Answer:
you will stc ohxoyxct txxtx xigigjjgjvvixiffjz,iffzikzfjvixii. hi h ohigiogooigoh
Explanation:
k jjvhvojvovvojivivivihvi hj
A steel bridge is 1000 m long at -20°C in winter. What is the change in length when the temperature rises to 40°C in summer? The average coefficient of linear expansion of this steel is 11 × 10-6 C-1.
Answer:
ΔL = 0.66 m
Explanation:
The change in length on an object due to rise in temperature is given by the following equation of linear thermal expansion:
ΔL = αLΔT
where,
ΔL = Change in Length of the bridge = ?
α = Coefficient of linear thermal expansion = 11 x 10⁻⁶ °C⁻¹
L = Original Length of the Bridge = 1000 m
ΔT = Change in Temperature = Final Temperature - Initial Temperature
ΔT = 40°C - (-20°C) = 60°C
Therefore,
ΔL = (11 x 10⁻⁶ °C⁻¹)(1000 m)(60°C)
ΔL = 0.66 m
Consider two parallel wires where the magnitude of the left currentis 2 I0(io) and that of the right current is I0(io). Point A is midway between the wires,and B is an equal distance on the other side of the wires.
The ratio ofthe magnitude of the magnetic field at point A to that at point Bis________
Answer:
Explanation:
At the point midway between wires
magnetic field due to wire having current 2I₀
= 10⁻⁷ x 2 x2I₀ / r where 2r is the distance between wires .
magnetic field due to wire having current I₀
= 10⁻⁷ x 4 I₀ / r
magnetic field due to wire having current I₀
= 10⁻⁷ x 2I₀ / r
= 10⁻⁷ x 2 I₀ / r where 2r is the distance between wires .
these fields are in opposite direction as direction of current is same in both .
net magnetic field = (4 - 2 )x 10⁻⁷ x I₀ / r
= 2 x 10⁻⁷ x I₀ / r
At point A net magnetic field = 2 x 10⁻⁷ x I₀ / r
At point B , we shall calculate magnetic field
magnetic field due to nearer wire having current 2 I₀ = 10⁻⁷ x 4 I₀ / r
magnetic field due to wire far away = 10⁻⁷ x 2 I₀ / 3r
These magnetic fields act in the same direction so they will add up
net magnetic field = [ (4 I₀ / r) + (2 I₀ / 3r) ] x 10⁻⁷
= (14 I₀ / 3r ) x 10⁻⁷
Magnetic field at point B = (14 I₀ / 3r ) x 10⁻⁷
Ratio of field at A and B
= 3 / 7 . Ans
The ratio of the magnitude of the magnetic field at point A to point B is :
3 / 7
Given data :
Magnitude of the left current is 2I₀
Magnitude of the right current is I₀
First step : Determine the magnetic field at point A
The magnetic field due to the left current ( 2I₀ )
10⁻⁷ * 2 * 2I₀ / r ( 2r = distance between wires )
The magnetic field due to the right current ( I₀ )
10⁻⁷ * 2 I₀ / r
From the expressions above the magnetic fields are in opposite direction
∴ Net magnetic field = (4 - 2 )* 10⁻⁷ * I₀ / r = 2 * 10⁻⁷ * I₀ / r
Hence The magnetic field at point A = 2 * 10⁻⁷ * I₀ / r
Next step : determine the magnetic field at point B
Magnetic field due to the closest wire to point B ( i.e.2I₀ ) = 10⁻⁷ * 4 I₀ / r
Magnetic field due to the wire away from point A = 10⁻⁷ * 2 I₀ / 3r
Since the fields acts in the same directions
The net magnetic field = (4 I₀ / r) + (2 I₀ / 3r) ] * 10⁻⁷ = ( 14 I₀ / 3r ) * 10⁻⁷
Hence The magnetic field at point A = ( 14 I₀ / 3r ) * 10⁻⁷
Therefore the ratio of the magnitude of the magnetic field at point A to point B = 3/ 7
Hence we can conclude that the ratio of the magnitude of the magnetic field at point A to point B = 3 / 7
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A train is approaching you at very high speed as you stand next to the tracks. Just as an observer on the train passes you, you both begin to play the same recorded version of a Beethoven symphony on identical MP3 players. (a) According to you, whose MP3 player finishes the symphony first?
