5)λ where m= 1,2.3....
Where given from the question the 4th dark Fringe which will take place at m= 3
θ=7/2y
Where y= 567nm
= 7/2(567)=1984.5nm
But
B)tan θ ≈ y/d
And sinθ = mλ/d
y=mλd when m= 1 which is the first bright we have
Then y=(1× 567.D)/d
But the distance from Central to the 4th dark Fringe is 1.83cm then
y= 7λD/2d= 1.83cm
D/d=(2)×(1.83×10^-2)/(7×567×10^-9)
=92221.5
y= (567×10^-9)× (92221.5)
=0.00523m
Therefore, the distance between the first and center is y1-y0= 523mm
Question 5 of 10
Heat is being transferred through currents within a liquid. When will this heat
transfer mostly end?
O A. When the substance changes state and becomes a gas
O B. When the entire liquid is a single temperature
O C. When the substance is very hot on top and cold beneath
O D. When the particles stop bumping into each other
SUBMIT
Answer:
When the entire liquid is a single temperature
Explanation:
When a liquid is heated, a convection current is set up. Convection is the movement of
fluid particles in response to a temperature gradient.
When you start heating a liquid, the particles near the base of the heating vessel increase in temperature, become less dense and rise upwards while the denser particles move downwards. This convection current will continue until an equilibrium temperature is obtained throughout the liquid.
Air flows through a converging-diverging nozzle/diffuser. A normal shock stands in the diverging section of the nozzle. Assuming isentropic flow, air as an ideal gas, and constant specific heats determine the state at several locations in the system. Solve using equations rather than with the tables.
Answer:
HELLO your question has some missing parts below are the missing parts
note: The specific heat ratio and gas constant for air are given as k=1.4 and R=0.287 kJ/kg-K respectively.
--Given Values--
Inlet Temperature: T1 (K) = 325
Inlet pressure: P1 (kPa) = 560
Inlet Velocity: V1 (m/s) = 97
Throat Area: A (cm^2) = 5.3
Pressure upstream of (before) shock: Px (kPa) = 207.2
Mach number at exit: M = 0.1
Answer: A) match number at inlet = 0.2683
B) stagnation temperature at inlet = 329.68 k
C) stagnation pressure = 588.73 kPa
D) ) Throat temperature = 274.73 k
Explanation:
Determining states at several locations in the system
A) match number at inlet
= V1 / C1 = 97/ 261.427 = 0.2683
C1 = sound velocity at inlet = [tex]\sqrt{K*R*T}[/tex] = [tex]\sqrt{1.4 *0.287*10^3}[/tex] = 361.427 m/s
v1 = inlet velocity = 97
B) stagnation temperature at inlet
= T1 + [tex]\frac{V1 ^2}{2Cp}[/tex] = 325 + [tex]\frac{97^2}{2 * 1.005*10^{-3} }[/tex]
stagnation temperature = 329.68 k
C) stagnation pressure
= [tex]p1 ( 1 + 0.2Ma^2 )^{3.5}[/tex]
Ma = match number at inlet = 0.2683
p1 = inlet pressure = 560
hence stagnation pressure = 588.73 kPa
D) Throat temperature
= [tex]\frac{Th}{T} = \frac{2}{k+1}[/tex]
Th = throat temperature
T = stagnation temp at inlet = 329.68 k
k = 1.4
make Th subject of the relation
Th = 329.68 * (2 / 2.4 ) = 274.73 k
Question 2.
In the US, lengths are often measured in inches, feet, yards and miles. Let's do
some conversions. The definition of the inch is: 1 inch = 25.4 mm, exactly. A foot is
12 inches and a mile is 5280 ft, exactly. A centimetre is exactly 0.01 m or 10 mm.
Sammy is 5 feet and 5.3 inches tall.
a). What is Sammy's height in Inches? (answer to 3 significant figures)
(3)
b). What is Sammy's height in Feet? (answer to 3 significant figures)
what is Sammy's hight in feet according to this statement
Explanation:
1 inch = 25.4 mm
1 foot = 12 inches
1 mile = 5260 feet
1 cm = 0.01 m or 10 mm
Now Sammy's height is 5 feet and 5.3 inches.
