Answer:
[tex]N = 208 \ turns[/tex]
Explanation:
From the question we are told that
The rate of current change is [tex]\frac{di }{dt} = 0.0200 \ A/s[/tex]
The magnitude of the induced emf is [tex]\epsilon = 12.7 \ mV = 12.7 *10^{-3} \ V[/tex]
The current is [tex]I = 1.50 \ A[/tex]
The average flux is [tex]\phi = 0.00458 \ Wb[/tex]
Generally the number of turns the number of turn the solenoid has is mathematically represented as
[tex]N = \frac{\epsilon_o * I}{ \phi * \frac{di}{dt} }[/tex]
substituting values
[tex]N = \frac{ 12.7*10^{-3} * 1.50 }{ 0.00458 * 0.0200 }[/tex]
[tex]N = 208 \ turns[/tex]
1. (I) If the magnetic field in a traveling EM wave has a peak magnitude of 17.5 nT at a given point, what is the peak magnitude of the electric field
Answer:
The electric field is [tex]E = 5.25 V/m[/tex]
Explanation:
From the question we are told that
The peak magnitude of the magnetic field is [tex]B = 17.5 nT = 17.5 *10^{-9}\ T[/tex]
Generally the peak magnitude of the electric field is mathematically represented as
[tex]E = c * B[/tex]
Where c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
So
[tex]E = 3.0 *10^{8} * 17.5 *10^{-9}[/tex]
[tex]E = 5.25 V/m[/tex]
The peak magnitude of the electric field will be "5.25 V/m".
Magnetic fieldAccording to the question,
Magnetic field's peak magnitude, B = 17.5 nT or,
= 17.5 × 10⁻⁹ T
Speed of light, c = 3.0 × 10⁸ m/s
We know the relation,
→ E = c × B
By substituting the values, we get
= 3.0 × 10⁸ × 17.5 × 10⁻⁹
= 5.25 V/m
Thus the above approach is appropriate.
Find out more information about magnetic field here:
https://brainly.com/question/26257705
The capacitor is originally charged. How does the current I in the ammeter behave as a function of time after the switch is closed?
1. I = 0 always.
2. I = constant, not equal to 0.
3. I increases, then is constant.
4. I instantly jumps up, then slowly decreases.
5. None of the above.
Answer:
The current in the ammeter is zero.
(1) is correct option.
Explanation:
Given that,
The capacitor is charged.
We need find the current after closed switched
We know that,
When switch is closed then the capacitor behave as a short circuit, and the all current flows through it. the current is zero.
Then, the ammeter reads zero.
Hence, The current in the ammeter is zero.
(1) is correct option.
IMPORTANT ANSWER ALL 3 PLEASE!
Answer:
4. Liters
5. Celsius
6. Grams
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
What is the difference between matter and energy
Answer:
Everything in the Universe is made up of matter and energy. Matter is anything that has mass and occupies space. ... Energy is the ability to cause change or do work. Some forms of energy include light, heat, chemical, nuclear, electrical energy and mechanical energy.
Explanation:
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.
Luz, who is skydiving, is traveling at terminal velocity with her body parallel to the ground. She then changes her body position to feet first toward the ground. What happens to her motion? She will continue to fall at the same terminal velocity because gravity has not changed. She will slow down because the air resistance will increase and be greater than gravity. She will speed up because air resistance will decrease and be less than gravity. She will begin to fall in free fall because she will have no air resistance acting on her.
Answer:
Option C - she will speed up because air resistance has reduced and be less than gravity
Explanation:
We are told that Luz is skydiving with terminal velocity and her body parallel to the ground. Now, at this point she will be experiencing a gravitational force acting downwards, and also air resistance as a result of the drag force on her body
Since the downward gravitational force on Luz is constant, she will fall with a net force of;
F_net = F_g - F_d
where;
F_net is the net force on Luz acting downwards
F_g is the gravitational force on Luz
F_d is the drag force on Luz
The drag force on her body is proportional to the surface area of attack.
We are now told that Luz changes her body position to feet first toward the ground. This means that the surface area of attack is reduced because the feet will consume less space than the frontal part of her body. Thus, the drag force will be lesser then before she changed her body position due to reduced air resistance on her body.
