Answer:
s=((vf+vi)/2)t vf is final velocity and vi is initial velocity
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°
A woman was told in 2020 that she had exactly 15 years to live. If she travels away from the Earth at 0.8 c and then returns at the same speed, the last New Year's Day the doctors expect her to celebrate is:
Answer:
2035
Explanation:
The doctor does not travel with the woman, and therefore, he won't experience any relativistic effect on his time. The doctor will judge time by the time here on earth. Technically, the last new year's day the doctor, who is here on earth, would expect the woman to celebrate will be in 2020 + 15 years = 2035
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.
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
A student holds a bike wheel and starts it spinning with an initial angular speed of 7.0 rotations per second. The wheel is subject to some friction, so it gradually slows down.
In the 10.0 s period following the inital spin, the bike wheel undergoes 60.0 complete rotations. Assuming the frictional torque remains constant, how much more time Δ????s will it take the bike wheel to come to a complete stop?
The bike wheel has a mass of 0.625 kg0.625 kg and a radius of 0.315 m0.315 m. If all the mass of the wheel is assumed to be located on the rim, find the magnitude of the frictional torque ????fτf that was acting on the spinning wheel.
Answer:
a) Δt = 24.96 s , b) τ = 0.078 N m
Explanation:
This is a rotational kinematics exercise
θ = w₀ t - ½ α t²
Let's reduce the magnitudes the SI system
θ = 60 rev (2π rad / 1 rev) = 376.99 rad
w₀ = 7.0 rot / s (2π rad / 1 rpt) = 43.98 rad / s
α = (w₀ t - θ) 2 / t²
let's calculate the annular acceleration
α = (43.98 10 - 376.99) 2/10²
α = 1,258 rad / s²
Let's find the time it takes to reach zero angular velocity (w = 0)
w = w₀ - alf t
t = (w₀ - 0) / α
t = 43.98 / 1.258
t = 34.96 s
this is the total time, the time remaining is
Δt = t-10
Δt = 24.96 s
To find the braking torque, we use Newton's law for angular motion
τ = I α
the moment of inertia of a circular ring is
I = M r²
we substitute
τ = M r² α
we calculate
τ = 0.625 0.315² 1.258
τ = 0.078 N m
The total time taken by the wheel to come to rest is 25.18 s and the magnitude of the frictional torque is 25.18 N-m.
Given data:
The initial angular speed of wheel is, [tex]\omega = 7.0 \;\rm rps[/tex] (rps means rotation per second).
The time interval is, t' = 10.0 s.
The number of rotations made by wheel is, n = 60.0.
The mass of bike wheel is, m = 0.625 kg.
The radius of wheel is, r = 0.315 m.
The problem is based on rotational kinematics. So, apply the second rotational equation of motion as,
[tex]\theta = \omega t-\dfrac{1}{2} \alpha t'^{2}[/tex]
Here, [tex]\theta[/tex] is the angular displacement, and its value is,
[tex]\theta =2\pi \times 60\\\\\theta = 376.99 \;\rm rad[/tex]
And, angular speed is,
[tex]\omega = 2\pi n\\\omega = 2\pi \times 7\\\omega = 43.98 \;\rm rad/s[/tex]
Solving as,
[tex]376.99 = 43.98 \times 10-\dfrac{1}{2} \alpha \times 10^{2}\\\\\alpha = 1.25 \;\rm rad/s^{2}[/tex]
Apply the first rotational equation of motion to obtain the value of time to reach zero final velocity.
[tex]\omega' = \omega - \alpha t\\\\0 = 43.98 - 1.25 \times t\\\\t = 35.18 \;\rm s[/tex]
Then total time is,
T = t - t'
T = 35.18 - 10
T = 25.18 s
Now, use the standard formula to obtain the value of braking torque as,
[tex]T = m r^{2} \alpha\\\\T = 0.625 \times (0.315)^{2} \times 1.25\\\\T = 0.0775 \;\rm Nm[/tex]
Thus, we can conclude that the total time taken by the wheel to come to rest is 25.18 s and the magnitude of the frictional torque is 25.18 N-m.
Learn more about the rotational motion here:
https://brainly.com/question/1388042
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.
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 .
