Answer:
Explanation:
a) Fx = F cos (θ)
= (200) cos(60)
= 100 N
b) FR = ma
Fx + Ff = ma
100 + Ff = (50)(1,5)
Ff = 75 - 100
= -25 N
c) Fy = F sin θ
= (200) sin(60)
= 173,2 N
A toy car of mass 600g moves through 6m in 2 seconds.The average kinetic energy og the toy car is?
Explanation:
kE =1/2mv²
1/2(0.6×(3m/s)²
1/2(0.6×9m/s)
2.7J I think this is the answer
The average kinetic energy of the toy car is 2.7 J.
What is kinetic energy?
The energy of the body by the virtue of its motion is known as the kinetic energy of the body. It is defined as the product of half of mass and square of the velocity.
Given data;
Mass of car is,m= 600 g = 0.6 kg
d is the distance travelled = 6 m
T is the time travelled = 2 sec
The velocity of the car is found as;
v = d /t
V = 6m / 2 sec
V = 3 m/sec
KE =1/2mv²
KE = 1/2 × 0.6 kg ×( 3 m/sec )²
KE = 2.7 J
Hence, the average kinetic energy of the toy car is 2.7 J.
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5. Steve is driving in his car to take care of some errands. The first errand has him driving to a location 2 km East and 6 km North of his starting location. Once he completes that errand, he drives to the second one which is 4 km East and 2 km South of the first errand. What is the magnitude of the vector that describes how far the car has traveled from its starting point, rounded to the nearest km?
Answer:
gshshs
Explanation:
hshsksksksbsbbshd
calculate the pressure of water having density 1000 kilo per metre square at a depth of 20 m inside the water
Answer:
the pressure of the water at the given depth is 196,200 N/m².
Explanation:
Given;
density of the water, ρ = 1000 kg/m³
depth of the water, h = 20 m
acceleration due to gravity, g = 9.81 m/s²
The pressure at the given depth of the water is calculated as;
P = ρgh
P = 1000 x 9.81 x 20
P = 196,200 N/m²
Therefore, the pressure of the water at the given depth is 196,200 N/m².
what is the energy of an electromagnetic wave that has a frequency of 8.0 x 10^15 Hz? Use the equation...
(C)
Explanation:
[tex]E = hf = (6.626×10^{-34}\:\text{J•s})(8.0×10^{15}\:\text{Hz})[/tex]
[tex]= 5.3×10^{-18}\:\text{J}[/tex]
Answer:
It's D
Explanation:
It's from alvs
State the law of conservation of momentum
Explanation:
Conservation of momentum, general law of physics according to which the quantity called momentum that characterizes motion never changes in an isolated collection of objects; that is, the total momentum of a system remains constant
15- A racehorse coming out of the gate accelerates from rest to a velocity f 15.0 m/s due west in 1.80 s. What is its average acceleration?
Answer: (15 - 0)/1.8 = 8. 33m/s^2
Explanation:
The acceleration of the racehorse is 8.33 m/s²
The given parameters;
initial velocity of the racehorse, u = 0
final velocity of the racehorse, v = 15 m/s
time of motion of the horse, t = 1.8 s
The acceleration of the racehorse is calculated from change in velocity per change in time of motion as shown below;
[tex]a = \frac{\Delta v}{\Delta t} = \frac{v-u}{t} \\\\a = \frac{15 - 0}{1.8} \\\\a = 8.33 \ m/s^2[/tex]
Thus, the acceleration of the racehorse is 8.33 m/s²
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If at a particular instant and at a certain point in space the electric field is in the x-direction and has a magnitude of 3.10 V/mV/m , what is the magnitude of the magnetic field of the wave at this same point in space and instant in time
Answer:
B = 1.03 10⁻⁸ T
Explanation:
For an electromagnetic wave, the electric and magnetic fields must oscillate in phase so that they remain between them at all times, otherwise the wave will extinguish
This relational is expressed by the relation
E /B = c
B = E / c
let's calculate
B = 3.10 / 3 10⁸
B = 1.03 10⁻⁸ T
Light from two lasers is incident on an opaque barrier with a single slit of width 4.0 x 10^-4 m. One laser emits light of wavelength 480 nm and the other is 640 nm. A screen to view the light intensity pattern is 2.0 m behind the barrier. The distance from the center of the pattern to the nearest completely dark spot (dark for both colors) is ____ cm. (include 2 digits after the decimal point)
Answer:
a) y = 2.4 x 10⁻³ m = 0.24 cm
b) y = 3.2 x 10⁻³ m = 0.32 cm
Explanation:
The formula of Young's Double Slit experiment will be used here:
[tex]y = \frac{\lambda L}{d}\\\\[/tex]
where,
y = distance between dark spots = ?