(a) We need to find Sammy's height in inches.
Since, 1 foot = 12 inches
5 feet = 5 × 12 inches = 60 inches
Now, 5 feet and 5.3 inches = 60 inches + 5.3 inches = 65.3 inches
Sammy's height is 65.3 inches.
(b) We need to find Sammy's height in feet.
Since, 1 foot = 12 inches
[tex]1\ \text{inch}=\dfrac{1}{12}\ \text{feet}[/tex]
So,
[tex]5.3\ \text{inch}=\dfrac{5.3}{12}\ \text{feet}=0.4416\ \text{feet}[/tex]
5 feet and 5.3 inches = 5 feet + 0.4416 feet = 5.44 feet
Sammy's height is 5.44 feet.
A hydraulic lift raises a 2 000-kg automobile when a 500-N force is applied to the smaller piston. If the smaller piston has an area of 10 cm2, what is the cross-sectional area of the larger piston
Answer:
The cross-sectional area of the larger piston is 392 cm²
Explanation:
Given;
output mass of the piston, m₀ = 2000 kg
input force of the piston, F₁ = 500 N
input area of the piston, A₁ = 10 cm² = 0.001 m²
The output force is given by;
F₀ = m₀g
F₀ = 2000 x 9.8
F₀ = 19600 N
The cross-sectional area of the larger piston or output area of the piston will be calculated by applying the following equations;
[tex]\frac{F_i}{A_i} = \frac{F_o}{A_o} \\\\A_o= \frac{F_o A_i}{F_i} \\\\A_o = \frac{19600*0.001}{500} \\\\A_o = 0.0392 \ m^2\\\\A_o = 392 \ cm^2[/tex]
Therefore, the cross-sectional area of the larger piston is 392 cm²
If a ray of light traveling in the liquid has an angle of incidence at the interface of 33.0 ∘, what angle does the refracted ray in the air make with the normal?
Answer:
29°
Explanation:
because the refracted ray angle is small than angle of incidence
Which statement belongs to Dalton’s atomic theory? Atoms have a massive, positively charged center. Atoms cannot be created or destroyed. Atoms can be broken down into smaller pieces. Electrons are located in energy levels outside of the nucleus.
Answer:
the correct statement is
* atoms cannot be created or destroyed
Explanation:
The Datlon atomic model was proposed in 1808 and represents atoms as the smallest indivisible particle of matter, they were the building blocks of matter and are represented by solid spheres.
Based on the previous descriptive, the correct statement is
* atoms cannot be created or destroyed
Answer:
the Answer is b hope it help
Explanation:
Can abnormality exist outside of a cultural context
A ball is thrown upward from a height of 432 feet above the ground, with an initial velocity of 96 feet per second. From physics it is known that the velocity at time t is v (t )equals 96 minus 32 t feet per second. a) Find s(t), the function giving the height of the ball at time t. b) How long will the ball take to reach the ground? c) How high will the ball go?
Answer;
A)S(t)=96t-16t² +432
B)it will take 9 seconds for the ball to reach the ground.
C)864feet
Explanation:
We were given an initial height of 432 feet.
And v(t)= 96-32t
A) we are to Find s(t), the function giving the height of the ball at time t
The position, or heigth, is the integrative of the velocity. So
S(t)= ∫(96-32)dt
S(t)=96t-16t² +K
S(t)=96t-16t² +432
In which the constant of integration K is the initial height, so K= 432
b) we need to know how long will the ball take to reach the ground
This is t when S(t)= 0
S(t)=96t-16t² +432
-16t² +96t +432=0
This is quadratic equation, if you solve using factorization method we have
t= -3 or t= 9
Therefore, , t is the instant of time and it must be a positive value.
So it will take 9 seconds for the ball to reach the ground.
C)V=s/t
Velocity= distance/ time
=96=s/9sec
S=96×9
=864feet
By applying the integrations,
(a) [tex]S = 96t-16t^2+432[/tex]
(b) Time will be "t = 9".
