Answer:In the case of two proton beams the protons repel one another because they have the same sign of electrical charge. There is also an attractive magnetic force between the protons, but in the proton frame of reference this force must be zero! Clearly then the attractive magnetic force that reduces the net force between protons in the two beams as seen in our frame of reference is relativistic. In particular the apparent magnetic forces or fields are relativistic modifications of the electrical forces or fields. As such modifications, they cannot be stronger than the electrical forces and fields that produce them. This follows from the fact that switching frames of reference can reduce forces, but it can’t turn what is attractive in one frame into a repulsive force in another frame.
In the case of wires the net charges in two wires are zero everywhere along the wires. That makes the net electrical forces between the wires very nearly zero. Yet the relativistic magnetic forces and fields will be of the same sort as in the case of two beams of charges of a single sign. This is true even in the frame of reference of what we think as the moving charges, that is, the electrons. In the frame of reference moving at the drift velocity of these current-carrying electrons, it is the protons or positively charged ions that are moving in the other direction. Consequently in any frame of reference for current-carrying wires in parallel, the net electrical force will be essentially zero, and there will be a net attractive magnetic force
Explanation:
Explanation:
Particles with similar charges (both positive or both negative) will always repel each other, regardless of their speed or direction.
If the x-position of a particle is measured with an uncertainty of 1.00×10-10 m, then what is the uncertainty of the momentum in this same direction? (Useful constant: h-bar = 1.05×10-34 Js.)
Answer:
The uncertainty in momentum is 5.25x 10^25Jsm
Explanation:
We know that
h bar = h/2π
So
1.05x 10^34=h/2pπ
h=1.05x 10^ 34(2π)=6.597x 10^-34Js
dp=(6.597x10^-34/4pπ)/(1x10^-10)
=5.25x10^-25 Jsm
We've seen that for thermal radiation, the energy is of the form AVT4, where A is a universal constant, V is volume, and T is temperature. 1) The heat capacity CV also is proportional to a power of T, Tx. What is x
Answer:
this raise the temperature is x = 3
Explanation:
Heat capacity is the relationship between heat and temperature change
C = Q / ΔT
if the heat in the system is given by the change in energy and we carry this differential formulas
[tex]c_{v}[/tex] = dE / dT
In this problem we are told that the energy of thermal radiation is
E = A V T⁴
Let's look for the specific heat
c_{v} = AV 4 T³
the power to which this raise the temperature is x = 3
Which one of the following actions would make the maxima in the interference pattern from a grating move closer together?
A. Increasing the number of lines per length.
B. Decreasing the number of lines per length.
C. Increasing the distance to the screen.
D. Increasing the wavelength of the laser.
Answer:
Answer:
A. Increasing the number of lines per length.
A fish appears to be 2.00 m below the surface of a pond when viewed almost directly above by a fisherman. What is the actual depth of the fish
Answer:
2,66
Explanation:
The refractive index= real depth/ apparent depth
real depth = refractive index * apparent depth
Let's assume index for water is 1.33
real depth = 2*1,33 = 2,66
Give an example of hypothesis for an experiment and then identify its dependent and independent variables. Write all the steps of the scientific method. Explain why it is good to limit an experiment to test only one variable at a time whenever possible ?
Please somebody !!!!
What is the frequency of the fundamental mode of vibration of a steel piano wire stretched to a tension of 440 N? The wire is 0.630 m long and has a mass of 5.69 g.
Answer:
220.698HzExplanation:
The fundamental frequency f₀ is expressed as f₀ =V/2L where;
V is the speed of the string = [tex]\sqrt{\frac{T}{M} }[/tex]
m is the mass of the string
L is the length of the string
T is the tension in the string
f₀ = [tex]\frac{1}{2L} \sqrt{\frac{T}{m} }[/tex]
Given datas
m = 5.69g = 0.00569 kg
T = 440N
L = 0.630 m
Required
Fundamental frequency of the steel piano wire f₀
[tex]f_0 = \frac{1}{2(0.630)}\sqrt{\frac{440}{0.00569} } \\ \\f_0 = \frac{1}{1.26}\sqrt{77,328.65 } \\\\f_0 = \frac{1}{1.26} * 278.08\\\\f_0 = 220.698Hz[/tex]
Hence the frequency of the fundamental mode of vibration of the steel piano wire stretched to a tension of 440N is 220.698Hz
A clown 2 m tall looks at himself in a full-length mirror (floor-to-ceiling). Where in the mirror must he look to see his feet?