A. your player,
B. the observer's player,
C. both finish at the same time. (b) According to the observer on the train, whose MP3 player finishes the symphony first?
A. your player,
B. the observer's player,
C. both finish at the same time. (c) Whose MP3 player actually finishes the symphony first?
A. your player,
B. the observer's player,
C. each observer measures his symphony as finishing first,
D. each observer measures the other's symphony as finishing first.
Answer:
a) Your player
b) Observer's player
c) Each measures their own first
Explanation:
Because given problem is having relative velocity to each other. The person sitting on the train is moving with a very high speed relative to the person standing next to the track.
In this case, the clock situated in the train will be running slow with respect to the stationary frame of reference
A goldfish bowl is spherical, 8.0 cm in radius. A goldfish is swimming 3.0 cm from the wall of the bowl. Where does the fish appear to be to an observer outside? The index of refraction of water is 1.33. Neglect the effect of the glass wall of the bowl.
Answer:
41.5 cm to the left of the observer
Explanation:
See attached file
If the distance from your eye's lens to the retina is shorter than for a normal eye, you will struggle to see objects that are
Answer:
far away
Explanation:
There are different types of eye defect ranging from short sightedness, longsighted, astigmatism, presbyopia etc.
If someone is only able to see close ranged object clearly but not far distant object, then such person is suffering from short sightedness or myopia. This occurs when the light rays entering the eye does not converge on the retina. Instead of converging on the retina, the light ray is formed on a point in front of the retina. This causes the distance from the eye's lens to the retina shorter compared to that of a normal eye. This eye defect is usually corrected using concave lens in order to diverge the rays thereby allowing it to focus on the retina.
Hence, if the distance from your eye's lens to the retina is shorter than for a normal eye, you will struggle to see objects that are far away (at a far distant).
A particle moves along line segments from the origin to the points (1, 0, 0), (1, 5, 1), (0, 5, 1), and back to the origin under the influence of the force field. F(x, y, z)= z^2i + 4xyj + 5y^2kFind the work done.
Answer:
0 J
Explanation:
Since work done W = ∫F.dr and F(x, y, z)= z²i + 4xyj + 5y²k and dr = dxi + dyj + dzk
F.dr = (z²i + 4xyj + 5y²k).(dxi + dyj + dzk) = z²dx + 4xydy + 5y²dz
W = ∫F.dr = ∫z²dx + 4xydy + 5y²dz = z²x + 2xy² + 5y²z
We now evaluate the work done for the different regions
W₁ = work done from (0,0,0) to (1,0,0)
W₁ = {z²x + 2xy² + 5y²z}₀₀₀¹⁰⁰ = 0²(1) + 2(1)(0)² + 5(0)²(0) - [(0)²(0) + 2(0)(0)² + 5(0)²(0)] = 0 - 0 = 0 J
W₂ = work done from (1,0,0) to (1,5,1)
W₂ = {z²x + 2xy² + 5y²z}₁₀₀¹⁵¹ = (1)²(1) + 2(1)(5)² + 5(5)²(1) - [0²(1) + 2(1)(0)² + 5(0)²(0)] = 1 + 50 + 125 - 0 = 176 J
W₃ = work done from (1,5,1) to (0,5,1)
W₃ = {z²x + 2xy² + 5y²z}₁₅₁⁰⁵¹ = 1²(0) + 2(0)(5)² + 5(5)²(1) - [(1)²(1) + 2(1)(5)² + 5(5)²(1)] = 125 - (1 + 50 + 125) = 125 - 176 = -51 J
W₄ = work done from (0,5,1) to (0,0,0)
W₄ = {z²x + 2xy² + 5y²z}₁₅₁⁰⁰⁰ = (0)²(0) + 2(0)(0)² + 5(0)²(0) - [1²(0) + 2(0)(5)² + 5(5)²(1)] = 0 - 125 = -125 J
The total work done W is thus
W = W₁ + W₂ + W₃ + W₄
W = 0 J + 176 J - 51 J - 125 J
W = 176 J - 176 J
W = 0 J
The total work done equals 0 J
In Young's experiment a mixture of orange light (611 nm) and blue light (471 nm) shines on the double slit. The centers of the first-order bright blue fringes lie at the outer edges of a screen that is located 0.497 m away from the slits. However, the first-order bright orange fringes fall off the screen. By how much and in what direction (toward or away from the slits) should the screen be moved, so that the centers of the first-order bright orange fringes just appear on the screen
Answer:
0.5639m
Explanation:
For a young double slit experiment the expression below gives the angular separation for m dark fringe having slit width d and wavelength λ
=sin⁻¹(mλ/d)
mλ /d =y/L
for the first order,
y= mλL/d
For ratio separation y₀/yD=1 and d= 1
y₀/yD= [mλ ₀L₀/d]/[mλD.LD./d]
1=λ ₀L₀/λD.LD.