λ = wavelength
L = distance of screen = 2 m
d = slit width = 4 x 10⁻⁴ m
a) FOR λ = 480 nm = 4.8 x 10⁻⁷ m:
[tex]y = \frac{(4.8\ x\ 10^{-7}\ m)(2\ m)}{4\ x\ 10^{-4}\ m}[/tex]
y = 2.4 x 10⁻³ m = 0.24 cm
a) FOR λ = 640 nm = 6.4 x 10⁻⁷ m:
[tex]y = \frac{(6.4\ x\ 10^{-7}\ m)(2\ m)}{4\ x\ 10^{-4}\ m}[/tex]
y = 3.2 x 10⁻³ m = 0.32 cm
A simple model of the human eye ignores its lens entirely. Most of what the eye does to light happens at the outer surface of the transparent cornea. Assume that this surface has a radius of curvature of 6.50 mm and that the eyeball contains just one fluid, with a refractive index of 1.41. Determine the distance from the cornea where a very distant object will be imaged.
Answer:
the distance from the cornea where a very distant object will be imaged is 23.35 mm
Explanation:
Given the data in the question;
For a spherical refracting surface;
[tex]n_i[/tex]/[tex]d_0[/tex] + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R
where [tex]n_i[/tex] is the index of refraction of the light of ray in the incident medium
[tex]d_0[/tex] is the object distance
[tex]n_t[/tex] is the index of refraction of light ray in the refracted medium
[tex]d_i[/tex] is the image distance
R is the radius of curvature
Now, let [tex]d_0[/tex] = ∞, such that;
[tex]n_i[/tex]/∞ + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R
0 + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R
we make [tex]d_i[/tex] subject of the formula
[tex]n_t[/tex]R = [tex]d_i[/tex]( [tex]n_t[/tex] - [tex]n_i[/tex] )
[tex]d_i[/tex] = ( [tex]n_t[/tex] × R ) / ( [tex]n_t[/tex] - [tex]n_i[/tex] )
given that; R = 6.50 mm, [tex]n_t[/tex] = 1.41, we know that [tex]n_i[/tex] = 1.00
so we substitute
[tex]d_i[/tex] = (1.41 × 6.50 mm ) / ( 1.41 - 1.00 )
[tex]d_i[/tex] = 9.165 / 0.41
[tex]d_i[/tex] = 23.35 mm
Therefore, the distance from the cornea where a very distant object will be imaged is 23.35 mm
Two round concentric metal wires lie on a tabletop, one inside the other. The inner wire has a diameter of 21.0 cm and carries a clockwise current of 16.0 A , as viewed from above, and the outer wire has a diameter of 32.0 cm.
Required:
a. What must be the direction (as viewed from above) of the current in the outer wire so that the net magnetic field due to this combination of wires is zero at the common center of the wires?
b. What must be the magnitude of the current in the outer wire so that the net magnetic field due to this combination of wires is zero at the common center of the wires?
Solution :
a). B at the center :
[tex]$=\frac{u\times I}{2R}$[/tex]
Here, one of the current is in the clockwise direction and therefore, the other current must be in the clockwise direction in order to cancel out the effect of the magnetic field that is produced by the other.
Therefore, the answer is ANTICLOCKWISE or COUNTERCLOCKWISE
b). Also, the sum of the fields must be zero.
Therefore,
[tex]$\left(\frac{u\times I_1}{2R_1}\right) + \left(\frac{u\times I_2}{2R_2}\right) = 0$[/tex]
So,
[tex]$\frac{I_1}{d_1}= \frac{I_2}{d_2}$[/tex]
[tex]$=\frac{16}{21}=\frac{I_2}{32}$[/tex]
[tex]$I_2=24.38 $[/tex] A
Therefore, the current in the outer wire is 24.38 ampere.