Now, from earlier, we saw that;
F_net = F_g - F_d
So, the lesser F_d is, the higher F_net becomes.
Thus, she will speed up because air resistance has reduced and be less than gravity.
Answer:
C
Explanation:
EDGE 2020
A jetboat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m 5, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction to the right when the driver turns on the motor. The boat speeds up for 6.0\,\text s6.0s6, point, 0, start text, s, end text with an acceleration of 4.0\,\dfrac{\text m}{\text s^2}4.0 s 2 m 4, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction leftward.
The question is incomplete. Here is the entire question.
A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?
Answer: Δx = - 42m
Explanation: The jetboat is moving with an acceleration during the time interval, so it is a linear motion with constant acceleration.
For this "type" of motion, displacement (Δx) can be determined by:
[tex]\Delta x = v_{i}.t + \frac{a}{2}.t^{2}[/tex]
[tex]v_{i}[/tex] is the initial velocity
a is acceleration and can be positive or negative, according to the referential.
For Referential, let's assume rightward is positive.
Calculating displacement:
[tex]\Delta x = 5(6) - \frac{4}{2}.6^{2}[/tex]
[tex]\Delta x = 30 - 2.36[/tex]
[tex]\Delta x[/tex] = - 42
Displacement of the boat for t=6.0s interval is [tex]\Delta x[/tex] = - 42m, i.e., 42 m to the left.
Find the total electric potential due to these charges at the point P, whose coordinates are (4.00, 0) m. SOLUTION
Answer:
Some parts of your question is missing attached below is the missing parts and the answer provided is pertaining to your question alone
answer : -6661.59 volts
Explanation:
The total electric potential can be calculated using this relation
V = k [tex](\frac{q1}{r1} + \frac{q2}{r2})[/tex]
q 1 = 1.62 uc
r1 = 4.00 m
q2 = -5.73 uc
r2 = 5.00 m
k = 8.99 * 10^9 N.m^2/c^2
insert the given values into the above equation
V = ( 8.99 * 10^9 ) * [tex](\frac{1.62*10^{-6} }{4} + \frac{-5.73*10^{-6} }{5})[/tex] = -6661.59 volts
A/An ____________________ is a small, flexible tube with a light and lens on the end that is used for examination. Question 96 options:
Answer:
"Endoscope" is the correct answer.
Explanation:
A surgical tool sometimes used visually to view the internal of either a body cavity or maybe even an empty organ like the lung, bladder, as well as stomach. There seems to be a solid or elastic tube filled with optics, a source of fiber-optic light, and sometimes even a sample, epidurals, suction tool, and perhaps other equipment for sample analysis or recovery.Two identical rooms in a house are connected by an open doorway. The temperatures in the two rooms are maintained at different values. Which room contains more air
Answer:
The room with the lower temperature
Explanation:
Using
PV=nRT
Since both the rooms same volume and are connected, so they will have same pressure
PV=nRT=constant
nT=Constant/R=constant
If T is more n has to be less
Thus, lower the temperature, more the number molecules.
A 23 cm tall object is placed in front of a concave mirror with a radius of 37 cm. The distance of the object to the mirror is 86 cm. Calculate the focal length of the mirror.
Answer:
18.5 cm
Explanation:
From;
1/u + 1/v = 1/f
Where;
u= object distance = 86cm
image height = 23 cm
Radius of curvature = 37 cm
The radius of curvature (r) is the radius of the sphere of which the mirror forms a part.