λD.LD= λ ₀L₀
L₀= λD.LD/ λ ₀..............(1)
Then substitute the given values into (1) we have
L₀=471 *0.497/611
= 0.3831m
Distance by which the screen has to be moved towards the slit is
LD- Lo
0.947-0.3831= 0.5639m
If the intensity of an electromagnetic wave is 80 MW/m2, what is the amplitude of the magnetic field of this wave
CHECK THE COMPLETE QUESTION BELOW;
the intensity of an electromagnetic wave is 80 MW/m2, what is the amplitude of the magnetic field of this wave? (c=3.0×108m/s, μ0=4π×10−7T⋅m/A, ε0=8.85×10−12C2/N⋅m2)
Answer:
2.4×10^5 N/C
Explanation:
the amplitude can be explained as the maximum field strength of the electric and magnetic fields. Wave energy is proportional to its amplitude squared (E2 or B2).
We were told to calculate the amplitude of the magnetic field, which can be done using expresion below
S=ε²/2cμ
Where S is the intensity intensity of an electromagnetic wave given as 80 MW/m2
ε² is the Amplitude which we are looking for
c= speed of light given as 3×10^8m/s
Substitute the values into above formula we have,
S=ε²/2cμ
Making Amplitude subject of formula
ε²=S×2cμ
ε²=[80×10^6)(2×3×10^8)(4Π×10^-7)
= 245598.44
ε²=2.4×10^5 N/C
Therefore, amplitude of the magnetic field of this wave is S=2.4×10^5 N/C
In state-of-the-art vacuum systems, pressures as low as 1.00 10-9 Pa are being attained. Calculate the number of molecules in a 1.90-m3 vessel at this pressure and a temperature of 28.0°C. molecules
Answer:
The number of molecules is 4.574 x 10¹¹ Molecules
Explanation:
Given;
pressure in the vacuum system, P = 1 x 10⁻⁹ Pa
volume of the vessel, V = 1.9 m³
temperature of the system, T = 28°C = 301 K
Apply ideal gas law;
[tex]PV= nRT = NK_BT[/tex]
Where;
n is the number of gas moles
R is ideal gas constant = 8.314 J / mol.K
[tex]K_B[/tex] is Boltzmann's constant, = 1.38 x 10⁻²³ J/K
N is number of gas molecules
N = (PV) / ([tex]K_B[/tex]T)
N = (1 x 10⁻⁹ X 1.9) / ( 1.38 x 10⁻²³ X 301)
N = 4.574 x 10¹¹ Molecules
Therefore, the number of molecules is 4.574 x 10¹¹ Molecules
An undiscovered planet, many light-years from Earth, has one moon, which has a nearly circular periodic orbit. If the distance from the center of the moon to the surface of the planet is 2.165×105 km and the planet has a radius of 4175 km and a mass of 6.70×1022 kg , how long (in days) does it take the moon to make one revolution around the planet? The gravitational constant is 6.67×10−11N·m2/kg2 .