(c) Height will be "576"
Given:
Height,
423 feetInitial velocity,
96 feet/secAccording to the question,
(a)
Integrate v:
[tex]S = 96t-16t^2+C[/tex]Initial Condition,
→ [tex]S = 96t-16t^2+432[/tex]
(b)
Hits the ground when,
S = 0→ [tex]0=96t-16t^2+432[/tex]
→ [tex]t =9[/tex]
(c)
Maximum height when,
v = 0→ [tex]0 = 96-32 t[/tex]
→ [tex]t = 3[/tex]
Now,
→ [tex]S = 96\times 3-16\times 3^2+432[/tex]
[tex]= 576[/tex]
Thus the answer above is correct.
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PLEASE HELP WILL GIVE BRAINLIEST In an experiment, the hypothesis is that if leaf color is related to temperature, then exposing the plant to low temperatures will result in a leaf color change . This hypothesis is _____. 1. testable 2.falsifiable 3.a and b above 4.none of the above
Answer:
testable
Explanation:
high heat can cause browning and or welting
Answer:
A
Explanation:
If it is tested, it will really changed when the plant into low temperature
You are holding on to one end of a long string that is fastened to a rigid steel light pole. After producing a wave pulse that was 5 mm high and 4 em wide, you want to produce a pulse that is 4 cm wide but 7 mm high. You must move your hand up and down once,
a. a smaller distance up, but take a shorter time.
b. the same distance up as before, but take a shorter time.
c. a greater distance up, but take a longer time.
d. the same distance up as before, but take a longer time.
e. a greater distance up, but take the same time.
Answer:
It will take. the same distance up as before, but take a longer time
Warm blooded animals are homeothermic; that is, they maintain an approximately constant body temperature. (Forhumans it's about 37 oC.) When they are in an environment that is below their optimum temperature, they use energy derived from chemical reactions within their bodies to warm them up. One of the ways that animals lose energy to their environment is through radiation. Every object emits electromagnetic radiation that depends on its temperature. For very hot objects like the sun, that radiation is visible light. For cooler objects, like a house or a person, that radiation is in the infrared and is invisible. Nonetheless, it still carries energy. Other ways that energy is lost by a warm animal to a cool environment includes conduction (direct touching of a cooler object) and convection (cooler air moving and carrying thermal energy away). See Heat Transfer for a discussion of all three.
For this problem, we'll just consider how much energy an animal needs to burn (obtain from internal chemical reactions) in order to stay warm just from radiation losses. The rate at which an object loses energy through radiation is given by the Stefan-Boltzmann equation:
Rate of energy loss = AεσT4
where T is the absolute (Kelvin) temperature, A is the area of the object, ε is the emissivity (unitless and =1 for a perfect emitter, less for anything else), and σ is the Stefan-Boltzmann constant:
σ = 5.67 x 10-8 J/(s m2 K4)
Consider a patient trying to sleep naked in a cool room (55 oF = 13 oC). Assume that the person being considered is a perfect emitter and absorber of radiation (ε = 1), has a surface area of about 2.5 m2, and a mass of 80 kg.
a. A person emits thermal radiation at a rate corresponding to a temperature of 37 oC and absorbs radiation at a rate (from the air and walls) corresponding to a temperature of 13 oC. Calculate the individual's net rate of energy loss due to radiation (in Watts = Joules/second).
net rate of energy loss = Watts
b. Assume the patient produces no energy to keep warm. If they have a specific heat about equal to that of water (1 Cal/kg-oC) how much would their temperature fall in one hour? (1 Cal = 1kcal = 103 cal)
ΔT = oC
c. Given that the energy density of fat is about 9 Cal/g, how many grams of fat would the person have to utilize to maintain their body temperature in that environment for one hour?