Answer:
Around the center of the mirror
Sammy is 5 feet and 5.3 inches tall.tall.what is sammy's height in metres?
Answer:
65.3
Explanation:
1 foot = 12 inches
Sammy is 5 feet tall.
5 feet = ? inches
Multiply the feet value by 12 to find in inches.
5 × 12
= 60
Add 5.3 inches to 60 inches.
60 + 5.3
= 65.3
Answer:
It will be 》》》》1.664716m
Suppose you exert a force of 185 N tangential to the outer edge of a 1.73-m radius 76-kg grindstone (which is a solid disk).
Required:
a. What torque is exerted?
b. What is the angular acceleration assuming negligible opposing friction?
c. What is the angular acceleration if there is an opposing frictional force of 20.0 N exerted 1.50 cm from the axis?
Answer:
a. 320.06 Nm b. 2.814 rad/s² c. 2.811 rad/s².
Explanation:
a. The torque exerted τ = Frsinθ where F = tangential force exerted = 185 N, r = radius of grindstone = 1.73 m and θ = 90° since the force is tangential to the grindstone.
τ = Frsinθ
= 185 N × 1.73 m × sin90°
= 320.05 Nm
So, the torque τ = 320.05 Nm
b. Since torque τ = Iα where I = moment of inertia of grindstone = 1/2MR² where M = mass of grindstone = 76 kg and R = radius of grindstone = 1.73 m
α = angular acceleration of grindstone
τ = Iα
α = τ/I = τ/(MR²/2) = 2τ/MR²
substituting the values of the variables, we have
α = 2τ/MR²
= 2 × 320.05 Nm/[76 kg × (1.73 m)²]
= 640.1 Nm/227.4604 kgm²
= 2.814 rad/s²
So, the angular acceleration α = 2.814 rad/s²
c. The opposing frictional force produces a torque τ' = F'r' where F' = frictional force = 20.0 N and r' = distance of frictional force from axis = 1.50 cm = 0.015 m.
So τ' = F'r' = 20.0 N × 0.015 m = 0.3 Nm
The net torque on the grindstone is thus τ'' = τ - τ' = 320.05 Nm - 0.3 Nm = 319.75 Nm
Since τ'' = Iα
α' = τ''/I where α' = its new angular acceleration
α' = 2τ/MR²
= 2 × 319.75 Nm/[76 kg × (1.73 m)²]
= 639.5 Nm/227.4604 kgm²
= 2.811 rad/s²
So, the angular acceleration α' = 2.811 rad/s²
A fireperson is 50 m from a burning building and directs a stream of water from a fire hose at an angle of 300 above the horizontal. If the initial speed of the stream is 40 m/s the height that the stream of water will strike the building is
Answer:
We can think the water stream as a solid object that is fired.
The distance between the fireperson and the building is 50m. (i consider that the position of the fireperson is our position = 0)
The angle is 30 above the horizontal. (yo wrote 300, but this has no sense because 300° implies that he is pointing to the ground).
The initial speed of the stream is 40m/s.
First, using the fact that:
x = R*cos(θ)
y = R*sin(θ)
in this case R = 40m/s and θ = 30°
We can use the above relation to find the components of the velocity:
Vx = 40m/s*cos(30°) = 34.64m/s
Vy = 20m/s.
First step:
We want to find the time needed to the stream to hit the buildin.
The horizontal speed is 34.64m/s and the distance to the wall is 50m
So we want that:
34.64m/s*t = 50m
t = 50m/(34.64m/s) = 1.44 seconds.