Answer:
364days
Explanation:
Pls see attached file
Explanation:
The moon will take 112.7 days to make one revolution around the planet.
What is Kepler's third law?The period of the satellite around any planet only depends upon the distance between the planet's center and satellite and also depends upon the planet's mass.
Given, the distance from the moon's center to the planet's surface,
h = 2.165 × 10⁵ km,
The radius of the planet, r = 4175 km
The mass of the planet = 6.70 × 10²² kg
The total distance between the moon's center to the planet's center:
a = r +h = 2.165 × 10⁵ + 4175
a = 216500 + 4175
a = 220675
a = 2.26750 × 10⁸ m
The period of the planet can be calculated as:
[tex]T =2\pi \sqrt{\frac{a^3}{Gm} }[/tex]
[tex]T =2\3\times 3.14 \sqrt{\frac{(2.20675 \times 10^8)^3}{(6.67\times 10^{-11}).(6.70\times 10^{22})} }[/tex]
T = 9738253.26 s
T = 112.7 days
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You spaceship has a snazzy lounge called Ten-Forward that needs a new spacecouch. Sadly the couch you bought is too long and won't fit into your car when it is stationary. The couch has a length Lc = 14.0 m and mass m = 49 kg, and your shuttlecraft has a length of L = 9.0 m How fast would you need to run, in terms of the speed of light c, in order to get the couch to fit inside the length of the shuttlecraft?
Answer:
v = speed
c = speed of light
using the equation
L = Lc Sqrt( 1 - (v/c)2)
9 = 14 Sqrt( 1 - (v/c)2)
v/c = 0.765
v = 0.765 c
g Calculate the maximum wavelength of light that will cause the photoelectric effect for potassium. Potassium has work function 2.29 eV = 3.67 x 10–19 J.
Answer:
λ = 5.4196 10⁻⁷m, λ = 541.96 nm this is green ligh
Explanation:
The photoelectric effect was explained by Eintein assuming that the light was made up of particles called photons and these collided with the electrons taking them out of the material.
K = h f -Ф
where K is the kinetic energy of the ejected electrons, hf is the energy of the light quanta and fi is the work function of the material.
The speed of light is related to wavelength and frequency
c = λ / f
f = c /λ
we substitute
K = h c / λ - Φ
for the case that they ask us the kinetic energy of the electons is zero (K = 0)
h c / λ = Ф
λ = h c / Ф
we calculate
λ = 6.63 10⁻³⁴ 3 10⁸ / 3.67 10⁻¹⁸
λ = 5.4196 10⁻⁷m
let's take nm
lam = 541.96 nm
this is green light
If a wave has speed of 235 m/s with a wavelength of 3 m, what is the frequency of the wave?
A mechanic wants to unscrew some bolts. She has two wrenches available: one is 35 cm long, and one is 50 cm long. Which wrench makes her job easier and why?
Answer:
50 cm long
When 35cm long wrench is compared to 50cm long wrench, we find that the 50cm long wrench produces more turning effect of force because it has longer distance between fulcrum and line of action of force. At conclusion, the more the turning effect of force the more it is easy to unscrew bolts.
1.An elevator is ascending with constant speed of 10 m/s. A boy in the elevator throws a ball upward at 20 m/ a from a height of 2 m above the elevator floor when the elevator floor when the elevator is 28 m above the ground.
a. What's the maximum height?
b. How long does it take for the ball to return to the elevator floor?
(a) The maximum height reached by the ball from the ground level is 75.87m
(b) The time taken for the ball to return to the elevator floor is 2.21 s
The given parameters include:
constant velocity of the elevator, u₁ = 10 m/sinitial velocity of the ball, u₂ = 20 m/sheight of the boy above the elevator floor, h₁ = 2 mheight of the elevator above the ground, h₂ = 28 mTo calculate:
(a) the maximum height of the projectile
total initial velocity of the projectile = 10 m/s + 20 m/s = 30 m/s (since the elevator is ascending at a constant speed)
at maximum height the final velocity of the projectile (ball), v = 0
Apply the following kinematic equation to determine the maximum height of the projectile.