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 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
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]
Water flows at 0.00027 m3/s through a 10-m long garden hose lying on the ground, with a radius of 0.01 m. Water has a viscosity of 1 mPa.s What is the magnitude of gauge pressure in Pa of the water entering the hose
Answer:
The gauge pressure is [tex]P = 687.4 \ Pa[/tex]
Explanation:
From the question we are told that
The rate of flow is [tex]Q = 0.00027 m^3 /s[/tex]
The height is h = 10 m
The radius is r = 0.01 m
The viscosity is [tex]\eta = 1mPa \cdot s = 1 *10^{-3} \ Pa\cdot s[/tex]
Generally the gauge pressure according to Poiseuille's equation is mathematically represented as
[tex]P = 8 \pi \eta * \frac{L * v }{ A}[/tex]
Here v is the velocity of the water which is mathematically represented according to continuity equation as
[tex]v = \frac{Q}{A }[/tex]
Where A is the cross-sectional area which is mathematically represented as
[tex]A = \pi r^2[/tex]
substituting values
[tex]A = 3.142 *(0.01)^2[/tex]
[tex]A = 3.142 *10^{-4} \ m^2[/tex]
So
[tex]v = \frac{ 0.00027}{3.142*10^{-4}}[/tex]
[tex]v = 0.8593 \ m/s[/tex]
So
[tex]P = 8 * 3.142 * 1.0*10^{-3}* \frac{10 * 0.8593 }{ 3.142*10^{-4}}[/tex]
[tex]P = 687.4 \ Pa[/tex]
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
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
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
At what frequency should a 200-turn, flat coil of cross sectional area of 300 cm2 be rotated in a uniform 30-mT magnetic field to have a maximum value of the induced emf equal to 8.0 V
Answer:
The frequency of the coil is 7.07 Hz
Explanation:
Given;
number of turns of the coil, 200 turn
cross sectional area of the coil, A = 300 cm² = 0.03 m²
magnitude of the magnetic field, B = 30 mT = 0.03 T
Maximum value of the induced emf, E = 8 V
The maximum induced emf in the coil is given by;
E = NBAω
Where;
ω is angular frequency = 2πf
E = NBA(2πf)
f = E / 2πNBA
f = (8) / (2π x 200 x 0.03 x 0.03)
f = 7.07 Hz
Therefore, the frequency of the coil is 7.07 Hz
A 0.500 H inductor is connected in series with a 93 Ω resistor and an ac source. The voltage across the inductor is V = −(11.0V)sin[(500rad/s)t]. What is the voltage across the resistor at 2.09 x 10-3 s? Group of answer choices 205 V 515 V 636 V 542 V
Answer:
205 V
V[tex]_{R}[/tex] = 2.05 V
Explanation:
L = Inductance in Henries, (H) = 0.500 H
resistor is of 93 Ω so R = 93 Ω
The voltage across the inductor is
[tex]V_{L} = - IwLsin(wt)[/tex]
w = 500 rad/s
IwL = 11.0 V
Current:
I = 11.0 V / wL
= 11.0 V / 500 rad/s (0.500 H)
= 11.0 / 250
I = 0.044 A
Now
V[tex]_{R}[/tex] = IR
= (0.044 A) (93 Ω)
V[tex]_{R}[/tex] = 4.092 V
Deriving formula for voltage across the resistor
The derivative of sin is cos
V[tex]_{R}[/tex] = V[tex]_{R}[/tex] cos (wt)
Putting V[tex]_{R}[/tex] = 4.092 V and w = 500 rad/s
V[tex]_{R}[/tex] = V[tex]_{R}[/tex] cos (wt)
= (4.092 V) (cos(500 rad/s )t)
So the voltage across the resistor at 2.09 x 10-3 s is which means
t = 2.09 x 10⁻³
V[tex]_{R}[/tex] = (4.092 V) (cos (500 rads/s)(2.09 x 10⁻³s))
= (4.092 V) (cos (500 rads/s)(0.00209))
= (4.092 V) (cos(1.045))
= (4.092 V)(0.501902)
= 2.053783
V[tex]_{R}[/tex] = 2.05 V
A neutron star has a mass of between 1.4-2.8 solar masses compressed to the size of:
A. Earth
B. The state of Oregon
C. North America
D. An average city
The correct answer is D. An average city
Explanation:
A neutron star differs from others due to its massive density, this means a lot of matter is compressed in a small area. Indeed, neutron stars have a mass of around 1.4 to 2.8 times the mass of the sun. But these are considerably small as they only measure around 20 kilometers, which is the size of an average city. Additionally, neutron stars are this dense because they are the result of a regular star exploding, which leads to a super-dense core, or neutron star. In this context, the mass of a neutron star is compressed to the size of an average city.