amount of fat needed = g
Answer:
a) 360.7 J/s
b) 16.23 °C
c) 34.48 g
Explanation:
The mass of the person = 80 kg
The person is a perfect emitter, ε = 1
surface area of the person = 2.5 m^2
a) If he emits radiation at 37 °C, [tex]T_{out}[/tex] = 37 + 273 = 310 K
and receives radiation at 13 °C, [tex]T_{in}[/tex] = 13 + 273 = 286 K
Rate of energy loss E = Aεσ([tex]T^{4} _{out}[/tex] - [tex]T^{4} _{in}[/tex] )
where σ = 5.67 x 10^-8 J/(s m^2 K^4)
substituting values, we have
E = 2.5 x 1 x 5.67 x 10^-8 x ([tex]310^{4}[/tex] - [tex]286^{4}[/tex]) = 360.7 J/s
b) If they have specific heat about equal to that of water = 1 Cal/kg-°C
but 1 Cal = 1 kcal = 10^3 cal
specific heat of person is therefore = 10^3 cal/kg-°C
heat loss = 360.7 J/s = 360.7 x 3600 = 1298520 J/hr
heat lost in 1 hour = 1 x 1298520 = 1298520 J
This heat lost = mcΔT
where ΔT is the temperature fall
m is the mass
c is the specific heat equivalent to that of water
the specific heat is then = 10^3 cal/kg-°C
equating, we have
1298520 = 80 x 10^3 x ΔT
1298520 = 80000ΔT
ΔT = 1298520/80000 = 16.23 °C
c) 1298520 J = 1298520/4184 = 310.35 Cal
density of fat = 9 Cal/g
gram of fat = 310.35/9 = 34.48 g
All household circuits are wired in parallel. A 1140-W toaster, a 270-W blender, and a 80-W lamp are plugged into the same outlet. (The three devices are in parallel when plugged into the same outlet.) Assume that this is the standard household 120-V circuit with a 15-A fuse.
a. What current is drawn by each device?
b. To see if this combination will blow the 15-A fuse, find the total current used when all three appliances are on.
Answer:
total current = 12.417 A
so it will not fuse as current is less than 15 A
Explanation:
given data
toaster = 1140-W
blender = 270-W
lamp = 80-W
voltage = 120 V
solution
we know that current is express as
current = power ÷ voltage ......................1
here voltage is same in all three device
so
current by toaster is
I = [tex]\frac{1140}{120}[/tex]
I = 9.5 A
and
current by blender
I = [tex]\frac{270}{120}[/tex]
I = 2.25 A
and
current by lamp is
I = [tex]\frac{80}{120}[/tex]
I = 0.667 A
so here device in parallel so
total current is = 9.5 A + 2.25 A + 0.667 A
total current = 12.417 A
so it will not fuse as current is less than 15 A
A collector that has better efficiency in cold weather is the:
flat-plate collector due to reduced heat loss
evacuated tube collector due to its larger size
flat-plate collector due to the dark-colored coating
O evacuated tube collector due to reduced heat loss
Question 23 (1 point) Saved
One of the following is not found in Thermosyphon systems
o
Answer:
D. evacuated tube collector due to reduced heat loss
Explanation:
Evacuated tube collectors has vacuum which reduces the loss of heat and increase the efficiency of the collector. It has a major application in solar collector, and converts solar energy to heat energy. It can also be used for heating of a definite volume of water majorly for domestic purpose.
During cold weather, the conservation and efficient use of heat is required. Therefore, evacuated tube collector is preferred so as to reduce heat loss and ensure the maximum use of heat energy.
¿Cómo podrías utilizar el sistema de posicionamiento global para proponer recorridos alternativos para llegar a un lugar específico (centro educativo, supermercado, el hogar, el parque, entre otros)?
Answer:
El sistema de posicionamiento global (conocido mundialmente como GPS) podría utilizarse para proponer recorridos alternativos para llegar a un lugar específico, como un parque, a través de la creación de un recorrido guiado por una aplicación móvil con diferentes rutas de acceso al lugar.
Así, por ejemplo, se crearían diferentes rutas de acceso desde un punto A hasta un punto B, teniendo en cuenta factores como: rapidez, congestión vehicular, pago o no de peajes, posibilidad de acceso a pie y determinados factores extra que influyan en la forma de llegar al lugar. Todo ello plasmado en un mapa interactivo en el cual se señalen las rutas disponibles mediante el marcado del mapa en cuestión.
Specific heat is a measurement of the amount of heat energy input required for one gram of a substance to increase its temperature by one degree Celsius. Solid lithium has a specific heat of 3.5 J/g·°C. This means that one gram of lithium requires 3.5 J of heat to increase 1°C. Plot the temperature of 1g of lithium after 3.5, 7, and 10.5 J of thermal energy are added.