[tex]v^2 = u^2 + 2(-g)h_3\\\\where;\\\\g \ is \ the \ acceleration \ due \ to\ gravity = 9.81 \ m/s^2\\\\h_3 \ is \ maximum \ height \ reached \ by \ the \ ball \ from \ the \ point \ of \ projection\\\\0 = u^2 -2gh_3\\\\2gh_3 = u^2 \\\\h_3 = \frac{u^2}{2g} \\\\h_3 = \frac{(30)^2}{2\times 9.81} \\\\h_3 = 45.87 \ m[/tex]
The maximum height reached by the ball from the ground level (h) = height of the elevator from the ground level + height of he boy above the elevator + maximum height reached by elevator from the point of projection
h = h₁ + h₂ + h₃
h = 28 m + 2 m + 45.87 m
h = 75.87 m
(b) The time taken for the ball to return to the elevator floor
Final height of the ball above the elevator floor = 2 m + 45.87 m = 47.87 m
Apply the following kinematic equation to determine the time to return to the elevator floor.
[tex]h = vt + \frac{1}{2} gt^2\\\\where;\\\\v \ is \ the \ initial \ velocity \ of \ the \ ball \ at \ the \ maximum \ height = 0\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g}} \\\\t = \sqrt{\frac{2\times 47.87}{9.81}} \\\\t = 2.21 \ s[/tex]
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Two coherent sources of radio waves, A and B, are 5.00 meters apart. Each source emits waves with wavelength 6.00 meters. Consider points along the line connecting the two sources.Required:a. At what distance from source A is there constructive interference between points A and B?b. At what distances from source A is there destructive interference between points A and B?
Answer:
a
[tex]z= 2.5 \ m[/tex]
b
[tex]z = (1 \ m , 4 \ m )[/tex]
Explanation:
From the question we are told that
Their distance apart is [tex]d = 5.00 \ m[/tex]
The wavelength of each source wave [tex]\lambda = 6.0 \ m[/tex]
Let the distance from source A where the construct interference occurred be z
Generally the path difference for constructive interference is
[tex]z - (d-z) = m \lambda[/tex]
Now given that we are considering just the straight line (i.e points along the line connecting the two sources ) then the order of the maxima m = 0
so
[tex]z - (5-z) = 0[/tex]
=> [tex]2 z - 5 = 0[/tex]
=> [tex]z= 2.5 \ m[/tex]
Generally the path difference for destructive interference is
[tex]|z-(d-z)| = (2m + 1)\frac{\lambda}{2}[/tex]
=> [tex]|2z - d |= (0 + 1)\frac{\lambda}{2}[/tex]
=> [tex]|2z - d| =\frac{\lambda}{2}[/tex]
substituting values
[tex]|2z - 5| =\frac{6}{2}[/tex]
=> [tex]z = \frac{5 \pm 3}{2}[/tex]
So
[tex]z = \frac{5 + 3}{2}[/tex]
[tex]z = 4\ m[/tex]
and
[tex]z = \frac{ 5 -3 }{2}[/tex]
=> [tex]z = 1 \ m[/tex]
=> [tex]z = (1 \ m , 4 \ m )[/tex]
What happens when a ray of light enters a glass slab ?
Explanation:
then the
speed decreases
Answer:
It gets refracted,
Explanation:
Because i did a couple different searches.
Sunlight strikes a piece of crown glass at an angle of incidence of 37.4o. Calculate the difference in the angle of refraction between a red (660 nm) and a blue (470 nm) ray within the glass. The index of refraction is n
Answer:
The difference in angle of refraction between the red and blue light is 0.2°
Explanation:
Here is the complete question
Sunlight strikes a piece of crown glass at an angle of incidence of 37.4°. Calculate the difference in the angle of refraction between a red (660 nm) and a blue (470 nm) ray within the glass. The index of refraction is n=1.520 for red and n=1.531 for blue light.