Now we need to calculate the height of the stream at t = 1.44s
Second step:
The only force acting on the water is the gravitational one, so the acceleration of the stream is:
a(t) = -g.
g = -9.8m/s^2
For the speed, we integrate over time and we get:
v(t) = -g*t + v0
where v0 is the initial speed: v0 = 20m/s.
The velocity equation is:
v(t) = -g*t + 20m/s.
For the position, we integrate again over time:
p(t) = -(1/2)*g*t^2 + 20m/s*t + p0
p0 is the initial height of the stream, this data is not known.
Now, the height at the time t = 1.44s is
p(1.44s) = -5.9m/s^2*(1.44s)^2 + 20m/s*1.44s + po
= 16.57m + p0
So the height at wich the stream hits the building is 16.57 meters above the initial height of the fire hose.
An electric heater draws 13 amperes of current when connected to 120 volts. If the price of electricity is $0.10/kWh, what would be the approximate cost of running the heater for 8 hours?
(A) $0.19
(B) $0.29
(C) $0.75
(D) $1.25
(E) $1.55
Answer:
C $0.75 my friend I wish it is right answer
Changing the speed of a synchronous generator changes A) the frequency and amplitude of the output voltage. B) only the frequency of the output voltage. C) only the amplitude of the output voltage. D) only the phase of the output voltage.
Answer:
A) the frequency and amplitude of the output voltag
Explanation:
Changing the speed of a synchronous generator changes both the output voltage (amplitude of the wave) and frequency as they tend to increase.
Changing the speed regulator will change the engine throttle setting to maintain the speed.
While the power, torque, current, fuel flow rate and torque angle will have decreased.
By what length will a slab of concrete that is originally 18 m long contract when the temperature drops from 24°C to -16°C? The coefficient of linear thermal expansion for this concrete is 1.0 × 10-5 C-1.
Explanation:
According to Thermal Expansion of solids:
[tex]dl = \alpha \times l \times dt[/tex]
[tex]dl = {10}^{ - 5} \times 18 \times 40 [/tex]
[tex]dl = 7.2 \times {10}^{ - 3} [/tex]
A light beam has a wavelength of 330 nm in a material of refractive index 1.50. In a material of refractive index 2.50, its wavelength will be In a material of refractive index 2.50, its wavelength will be:_________
a. 495 nm .
b. 330 nm .
c. 220 nm .
d. 198 nm .
e. 132 nm .
Answer:
The wavelength of the ligt beam in a material of refractive index 2.50 is 198 mm
d. 198 mm
Explanation:
Refractive index is given by;
[tex]\mu= \frac{\lambda_{vacuum}}{\lambda _{medium}}[/tex]
where;
[tex]\lambda_{vacuum}[/tex] is the wavelength of the light beam in vacuum
[tex]\lambda_{medium}[/tex] is the wavelength of the beam in a material
[tex]\mu= \frac{\lambda_{vacuum}}{\lambda _{medium}} \\\\\lambda_{vacuum} = \mu *\lambda _{medium}\\\\\ the \ wavelength \ of \ the \ light \ beam \ is \ constant \ in \ a \ vacuum\\\\ \mu_1 *\lambda _{medium}_1 = \mu_2 *\lambda _{medium}_2\\\\\lambda _{medium}_2 = \frac{ \mu_1 *\lambda _{medium}_1 }{ \mu_2} \\\\\lambda _{medium}_2 =\frac{1.5*330}{2.5} \\\\\lambda _{medium}_2 = 198 \ mm[/tex]
Therefore, the wavelength of the ligt beam in a material of refractive index 2.50 is 198 mm.
d. 198 mm
A Van de Graaff generator produces a beam of 2.02-MeV deuterons, which are heavy hydrogen nuclei containing a proton and a neutron.
A) If the beam current is 10.0 μA, how far apart are the deuterons?
B) Is the electrical force of repulsion among them a significant factor in beam stability? Explain.