Answer:
ΔT = 1ºC , 2ºCand 3ºC
Explanation:
In this exercise they indicate the specific heat of lithium
let's calculate the temperature increase as a function of the heat introduced
Q = m [tex]c_{e}[/tex] ΔT
ΔT = Q / m c_{e}
calculate
for Q = 3.5 J
ΔT = 3.5 / (1 3.5)
ΔT = 1ºC
For Q = 7.0 J
ΔT = 7 / (1 3.5)
ΔT = 2ºC
for Q = 10.5 J
ΔD = 10.5 / (1 3.5)
ΔT = 3ºC
we see that this is a straight line, see attached
A diffraction grating 19.2 mm wide has 6010 rulings. Light of wavelength 337 nm is incident perpendicularly on the grating. What are the (a) largest, (b) second largest, and (c) third largest values of θ at which maxima appear on a distant viewing screen?
Answer:
(a). The largest value of θ is 71.9°.
(b). The second largest value of θ is 57.7°.
(c). The third largest value of θ is 47.7° .
Explanation:
Given that,
Width of diffraction grating [tex]w= 19.2\ mm[/tex]
Number of rulings[tex]N=6010[/tex]
Wavelength = 337 nm
We need to calculate the distance between adjacent rulings
Using formula of distance
[tex]d=\dfrac{w}{N}[/tex]
Put the value into the formula
[tex]d=\dfrac{19.2\times10^{-3}}{6010}[/tex]
[tex]d=3.19\times10^{-6}\ m[/tex]
We need to calculate the value of m
Using formula of constructive interference
[tex]d \sin\theta=m\lambda[/tex]
[tex]\sin\theta=\dfrac{m\lambda}{d}[/tex]
Here, m = 0,1,2,3,4......
[tex]\lambda[/tex]=wavelength
For largest value of θ
[tex]\dfrac{m\lambda}{d}>1[/tex]
[tex]m>\dfrac{d}{\lambda}[/tex]
Put the value into the formula
[tex]m>\dfrac{3.19\times10^{-6}}{337\times10^{-9}}[/tex]
[tex]m>9.46[/tex]
[tex]m = 9[/tex]
(a). We need to calculate the largest value of θ
Using formula of constructive interference
[tex]\theta=\sin^{-1}(\dfrac{m\lambda}{d})[/tex]
Now, put the value of m in to the formula
[tex]\theta=\sin^{-1}(\dfrac{9\times337\times10^{-9}}{3.19\times10^{-6}})[/tex]
[tex]\theta=71.9^{\circ}[/tex]
(b). We need to calculate the second largest value of θ
Using formula of constructive interference
[tex]\theta=\sin^{-1}(\dfrac{m\lambda}{d})[/tex]
Now, put the value of m in to the formula
[tex]\theta=\sin^{-1}(\dfrac{8\times337\times10^{-9}}{3.19\times10^{-6}})[/tex]
[tex]\theta=57.7^{\circ}[/tex]
(c). We need to calculate the third largest value of θ
Using formula of constructive interference
[tex]\theta=\sin^{-1}(\dfrac{m\lambda}{d})[/tex]
Now, put the value of m in to the formula
[tex]\theta=\sin^{-1}(\dfrac{7\times337\times10^{-9}}{3.19\times10^{-6}})[/tex]
[tex]\theta=47.7^{\circ}[/tex]
Hence, (a). The largest value of θ is 71.9°.
(b). The second largest value of θ is 57.7°.
(c). The third largest value of θ is 47.7° .
If 50 mL of each of the liquids in the answer choices were poured into a 250 mL beaker, which layer would be directly above a small rubber ball with a density of 0.960 g/mL? A. sea water – density of 1.024 g/mL B. mineral oil – density of 0.910 g/mL C. distilled water – density of 1.0 g/mL D. petroleum oil – density of 0.820 g/mL
Answer:
B. mineral oil – density of 0.910 g/mL.
Explanation:
Hello,
In this case, since the density is known as the degree of compactness a body has (mass in the occupied volume), the higher the density, the higher the weight of the body, therefore, if submerged into a liquid it could float if less dense than the liquid or sink if more dense than the liquid.