Solution
From Snell's law refractive index n = sini/sinr where i = angle of incidence and r = angle of refraction.
Now for the red light n₁ = 1.520, i = 37.4° and r₁ = angle of refraction of red light
So, n₁ = sini/sinr₁
n₁sinr₁ = sini
sinr₁ = sini/n₁
r₁ = sin⁻¹(sini/n₁) = sin⁻¹(sin37.4°/1.52) = sin⁻¹(0.6074/1.52) = sin⁻¹(0.3996) = 23.55°
Now for the blue light n₂ = 1.531, i = 37.4° and r₂ = angle of refraction of blue light
So, n₂ = sini/sinr₂
n₂sinr₂ = sini
sinr₂ = sini/n₂
r₂ = sin⁻¹(sini/n₂) = sin⁻¹(sin37.4°/1.531) = sin⁻¹(0.6074/1.531) = sin⁻¹(0.3967) = 23.37°
So the difference in angle of refraction between the red and blue light is r₁ - r₂ = 23.55° - 23.37° = 0.18° ≅ 0.2°
The heat of fusion of water is 79.5 cal/g. This means 79.5 cal of energy are required to:_________
A) raise the temperature of 1 g of water by 1K
B) turn 1 g of water to steam .
C) raise the temperature of 1 g of ice by 1 K .
D) melt 1 g of ice .
E) increase the internal energy of 1 g of water by 1 J .
Answer:
D) melt 1 g of ice
Explanation:
Heat of fusion is the energy required to change the state of a substance from solid state to liquid state at a constant pressure.
Heat of fusion of water occurs when solid ice acquires energy and turn into liquid water through a process known as melting.
Thus, if the heat of fusion of water is 79.5 cal/g, it means 79.5 cal of energy are required to melt 1 g of ice.
D) melt 1 g of ice
I MIND TRICK PLZ HELP LOL
Troy and Abed are running in a race. Troy finishes the race in 12 minutes. Abed finishes the race in 7 minutes and 30 seconds. If Troy is running at an average speed of 3 miles per hour and speed varies inversely with time, what is Abed’s average speed for the race?
Answer:
Explanation:
Let the race be of a fixed distance x
[tex]Average Speed = \frac{Total Distance}{Total Time}[/tex]
Troy's Average speed = 3 miles/hr = x / 0.2 hr
x = 0.6 miles
Abed's Average speed = 0.6 / 0.125 = 4.8 miles/hr
2. The vector sums of and the Ark witar must se rue our the directions and maintedes at an Bit CB? What meast le tue about the directions and magnitudes and it cor
Check attached photo
Check attached photo
Terms to describe the opposition by a material.to being magnetised is
Answer:
Repulsion
Explanation:
The near point (the smallest distance at which an object can be seen clearly) and the far point (the largest distance at which an object can be seen clearly) are measured for six different people.
Near Point(cm) Far Point(cm)
Avishka 40 [infinity]
Berenice 30 300
Chadwick 25 500
Danya 25 [infinity]
Edouard 80 200
Francesca 50 [infinity]
Of the farsighted people, rank them by the power of the lens needed to correct their hyperopic vision. Rank these from largest to smallest power required.
1. Berenice
2. Avishka
3. Francesca
4. Edouard
Answer:
1. Berenice = 0.67 D
2. Avishka = 1.50 D
3. Francesca = 2.00 D
4. Edouard = 2.75 D
Explanation:
The farsighted people are those with near point greater than 25 cm.
They include Avishka, Berenice, Edouard and Francisca.
A converging lens is needed to correct farsightedness, or hyperopia, therefore, the focal length, f, is positive. The image formed is virtual and on the same side of the lens. Thus the image distance is negative
From the lens formula, 1/f = 1/v 1/u; but v is negative
Therefore, 1/f = 1/u - 1/v
But, power of a lens = 1/f in meters.