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
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
beam of white light goes from air into water at an incident angle of 58.0°. At what angles are the red (660 nm) and blue (470 nm) parts of the light refracted? (Enter your answer to at least one decimal place.) red ° blue °
Answer:
For red light= 39.7°
Blue light 39.2°
Explanation:
Given that refractive index for red light is 1.33 and that of blue light is 1.342
So angle of refraction for red light will be
Sinစi/ (sinစ2) =( nw)r/ ni
Sin 58° x 1.000293/1.33. =( sinစ2)r
0.64= sinစ2r
Theta2r = 39.7°
For blue light
Sinစi/ (sinစ2) =( nw)b/ ni
Sin 58° x 1.000293/1.342 =( sinစ2)b
0.632= sinစ2r
Theta2b= 39.19°
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
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
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 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
An object is placed in a room where the temperature is 20 degrees C. The temperature of the object drops by 5 degrees C in 4 minutes and by 7 degrees C in 8 minutes. What was the temperature of the object when it was initially placed in the room
Answer:
28.3°C
Explanation:
Using
T(t) = (T(0) - 20)*(e^(-k*t)) + 20
for some positive number k, and some initial temperature T(0).
Boundary conditions:
T(4) = T(0) - 5 _______ (i)
T(8) = T(0) - 7 _______ (ii)
==> solving for T(0) and k :
(i):
(T(0) - 20)*(e^(-k*4)) + 20 = T(0) - 5 ==>
(T(0) - 20)*(e^(-k*4)) = T(0) - 20 - 5
(T(0) - 20)*(e^(-k*4)) = (T(0) - 20) - 5
5 = (T(0) - 20) - (T(0) - 20)*(e^(-k*4))
5 = (T(0) - 20) * ( 1 - e^(-k*4) )
(ii):
(T(0) - 20)*(e^(-k*8)) + 20 = T(0) - 7
(T(0) - 20)*(e^(-k*8)) = (T(0) - 20) - 7
7 = (T(0) - 20) - (T(0) - 20)*(e^(-k*8))
7 = (T(0) - 20) * (1 - e^(-k*8))
In both results, subsitute x = e^(-4k) and C = (T(0) - 20)
(i): 5 = C * (1 - x)
(ii): 7 = C * (1 - x^2) = C * (1-x)*(1+x)
Substitute C*(1-x) from (i) into (ii):
(ii): 7 = 5*(1+x) ==> (1+x) = 7/5 ==> x = 2/5
back into (i):
(i): 5 = C * (1 - 2/5) ==> 5 = C * 3/5 ==> C = 25/3
C = T(0) - 20 ==>
T(0) = C + 20 = 25/3 + 20 = 25/3 + 60/3 = 85/3
= 28.3°C
Terms to describe the opposition by a material.to being magnetised is
Answer:
Repulsion
Explanation:
Two identical trucks have mass 5500 kg when empty, and the maximum permissible load for each is 8000 kg. The first truck, carrying a 3900 kg, is at rest. The second truck plows into it at 64 km/h, and the pair moves away at 44 km/h. As an expert witnes, you're asked to determine whether the second truck was overloaded. What do you report? Yes the truck is overloaded, or no, the truck is not overloaded?
Answer:
no, the truck is not overloaded
Explanation:
The computation is shown below;
Let us assume the mass of the loan in the second truck be M
So, the equation is as follows
{(Mass + M) × second truck × 1000 ÷ 3,600} = {(Mass + M + mass + first truck) × Pair moves away × 1,000 ÷ 3,600}
{(5500 + M) × 64 × 1,000 ÷ 3,600 = {(5,500 + M + 5,500 + 3,900) × 44 × 1,000 ÷ 3,600}
(5500 + M) × 64 = (14,900 + M) × 44
352,000 + 64 M = 655,600 + 44 M
After solving this
M = 15,180 kg
Therefore the second truck is not overloaded
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. Suppose that a solid ball, a solid disk, and a hoop all have the same mass and the same radius. Each object is set rolling without slipping up an incline with the same initial linear (translational) speed. Which goes farthest up
the incline?
a. the ball
b. the disk
c. the hoop
d. the hoop and the disk roll to the same height, farther
than the ball
e. they all roll to the same height
2. Suppose that a solid ball, a solid disk, and a hoop all have the same mass and the same radius. Each object is set rolling with slipping up an incline with the same initial linear (translational) speed. Which goes farthest up
the incline?
a. the ball
b. the disk
c. the hoop
d. the hoop and the disk roll to the same height, farther
than the ball
e. they all roll to the same height
Answer:
The hoop
Explanation:
Because it has a smaller calculated inertia of 2/3mr² compares to the disc