Answer:
A) The distance of the deuterons from one another = 2.224× 10⁻⁷ m
B) The electrical force of repulsion among them shows a small effect in beam stability.
Explanation:
Given that:
A Van de Graaff generator produces a beam of 2.02-MeV deuterons
If the beam current is 10.0 μA, the distance of the deuterons from one another can be determined by using the concept of kinetic energy of the generator.
[tex]\mathtt{K.E = \dfrac{1}{2}mv^2}[/tex]
2 K.E = mv²
[tex]\mathtt{v^2 = \dfrac{2 K.E }{m}}[/tex]
[tex]\mathtt{v =\sqrt{ \dfrac{2 K.E }{m}}}[/tex]
so, v is the velocity of the deuterons showing the distance of the deuterons apart from one another.
[tex]\mathtt{v =\sqrt{ \dfrac{2 (2.02 \ MeV) \times \dfrac{10^6 \ eV}{ 1 \ MeV} \times \dfrac{1.60 \times 10^{-19} \ J }{1 \ eV} }{ 3.34 \times 10^ {-27} \ kg}}}[/tex]
[tex]\mathtt{v =\sqrt{ \dfrac{6.464 \times 10^{-13} \ J }{ 3.34 \times 10^ {-27} \ kg}}}[/tex]
v = 13911611.49 m/s
v = 1.39 × 10⁷ m/s
So, If the beam current is 10.0 μA.
We all know that:
[tex]I = \dfrac{q}{t}[/tex]
[tex]t = \dfrac{q}{I}[/tex]
[tex]\mathtt{ t = \dfrac{1.6 * 10 ^{-19} \ C}{10.0 * 10^{-6} \ A}}[/tex]
t = 1.6 × 10⁻¹⁴ s
Finally, the distance of the deuterons from one another = v × t
the distance of the deuterons from one another = (1.39 × 10⁷ m/s × 1.6 × 10⁻¹⁴ s)
the distance of the deuterons from one another = 2.224× 10⁻⁷ m
B) Is the electrical force of repulsion among them a significant factor in beam stability? Explain.
The electrical force of repulsion among them shows a small effect in beam stability. This is because, one nucleus tends to put its nearest neighbor at potential V = (k.E × q) / r = 7.3e⁻⁰³ V. This is very small compared to the 2.02-MeV accelerating potential, Thus, repulsion within the beam is a small effect.
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 ✌️
A 70 kg man floats in freshwater with 3.2% of his volume above water when his lungs are empty, and 4.85% of his volume above water when his lungs are full.
Required:
a. Calculate the volume of air he inhales - called his lung capacity - in liters.
b. Does this lung volume seem reasonable?
Answer:
Explanation:
A) Vair = 1.3 L
B) Volume is not reasonable
Explanation:
A)
Assume
m to be total mass of the man
mp be the mass of the man that pulled out of the water
m1 be the mass above the water with the empty lung
m2 be the mass above the water with full lung
wp be the weight that the buoyant force opposes as a result of the air.
Va be the volume of air inside man's lungs
Fb be the buoyant force due to the air in the lung
given;
m = 78.5 kg
m1 = 3.2% × 78.5 = 2.5 kg
m2 = 4.85% × 78.5 = 3.8kg
But, mp = m2- m1
mp = 3.8 - 2.5
mp = 1.3kg
So using
Archimedes principle, the relation for formula for buoyant force as;
Fb = (m_displaced water)g = (ρ_water × V_air × g)
Where ρ_water is density of water = 1000 kg/m³
Thus;
Fb = wp = 1.3× 9.81
Fb = 12.7N
But
Fb = (ρ_water × V_air × g)
So
Vair = Fb/(ρ_water × × g)
Vair = 12.7/(1000 × 9.81)
V_air = 1.3 × 10^(-3) m³
convert to litres
1 m³ = 1000 L
Thus;
V_air = 1.3× 10^(-3) × 1000
V_air = 1.3 L
But since the average lung capacity of an adult human being is about 6-7litres of air.