Therefore, P = 1/u - 1/v
For Avishka, u = 25 cm or 0.25 m, v = 0.4 m
P = 1/ 0.25 - 1/0.4 = 1.5 D
For Berenice, u = 0.25 m, v = 0.3 m
P = 1/0.25 - 1/0.30 =0.7 D
For Edouard, u = 0.25 m, v = 0.80 m
P = 1/0.25 - 1/0.80 =2.75 D
For Francesca, u = 0.25 m, v = 0.50 m
P = 1/0.25 - 1/0.5 = 2.0 D
When The farsighted people are those with a near point greater than 25cm.
Berenice is = 0.67 D
Avishka is = 1.50 D
Francesca is = 2.00 D
Edouard is = 2.75 D
What is Hyperopic Vision?
When The farsighted people are those with a near point greater than 25 cm. Then, They include Avishka, Berenice, Edouard, and also Francisca.
Also, A converging lens is needed to correct farsightedness, or hyperopia, thus, When the focal length, f, is positive. Also, The image formed is virtual and also on the same side of the lens. hence the image distance is negative.
Also, From the lens formula, 1/f = 1/v 1/u; but v is negative
Thus, 1/f = 1/u - 1/v
But, when the power of a lens = 1/f in meters.
Thus, P = 1/u - 1/v
For Avishka, u = 25 cm or 0.25 m, v = 0.4 m
After that, P = 1/ 0.25 - 1/0.4 = 1.5 D
Then, For Berenice, u = 0.25 m, v = 0.3 m
Now, P = 1/0.25 - 1/0.30 =0.7 D
For Edouard, u = 0.25 m, v is = 0.80 m
Then, P = 1/0.25 - 1/0.80 is =2.75 D
Now, For Francesca, u = 0.25 m, v is = 0.50 m
Therefore, P = 1/0.25 - 1/0.5 = 2.0 D
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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
In a front-end collision, a 1500 kg car with shock-absorbing bumpers can withstand a maximumforce of 80 000 N before damage occurs. If the maximum speed for a non-damaging collision is4.0 km/h, by how much must the bumper be able to move relative to the car
Answer:
The bumper will be able to move by 0.01155m.
Explanation:
The magnitude of deceleration of the car in the front end collision.
[tex]a = \frac{F_m}{m} \\[/tex]
[tex]a = \frac{80000}{1500} \\[/tex]
[tex]a = 53.33[/tex]
This is the deceleration of the car that is generated to stop due to a front end collision.
4 km/h = 1.11 m/s
Now, the initial speed of the bumper in the relation of car, Vi = 0
Now, the initial speed of the bumper in the relation of car, Vf = 1.11 m/s
Use the below equation:
[tex]s = \frac{(Intitial \ speed)^2 – (Final \ speed)^2}{2a} \\[/tex]
[tex]s = \frac{(1.11)^2 – (0)}{2 \times 53.33} \\[/tex]
[tex]s = 0.01155 \\[/tex]
Thus, the bumper can move relative to the car is 0.01155 m .
Please help wil give brainiest & 40p.
Answer:
This is the answer
Explanation:
You can find the ans in the photo I attached.
1) Un objeto realiza un movimiento circular uniforme en una circunferencia de 10 metros de diámetro y efectúa 20 vueltas por minuto. Se pide hallar:
a) El periodo.
b) La frecuencia en Hertz.
c) La velocidad tangencial. d) La velocidad angular.
e) La aceleración centrípeta.
Answer:
A RECIPE NEEDS A COMBINED WEIGHT OF 720G OF FLOUR AND SUGAR. IF THE RECIPE NEEDS 5TIME FLOUR THAN SUGAR,HOW MUCH OF EACH IS NEEDED
You are trying to push a 14 kg canoe across a beach to get it to a lake. Initially, the canoe is at rest, and you exert a force over a distance of 5 m until it has a speed of 2.1 m/s.
a. How much work was done on the canoe?
b. The coefficient of kinetic friction between the canoe and the beach is 0.2. How much work was done by friction on the canoe?
c. How much work did you perform on the canoe?
d. What force did you apply to the canoe?