Answer:
(a) counter clockwise
(b) 24.38 A
Explanation:
inner diameter, d = 21 cm
inner radius, r = 10.5 cm
Current in inner loop, I = 16 A clock wise
Outer diameter, D = 32 cm
Outer radius, R = 16 cm
(a) The magnetic filed due to the inner wire is inwards to the plane of paper. According to the Maxwell's right hand thumb rule, the direction of magnetic field in outer wire should be outwards so that the net magnetic field is zero at the center.
So, the direction of current in outer wire is counter clock wise in direction.
(b) Let the current in outer wire is I'.
The magnetic field due to the inner wire is balanced by the magnetic field due to the outer wire.
[tex]\frac{ \mu 0}{4\pi}\times \frac{2 I}{r}=\frac{\mu 0}{4\pi}\times \frac{2 I'}{R}\\\frac{16}{10.5}=\frac{I'}{16}\\\\I' = 24.38 A[/tex]
What is the largest known star?
Answer:
UY Scuti is slightly larger than VY Canis Majoris
Explanation:
These stars are millions of miles away and cannot be seen by the naked eye.
Beetlejuice is another large star that can be seen by the eye.
What type of potential energy is a 9 volt battery an example of?
Gravitational potential energy
Elastic potential energy
Electrical potential energy
chemical potential energy
Answer:
chemical potential energy
Explanation:
A 9v battery comes in different formats, such that the most common one is the carbon-zinc and alkaline chemistry, so these are alkaline batteries (there are also rechargeable or lithium batteries, these also depend on chemical interactions).
These batteries "draw" the energy from chemical interactions of the materials inside of it, so the type of potential energy that is stored in a battery is actually chemical (regardless of the fact that the energy can be transformed into electrical energy later) the "potential" refers to how the energy is stored.
Then the correct option is chemical potential energy
Answer:
Chemical Potential Energy
Explanation:
Hope this helps!!
Have a blessed day/night!! <33
herical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.010-3 kg/m/s. The diffusion coefficient of the candy solute in water is 2.010-9 m2/s, and the solubility of the candy solute in water is 2.0 kg/m3. Calculate the mass tran
Answer: Below is the complete question
A spherical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.0x10-3 kg/m/s. The diffusion coefficient of the candy solute in water is 2.0x10-9 m2/s, and the solubility of the candy solute in water is 2.0 kg/m3. Calculate the mass transfer coefficient (m/s)
answer:
mass transfer coefficient = 9.56 * 10^-5 m/s
Explanation:
Candy density = 1950 kg/m^3
Candy diameter = 1 cm
Velocity of water = 1 m/s
water density = 1000 kg/m^3
Viscosity of water = 1 * 10^-3 kg/m/s
diffusion coefficient of candy in water = 2 * 10^-9 m^2/s
solubility of candy = 2 kg/m^3
Determine the mass transfer coefficient ( m/s )
( Sh) mass transfer coefficient ( flow across sphere ) = 2 + 0.6Re^1/2 * SC^1/3
where : Re = vdp / μ , Sh = KLd / Deff
attached below is the remaining solution .
mass transfer coefficient = 9.56 * 10^-5 m/s
A man is pulling a 20 kg box with a rope that makes an angle of 60 with the horizontal.If he applies a force of 150 N and a frictional force of 15 N is present, calculate the acceleration of the box.
∑ F (horizontal) = (150 N) cos(60°) - 15 N = (20 kg) a
==> a = ((150 N) cos(60°) - 15 N)/(20 kg) = 3 m/s²
To calculate the acceleration of the box, we need to consider the net force acting on it. So, the acceleration of the box is 3 m/s².