Thus, the calculated lung volume is not reasonable
Explanation:
Calculate the density of the following material.
1 kg helium with a volume of 5.587 m³
700 kg/m³
5.587 kg/m³
0.179 kg/m³
Answer:
[tex]density \: = \frac{mass}{volume} [/tex]
1 / 5.587 is equal to 0.179 kg/m³
Hope it helps:)
Answer:
The answer is
0.179 kg/m³Explanation:
Density of a substance is given by
[tex]Density \: = \frac{mass}{volume} [/tex]
From the
mass = 1 kg
volume = 5.583 m³
Substitute the values into the above formula
We have
[tex]Density \: = \frac{1 \: kg}{5.583 \: {m}^{3} } [/tex]
We have the final answer as
Density = 0.179 kg/m³Hope this helps you
A circular conducting loop of radius 31.0 cm is located in a region of homogeneous magnetic field of magnitude 0.700 T pointing perpendicular to the plane of the loop. the loop is connected in series with a resistor of 265 ohms. The magnetic field is now increased at a constant rate by a factor of 2.30 in 29.0 s.
Calculate the magnitude of induced emf in the loop while the magnetic field is increasing.
With the magnetic field held constant a ts its new value of 1.61 T, calculate the magnitude of its induced voltage in the loop while it is pulled horizontally out of the magnetic field region during a time interval of 3.90s.
Answer:
(a) The magnitude of induced emf in the loop while the magnetic field is increasing is 9.5 mV
(b) The magnitude of the induced voltage at a constant magnetic field is 124.7 mV
Explanation:
Given;
radius of the circular loop, r = 31.0 cm = 0.31 m
initial magnetic field, B₁ = 0.7 T
final magnetic field, B₂ = 2.3B₁ = 2.3 X 0.7 T = 1.61 T
duration of change in the field, t = 29
(a) The magnitude of induced emf in the loop while the magnetic field is increasing.
[tex]E = A*\frac{\delta B}{\delta t} \\\\[/tex]
[tex]E = A*\frac{B_2 -B_1}{\delta t}[/tex]
Where;
A is the area of the circular loop
A = πr²
A = π(0.31)² = 0.302 m²
[tex]E = A*\frac{B_2 -B_1}{\delta t} \\\\E = 0.302*\frac{1.61-0.7}{29} \\\\E = 0.0095 \ V\\\\E = 9.5 \ mV[/tex]
(b) the magnitude of the induced voltage at a constant magnetic field
E = A x B/t
E = (0.302 x 1.61) / 3.9
E = 0.1247 V
E = 124.7 mV
Therefore, the magnitude of the induced voltage at a constant magnetic field is 124.7 mV
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
Learn more : https://brainly.com/question/22403676
An inductor is connected to the terminals of a battery that has an emf of 12.0 VV and negligible internal resistance. The current is 4.96 mAmA at 0.800 msms after the connection is completed. After a long time the current is 6.60 mAmA. Part A What is the resistance RR of the inductor
i
CHECK COMPLETE QUESTION BELOW
inductor is connected to the terminals of a battery that has an emf of 12.0 VV and negligible internal resistance. The current is 4.96 mAmA at 0.800 msms after the connection is completed. After a long time the current is 6.60 mAmA.
Part A)What is the resistance RR of the inductor
PART B) what is inductance L of the conductor
Answer:
A)R=1818.18 ohms
B)L=1.0446H
Explanation:
We were given inductor L with resistance R , there is a connection between the battery and the inductor with Emf of 12V, we can see that the circuit is equivalent to a simple RL circuit.
There is current of 4.96mA at 0.8ms, at the end of the connection the current increase to 6.60mA,
.
a)A)What is the resistance RR of the inductor?