Answer:
I hope this is correct!
Explanation:
a) work = (force)(displacement)
We know that d = 5m, now we just need to find the force
We need to calculate the acceleration first using a = v^2 - u^2 / 2d
final velocity (v) = 2.1 m/s , initial velocity (u) = 0 m/s , displacement (d) = 5m
a = 2.1^2 - 0^2 / 2(5)
= 0.441 m/s^2
Now we can find force using f = ma
f = (14kg)(0.441m/s^2)
= 6.174 newtons
Finally, you can calculate work now that you have force!
work = (6.174n)(5m)
= 30.87 J (you may need to round sig figs!)
b) Work done by friction = (coefficient of kinetic friction)(mg)(v)
m = 14kg, g = 9.8, coefficient of friction = 0.2
force due to friction = (0.2)(14kg x 9.8)(2.1)
= 57.624 J (again you may need to round to sig figs)
c) Work you perform = total work in direction of motion + work done by friction
total work = 30.87 J, work done by friction = 57.624 J
Work you perform = 30.84 + 57.624
= 88.464 J
d) Force you applied = work you performed / distance
work you performed = 88.464 / 5 m
= 17.6929 N (again you may need to round to sig figs)
Hope this helps ya! Best of luck <3
Rod cells in the retina of the eye detect light using a photopigment called rhodopsin. 1.8 eV is the lowest photon energy that can trigger a response in rhodopsin. Part A What is the maximum wavelength of electromagnetic radiation that can cause a transition
Answer:
The maximum wavelength of the e-m wave is 6.9 x 10^-7 m
Explanation:
Energy required to trigger a response = 1.8 eV
we convert to energy in Joules.
1 eV = 1.602 x 10^-19 J
1.8 eV = [tex]x[/tex] J
[tex]x[/tex] = 1.8 x 1.602 x 10^-19 = 2.88 x 10^-19 J
The energy of an electromagnetic wave is gotten as
E = hf
where
h is the Planck's constant = 6.63 x 10^-34 J-s
and f is the frequency of the wave.
substituting values, we have
2.88 x 10^-19 = 6.63 x 10^-34 x f
f = (2.88 x 10^-19)/(6.63 x 10^-34)
f = 4.34 x 10^14 Hz
We know that the frequency of an e-m wave is given as
f = c/λ
where
c is the speed of light = 3 x 10^8 m/s
λ is the wavelength of the e-m wave
From this we can say that
λ = c/f
λ = (3 x 10^8)/(4.34 x 10^14)
λ = 6.9 x 10^-7 m
Select from the following for the next two questions:
A virtual, inverted and smaller than the object
B real, inverted and smaller than the object
C virtual, upright and smaller than the object
D real, upright and larger than the object
E virtual, upright and larger than the object
F real, inverted and larger than the object
G virtual, inverted and larger than the object
H real, upright and smaller than the object
An object is placed 46.9 cm away from a converging lens. The lens has a focal length of 10.0 cm. Select the statement from the list above which best describes the image an objesthse place 46.9 cm away from a spherical convex mirror. The radius of curvature of the mirror is 20.0 cm. Select the statement from the An object is placed 46.9 cm away from a spherical convex mirror. The radius of curvature of the mirror is 20.0 cm. Select the statement from the list above which best describes the image.
Answer:
Explanation:
1 )
An object is placed 46.9 cm away from a converging lens. The lens has a focal length of 10.0 cm.
Since the object is placed at a distance more than twice the focal length , its image will be inverted , real and will be of the size less than the size of object . So option B is applied .
B) real, inverted and smaller than the object.
2 )
An object is placed 46.9 cm away from a spherical convex mirror. The radius of curvature of the mirror is 20.0 cm.
The object is placed at a point beyond its radius of curvature, its image will be formed at a point between f and C or between focal point and centre of curvature . Its size will be smaller than size of object and it will be real and inverted .
B) real, inverted and smaller than the object.