The net force is the vector sum of the applied force and the force of friction. First, let's find the horizontal and vertical components of the applied force:
Horizontal component of the applied force (F[tex]_{horizontal}[/tex]) = F[tex]_{applied}[/tex] × cos(θ)
F[tex]_{horizontal}[/tex] = 150 N × cos(60°)
F[tex]_{horizontal}[/tex] = 150 N × 0.5
F[tex]_{horizontal}[/tex] = 75 N
Vertical component of the applied force (F[tex]_{vertical}[/tex]) = F[tex]_{applied}[/tex] × sin(θ)
F[tex]_{vertical}[/tex] = 150 N × sin(60°)
F[tex]_{vertical}[/tex] = 150 N × (√3 / 2)
F[tex]_{vertical}[/tex] ≈ 129.9 N
Now, let's calculate the net force in the horizontal direction:
Net Force in the horizontal direction (F[tex]_{net horizontal}[/tex]) = F[tex]_{horizontal}[/tex] - F[tex]_{friction}[/tex]
F[tex]_{net horizontal}[/tex] = 75 N - 15 N
F[tex]_{net horizontal}[/tex] = 60 N
Now, we can calculate the acceleration (a) using Newton's second law of motion, F = ma:
F[tex]_{net horizontal}[/tex] = m × a
60 N = 20 kg × a
Now, solve for acceleration (a):
a = 60 N / 20 kg
a = 3 m/s²
So, the acceleration of the box is 3 m/s².
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1. Una pelota rueda hacia la derecha siguiendo una trayectoria en línea recta de modo que recorre una distancia de 10m en 5 s , después cambia su trayectoria cuando es lanzada hacia arriba 25m durante 7 s. Calcular la velocidad y la rapidez al punto final (altura maxima) al que llegó la pelota.
2. Una mariposa vuela en línea recta hacia el sur recorriendo una distancia de 15 m durante 28 s, después cambia de dirección hacia el Oeste recorriendo una distancia de 50 m en un tempo de 80 s ¿cuál es la velocidad y rapidez de la mariposa?
3.- Una persona camina durante 21 minutos hacia el este de su casa una distancia de 1500 m y después cambia su dirección hacia el Norte recorriendo una distancia de 3350 m en un tiempo 32 minutos llegando al supermercado. ¿Calcula la velocidad y rapidez de la persona?
4.- Un automóvil se mueve al Oeste recorriendo una distancia de 80 km en 1.2 horas, posteriormente cambia su trayectoria hacia el Sur, recorriendo una distancia de 120 km en un tiempo 1.6 hora. ¿Calcula la velocidad y rapidez del automóvil?
Answer:
https://youtu.be/ymHHdoCGJOU
Observe: Air pressure is equal to the weight of a column of air on a particular location. Air pressure is measured in hectopascals (hPa). Note how the air pressure changes as you move Station B towards the center of the high-pressure system.
a. What do you notice?
b. Why do you think this is called a high-pressure system?
Answer:
A. When moving towards a high pressure center the pressure values increase in the equipment
B. This area is called high prison since the weight of the atmosphere on top is maximum
Explanation:
A) A high atmospheric pressure system is an area where the pressure is increasing the maximum value is close to 107 Kpa, the other side as low pressure can have small values 85.5 kPa.
When moving towards a high pressure center the pressure values increase in the equipment
B) This area is called high prison since the weight of the atmosphere on top is maximum
in general they are areas of good weather
A scientist who studies insects, spiders, snails, and other bugs of an environment .
Botanist
Chemist
Ecologist
Entomologist
Question:- A scientist who studies insects, spiders, snails, and other bugs of an environment
Answer:- EntomologistExplanation:-
Entomologist word comes from two words Entomon and biologist
Entomon which means insectbiologist which means the person who study living formsFind the intensity of the electromagnetic wave described in each case. (a) an electromagnetic wave with a wavelength of 655 nm and a peak electric field magnitude of 1.5 V/m. 0.002984 W/m2 (b) an electromagnetic wave with an angular frequency of 6.5 ✕ 1018 rad/s and a peak magnetic field magnitude of 10−10 T. 1.19366E-6 W/m2
The intensity of the electromagnetic wave in terms of the electric field is 0.00298 W/m² and the intensity of the electromagnetic wave in terms of the magnetic field is 1.193x10⁻⁶ W/m².
The intensity of the electromagnetic wave is related to the electric field as well as to the magnetic field.
a) Intensity of the electromagnetic wave for the electromagnetic field.