The current flowing into RL circuit can be calculated using below expresion
i=ε/R[1-e⁻(R/L)t]
at t=∞ there is maximum current
i(max)= ε/R
Where ε emf of the battery
R is the resistance
R=ε/i(max)
= 12V/(6.60*10⁻³A)
R=1818.18 ohms
Therefore, the resistance R=1818.18 ohms
b)what is inductance L of the conductor?
i(t=0.80ms and 4.96mA
RT/L = ⁻ln[1- 1/t(max)]
Making L subject of formula we have
L=-RT/ln[1-i/i(max)]
If we substitute the values into the above expresion we have
L= -(1818.18 )*(8.0*10⁻⁴)/ln[1-4.96/6.60)]
L=1.0446H
Therefore, the inductor L=1.0446H
A 180-km-long high-voltage transmission line 2.00 cm in diameter carries a steady current of 1,050 A. If the conductor is copper with a free charge density of 8.50 1028 electrons per cubic meter, how many years does it take one electron to travel the full length of the cable
Answer:
27yrs
Explanation:
h= difference in height between the initial position and the bottom position
We are told that the rope is L = 2.30 m long and inclined at 45.0° from the vertical
h=L-Lcos(x)= L(1-cosx)=2.30(1-cos45)
=0.674m
Potential Energy = 28× 9.8×0.674
=184.9J
B)we can see that at the bottom of the motion, all the initial potential energy of the child has been converted into kinetic energy:
E= 0.5mv^2
You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. You set your slit spacing at 1.11 mm and place your screen 8.63 m from the slits. Then, you illuminate the slits with your new toy and find on the screen that the tenth bright fringe is 4.71 cm away from the central bright fringe (counted as the zeroth bright fringe). What is your laser's wavelength lambda expressed in nanometers?
Answer:
λ = 605.80 nm
Explanation:
These double-slit experiments the equation for constructive interference is
d sin θ = m λ
where d is the distance between the slits, λ the wavelength of light and m an integer that determines the order of interference.
In this case, the distance between the slits is d = 1.11 mm = 1.11 10⁻³ m, the distance to the screen is L = 8.63 m, the range number m = 10 and ay = 4.71 cm
Let's use trigonometry to find the angle
tan θ = y / L
as the angles are very small
tan θ = sin θ / cos θ = sin θ
we substitute
sin θ = y / L
we substitute in the first equation
d y / L = m λ
λ = d y / m L
let's calculate
λ = 1.11 10⁻³ 4.71 10⁻²/ (10 8.63)
λ = 6.05805 10⁻⁷ m
let's reduce to nm
λ = 6.05805 10⁻⁷ m (10⁹ nm / 1m)
λ = 605.80 nm
A lens is made with a focal length of -40 cm using a material with index of refraction 1.50. A second lens is made with the SAME GEOMETRY as the first lens, but using a material having refractive index of 2.00. What is the focal length of the second lens
Answer:
f = - 20 cm
Explanation:
This exercise asks us for the focal length, which for a lens in air is
1 / f = (n₂-n₁) (1 / R₁ - 1 / R₂)
where n₂ is the refractive index of the material, n₁ is the refractive index of the medium surrounding the lens, R₁ and R₂ are the radii of the two surfaces.
In this exercise the medium that surrounds the lens is air n₁ = 1 and the lens material has an index of refraction n₂ = n = 1.50, let's substitute in the expression
- 1/40 = (n-1) (1 / R₁ -1 / R₂)
(1 / R₁ - 1 / R₂) = - 1/40 (n-1)
let's calculate
(1 / R₁ -1 / R₂) = - 1/40 (1.50 -1)
(1 / R₁ -1 / R₂) = -1/20
Now we change the construction material for one with refractive index
n = 2, keeping the radii,
1 / f = (n-1) (1 / R₁-1 / R₂)
1 / f = (n-1) (-1/20)
let's calculate
1 / f = (2.00-1) (-1/20)
1 / f = -1/20
f = - 20 cm
A laser emits photons having an energy of 3.74 × 10–19 J. What color would be expected for the light emitted by this laser? (c = 3.00 × 108 m/s, h = 6.63 × 10–34 J ⋅ s)
Answer:
The wavelength of the emitted photons 532 nm, corresponds to a visible light having GREEN color.