In such a way, since the rubber is more dense than mineral (0.960 g/mL > 0.910 g/mL) oil but less dense than distilled water (0.960 g/mL < 1.0 g/mL) we can say that B. mineral oil – density of 0.910 g/mL is directly above it when submerged.
Best regards.
During the spin cycle of your clothes washer, the tub rotates at a steady angular velocity of 31.7 rad/s. Find the angular displacement Δθ of the tub during a spin of 98.3 s, expressed both in radians and in revolutions.
Answer:
[tex]\Delta \theta = 3116.11\,rad[/tex] and [tex]\Delta \theta = 495.944\,rev[/tex]
Explanation:
The tub rotates at constant speed and the kinematic formula to describe the change in angular displacement ([tex]\Delta \theta[/tex]), measured in radians, is:
[tex]\Delta \theta = \omega \cdot \Delta t[/tex]
Where:
[tex]\omega[/tex] - Steady angular speed, measured in radians per second.
[tex]\Delta t[/tex] - Time, measured in seconds.
If [tex]\omega = 31.7\,\frac{rad}{s}[/tex] and [tex]\Delta t = 98.3\,s[/tex], then:
[tex]\Delta \theta = \left(31.7\,\frac{rad}{s} \right)\cdot (98.3\,s)[/tex]
[tex]\Delta \theta = 3116.11\,rad[/tex]
The change in angular displacement, measured in revolutions, is given by the following expression:
[tex]\Delta \theta = (3116.11\,rad)\cdot \left(\frac{1}{2\pi} \frac{rev}{rad} \right)[/tex]
[tex]\Delta \theta = 495.944\,rev[/tex]
A 384 Hz tuning fork produces standing waves with a wavelength of 0.90 m inside a resonance tube. The speed of sound at experimental conditions is
Answer:
v = 345.6m/s
Explanation:
v = 384 x 0.9 = 345.6
v = 345.6m/s
A double-convex thin lens is made of glass with an index of refraction of 1.52. The radii of curvature of the faces of the lens are 60 cm and 72 cm. What is the focal length of the lens
Answer:
63 cm
Explanation:
Mathematically;
The focal length of a double convex lens is given as;
1/f = (n-1)[1/R1 + 1/R2]
where n is the refractive index of the medium given as 1.52
R1 and R2 represents radius of curvature which are given as 60cm and 72cm respectively.
Plugging these values into the equation, we have:
1/f = (1.52-1)[1/60 + 1/72)
1/f = 0.0158
f = 1/0.0158
f = 63.29cm which is approximately 63cm
A foot is 12 inches and a mile is 5280I ft exactly. A centimeter is exactly 0.01m or mm. Sammy is 5 feet and 5.3tall. what is Sammy's height in inches?
Answer:
65.3 Inches tall
Explanation:
If Sammy is 5 feet and 5.3 inches tall, we simply need to convert the feet to inches, and sum the remaining inches from his height to determine his overall height in inches.
So, 5 feet = (12 inches/1foot) * (5 feet) = 60 inches
And 60 inches + 5.3 inches = 65.3 inches.
Hence, Sammy is 65.3 inches tall.
Cheers.
hi guys!!! i have no more points, can someone nice guess all of these for me? :)
1.What happens to the ocean water before the precipitation part of the water cycle
2.During which stage of the water cycle does water from the ocean form clouds?
3.what is a runoff??
4.Which statement about oceans is incorrect? A.Evaporation occurs when water is warmed by the sun. B.Most evaporation and precipitation occur over the ocean. C.97 percent of Earth's water is fresh water from the ocean. D.Water leaves the ocean by the process of evaporation
5.How does most ocean water return to the ocean in the water cycle
tysm to u who answers :)
1. The ocean water collects back in the ocean.
2. Condensation is the process by which water vapor in the air is changed into liquid water. Condensation is crucial to the water cycle because it is responsible for the formation of clouds.
3. an excessive amount of water flowing from downslope along earths surface
4. A.Evaporation occurs when water is warmed by the sun.
5. The water returns into the ocean by the water cycle . It evaporates , then it condensates , then it participates ( Rains ) and then goes back into the ocean.
Hope this answer correct ✌️