The intensity of the electromagnetic wave (I) in terms of the electromagnetic field is given by:
[tex] I = \frac{E^{2}*c*\epsilon_{0}}{2} [/tex] (1)
Where:
c: is the speed of light = 3.00*10⁸ m/s
E: is the magnitude of the electric field = 1.5 V/m
ε₀: is the permittivity of free space = 8.85*10⁻¹² C²/Nm²
Hence, the intensity of the electromagnetic wave (eq 1) is:
[tex] I = \frac{(1.5 V/m)^{2}*3.00 \cdot 10^{8} m/s*8.85 \cdot 10^{-12} C^{2}/(N*m^{2})}{2} = 0.00298 W/m^{2} [/tex]
b) Intensity of the electromagnetic wave for the magnetic field
We can calculate the intensity of the electromagnetic wave (I) in terms of the magnetic field with the following equation:
[tex] I = \frac{cB^{2}}{2\mu_{0}} [/tex] (2)
Where:
B: is the magnitude of the magnetic field = 10⁻¹⁰ T
μ₀: is the vacuum permeability = 4π*10⁻⁷ m*T/A
Therefore, the intensity of the electromagnetic wave (eq 2) is:
[tex] I = \frac{3.00 \cdot 10^{8} m/s*(1\cdot 10^{-10} m*T/A)^{2}}{2*4\pi \cdot 10^{-7} T/A} = 1.193 \cdot 10^{-6} W/m^{2} [/tex]
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A physics student likes to study while listening to loud music. If electricity costs 12.00$/kWh (kilowatt-hour), how much would it cost the student to run a 220 W stereo system 8.0 hours per day for 10 days of studying?
Answer:
the cost of running the stereo is $211.2
Explanation:
Given;
cost of electricity, = 12.00$/kWh
power consumed by the stereo system, P = 220 W
duration of the power consumption, t = 8 hours
number of days, = 10 days
total time of the power consumption = 8 hours x 10 = 80 hours
power consumed in kW = 220 W / 1000 = 0.22 kW
Energy consumed = 0.22 kW x 80 h = 17.6 kWh
The cost of using 17.6 kWh
= 17.6 x $12
= $211.2
Therefore, the cost of running the stereo is $211.2
which watch is more preferable for the measurement of time among pendulum, quartz and atomic watch
Answer:
pendulum, quartz
Explanation:
What is cubical expansivity of liquid while freezing
Answer:
"the ratio of increase in the volume of a solid per degree rise of temperature to its initial volume" -web
Explanation:
tbh up above ✅
Answer:
cubic meter
Explanation:
Increase in volume of a body on heating is referred to as volumetric expansion or cubical expansion
Which of the following groups is the largest ?
population
community
ecosystem
biome
Answer:
B. Community
Took science classes for 6 years now
Which of the following is not true about Triton, the large moon of Neptune? It is more reflective than Earth's Moon. It is larger than Earth's Moon. It is in a retrograde orbit. It has a thin atmosphere. It has nitrogen geysers.
Answer:
Triton is the largest of Neptune's 13 moons. It is unusual because it is the only large moon in our solar system that orbits in the opposite direction of its planet's rotation―a retrograde orbit. ... Like our own moon, Triton is locked in synchronous rotation with Neptune―one side faces the planet at all times.
How can I solve this?
You have three capacitors of values 40 F, 10 F and 50 F. What would their equivalent capacitance (in F) be if they were connected in parallel with each other? Enter your answer as a number only, to one decimal place.
Explanation:
The equivalent capacitance of capacitors in parallel can be determined as
[tex]C_{eq} = C_1 + C_2 + C_3[/tex]
[tex]\:\:\:\:\:= 40\:\text{F} + 10\:\text{F} + 50\:\text{F} = 100\:\text{F}[/tex]
what are three effects of gravity
Answer:
effect on motation.effect on direction
A pendulum is constructed from a 6 kg mass attached to a strong cord of length 1.7 m also attached to a ceiling. Originally hanging vertically, the mass is pulled aside a small distance of 7.6 cm and released from rest. While the mass is swinging the cord exerts an almost-constant force on it. For this problem, assume the force is constant as the mass swings. How much work in J does the cord do to the mass as the mass swings a distance of 8.0 cm
Answer:
work done is -2.8 × 10⁻⁶ J
Explanation:
Given the data in the question;
mass of the pendulum m = 6 kg
Length of core = 1.7 m
Now, case1, mass is pulled aside a small distance of 7.6 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.
so, θ₁ = ( 7.6 × 10⁻² m / 1.7 m ) = 0.045 rad
In case2, mass is pulled aside a small distance of 8 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.
so, θ₂ = ( 8 × 10⁻² m / 1.7 m ) = 0.047 rad.