Explanation:
Given;
energy of the emitted photons, E = 3.74 x 10⁻¹⁹ J
speed of light, c = 3 x 10⁸ m/s
Planck's constant, h = 6.63 x 10⁻³⁴ J.s
The wavelength of the emitted light will be calculated by applying energy of photons;
[tex]E = hf[/tex]
where;
E is the energy emitted light
h is Planck's constant
f is frequency of the emitted photon
But f = c / λ
where;
λ is the wavelength of the emitted photons
[tex]E = \frac{hc}{\lambda} \\\\\lambda = \frac{hc}{E} \\\\\lambda = \frac{6.63*10^{-34} *3*10^{8}}{3.74*10^{-19}} \\\\\lambda = 5.318 *10^{-7} \ m\\\\\lambda = 531.8 *10^{-9} \ m\\\\\lambda = 531.8 \ nm[/tex]
λ ≅ 532 nm
the wavelength of the emitted photons is 532 nm.
Therefore, the wavelength of the emitted photons 532 nm, corresponds to a visible light having GREEN color.
Experiment to find ways to make rainbows.
a) Insert at least one setup where light passing through a prism gives a rainbow and describe why a rainbow is formed.
b) Explain why only some types of light will yield rainbows.
Answer:
Explanation:
a) To get a rainbow from a prism arrangement, we will need
A triangular prismA black cardboard boxA source of white light (light from the window will suffice)A pocket knifeFirst, you cut a slit in one end of the cardboard with the pen knife.
Next you open up a space on top of the cardboard through which you can observe the experiment and its result.
Next, you place the triangular prism with its slant face facing the the cut slit.
Finally, position the slit to face the light from the open window, and adjust the prism till the projected bands of colored light (rainbow) is very much obvious on the other end of the box, opposite the slit.
b) For a light to yield rainbow, it most be composed of different component colors of light. The colors of light is due to the difference in wavelength, and dispersion is due to the different in the wavelengths of the component light. So to get rainbow from a light source, the light must not be monochromatic. This means that only light composed of component light of different colors can produce rainbow. Light from the sun for example is composed of 7 distinct colors of light, and white light can be created with just three colors; blue, green, and red light.
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 .
Which notation is better to use? (Choose between 4,000,000,000,000,000 m and 4.0 × 1015 m)
Answer:
4 x 10¹⁵
Explanation:
Types of friction in physics
-- static friction
-- kinetic friction
-- fluid friction
-- sliding friction
-- air resistance
-- drag
-- professional debate
A pair of narrow, parallel slits separated by 0.230 mm is illuminated by green light (λ = 546.1 nm). The interference pattern is observed on a screen 1.50 m away from the plane of the parallel slits.
A) Calculate the distance from the central maximum to the first bright region on either side of the central maximum.
B) Calculate the distance between the first and second dark bands in the interference pattern.
Answer:
A) y = 3.56 mm
B) y = 3.56 mm
Explanation:
A) The distance from the central maximum to the first bright region can be found using Young's double-slit equation:
[tex] y = \frac{m\lambda L}{d} [/tex]
Where:
λ: is the wavelength = 546.1 nm
m: is first bright region = 1
L: is the distance between the screen and the plane of the parallel slits = 1.50 m
d: is the separation between the slits = 0.230 mm
[tex] y = \frac{m\lambda L}{d} = \frac{1*546.1 \cdot 10^{-9} m*1.50 m}{0.230 \cdot 10^{-3} m} = 3.56 \cdot 10^{-3} m [/tex]
B) The distance between the first and second dark bands is:
[tex] \Delta y = \frac{\Delta m*\lambda L}{d} [/tex]
Where:
[tex] \Delta m = m_{2} - m_{1} = 2 - 1 = 1 [/tex]
[tex] \Delta y = \frac{1*546.1 \cdot 10^{-9} m*1.50 m}{0.230 \cdot 10^{-3} m} = 3.56 \cdot 10^{-3} m [/tex]
I hope it helps you!