Now, the required work done will be;
[tex]W = \int\limits^{\theta_2} _{\theta_1} {r} \, d\theta[/tex]
[tex]W = \int\limits^{\theta_2} _{\theta_1} {-mgl sin\theta } \, d\theta[/tex]
[tex]W = -mgl \int\limits^{0.047 } _{0.045 } {sin\theta } \, d\theta[/tex]
W = [tex]-mgl[[/tex] -cosθ [tex]]^{0.047}_{0.045 }[/tex]
W = 6 × 9.8 × 1.7 × [ cos( 0.047 ) - cos( 0.045 ) ]
W = 6 × 9.8 × 1.7 × [ -2.8 × 10⁻⁸ ]
W = -2.8 × 10⁻⁶ J
Therefore, work done is -2.8 × 10⁻⁶ J
.
A mass of 8.72 kg gains 446 J of gravitational potential energy. To what height was it lifted?
Answer:
[tex]\boxed {\boxed {\sf 5.22 \ m}}[/tex]
Explanation:
Gravitational potential energy is the energy an object possesses due to its position. It is calculated using the following formula:
[tex]E_P=mgh[/tex]
Where m is the mass, g is the acceleration due to gravity, and h is the height.
The object has a mass of 8.72 kilograms. Assuming this occurs on Earth, the acceleration due to gravity is 9.8 meters per second squared. The object gains 446 Joules of potential energy.
Let's convert the units of Joules. This makes the process of canceling units simpler later on. 1 Joule is equal to 1 kilogram meter squared per second squared. The object gains 446 J, which is equal to 446 kg *m²/s².
EP= 446 kg*m²/s²m= 8.72 kg g= 9.8 m/s²Substitute the values into the formula.
[tex]446 \ kg*m^2/s^2 = 8.72 \ kg * 9.8 \ m/s^2 *h[/tex]
Multiply on the right side of the equation.
[tex]446 \ kg*m^2/s^2 = 85.456 kg*m/s^2 *h[/tex]
We are solving for the height, so we must isolate the variable h. It is being multiplied by 85.456 kg*m/s². The inverse operation of multiplication is division, so we divide both sides by this value.
[tex]\frac{ 446 \ kg*m^2/s^2}{85.456 kg*m/s^2} = \frac{85.456 kg*m/s^2 *h}{85.456 kg*m/s^2}[/tex]
[tex]\frac{ 446 \ kg*m^2/s^2}{85.456 kg*m/s^2} =h[/tex]
The units of kg*m/s² cancel, leaving meters as our unit.
[tex]\frac{ 446 }{85.456 } \ m =h[/tex]
[tex]5.2190601011 \ m =h[/tex]
The original measurements of mass and potential energy have 3 significant figures, so our answer must have the same.
For the number we calculated, that is the hundredths place. The 9 in the thousandths place to the right tells us to round the 1 up to a 2.
[tex]5.22 \ m \approx h[/tex]
The object was lifted to a height of approximately 5.22 meters.
A car starting at rest accelerates at 3m/s² How far has the car travelled after 4s?
Answer:
24m
Explanation:
you can use the formula
s=ut+1/2at²
s=0+1/2(3)(4)²
=1/2(3)(8)
=24m
I hope this helps
Assume that the far point of a myopic (nearsighted) eye is 5.04 m in front of the eye. A lens is used to correct the vision, such that it can focus sharply an object at infinity. What is the power of the lens (in diopters; answer sign and magnitude)?
Answer:
[tex]P=-0.2D[/tex]
Explanation:
From the question we are told that:
Far-point [tex]v=-5.04m[/tex]
Where
u=-\alpha
Generally the equation for Lens is mathematically given by
[tex]\frac{1}{f}=\frac{1}{v}-\frac{1}{u}[/tex]
[tex]P=\frac{1}{-5.04}-\frac{1}{\alpha}[/tex]
[tex]P=-0.2D[/tex]
.. Solve: 91
Find the half angular width of the central bright maximum in the Fraunhofer diffraction pattern of
a slit of width 12x10^-5 cm when the slit illuminated by monochromatic light of wave length
6000 A
[KUET’10-11)
(a) 30°
(b) 60°
(c) 15°
(d) None of these
Solution
Explanation:
bro I have no idea fam......