Focal length (f) = radius of curvature (r)/2 = 37cm/2 = 18.5 cm
Therefore, the focal length of the mirror is 18.5 cm
The orbital motion of Earth around the Sun leads to an observable parallax effect on the nearest stars. For each star listed, calculate the distance in parsecs before converting that distance to astronomical units. A. Sirius (0.38") B. Alpha Centauri A (0.75") C. Procyon (0.28") D. Wolf 359 (0.42") E. Epsilon Eridani (0.31") D(pc) = 1/parallax(arcsecs), D(a.u.) = D(pc) * 206265 (arcsecs per radian)
Answer:
Following are the answer to this question:
Explanation:
Formula:
[tex]D(PC) =\frac{1}{parallax}\\\\D(av)=D(PC) \times 20.626\ J[/tex]
Calculating point A:
when the value is [tex]0.38[/tex]
[tex]\to 0.38 \toD(PC)= \frac{1}{0.38}\\\\[/tex]
[tex]=2.632[/tex]
[tex]\to D(a.v) = \frac{1}{0.38} \times 206265\\[/tex]
[tex]=542,802.6[/tex]
Calculating point B:
when the value is [tex]0.75[/tex]
[tex]\to D(PC)=\frac{1}{0.75}[/tex]
[tex]=1.33[/tex]
[tex]\to D(a.v) = \frac{1}{0.75} \times 206265\\[/tex]
[tex]=275,020[/tex]
Calculating point C:
when the value is [tex]0.28[/tex]
[tex]\to D(PC)=\frac{1}{0.28}[/tex]
[tex]=3.571[/tex]
[tex]\to D(a.v) = \frac{1}{0.28} \times 206265\\[/tex]
[tex]=736660.7[/tex]
Calculating point D:
when the value is [tex]0.42[/tex]
[tex]\to D(PC)=\frac{1}{0.42}[/tex]
[tex]=2.38[/tex]
[tex]\to D(a.v) = \frac{1}{0.42} \times 206265\\[/tex]
[tex]=490910.7[/tex]
Calculating point E:
when the value is [tex]0.31[/tex]
[tex]\to D(PC)=\frac{1}{0.31}[/tex]
[tex]=3.226[/tex]
[tex]\to D(a.v) = \frac{1}{0.31} \times 206265\\[/tex]
[tex]=665370.97[/tex]
Some radar systems detect the size and shape of objects such as aircraft and geological terrain. Approximately what is the smallest observable detail (in m) utilizing 495 MHz radar?
Answer:
0.61 m
Explanation:
The smallest observable length by the radar must be at least equal to or greater than the wavelength of the radar.
using the relationship
c = fλ
where
c is the speed of light in vacuum = 3 x 10^8 m/s
f is the frequency of the wave = 495 MHz = 4.95 x 10^8 Hz
λ is the wavelength = ?
λ = c/f = (3 x 10^8)/(4.95 x 10^8) = 0.61 m
answer to your question is 0.6m
A simple arrangement by means of which e.m.f,s. are compared is known
Answer:
A simple arrangement by means of which e.m.f,s. are compared is known as?
(a)Voltmeter
(b)Potentiometer
(c)Ammeter
(d)None of the above
Explanation:
Improved balance is a primary benefit of regular cardiovascular exercise .
Answer:
Cardiovascular exercise is the activity use that aerobic metabolism and cellular reaction.
Explanation:
Cardiovascular exercise is activity increase heart rate and raises oxygen large muscle group of the body.Cardiovascular exercise is that contain cardio improve to the health mental health, heart health.Cardiovascular exercise such as walking, swimming, running is that exercise is benefit to the health.Cardiovascular exercise to the internal body organs that the healthy heart for the function and performance of the heart.Cardiovascular exercise that having involve feet of the ground this type of activity is called high impact of cardio.Cardio is a good and maintaining exercise for the lungs and heart or healthy bones.Cardio exercise is performed that to a water in reduce to the gravity of that pull on the body weight.Cardiovascular daily to build the stronger muscle and that control the blood pressure.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!
Which is a “big idea” for space and time? Energy can be transferred but not destroyed. Forces describe the motion of the universe. The universe is very big and very old. The universe consists of matter.
Answer:
Explanation:
That Universe Consists of Matter
If we compare the force of gravity to strong nuclear force, we could conclude that
O gravity is the weaker force; it is related to mass
O gravity is the stronger force; it is related to distance
strong nuclear is the stronger force; it is related to mass
O strong nuclear is the weaker force; it is related to distance
Answer:
strong nuclear is the stronger force; it is related to mass
Explanation:
If we compare the force of gravity to strong nuclear force, we could conclude that strong nuclear is the stronger force; it is related to mass, therefore the correct answer is option C
What are nuclear forces?The nuclear force is the interaction between the subatomic particles that make up a nucleus. There are two types of nuclear forces: the strong nuclear force and the weak nuclear force. Depending on the separation between the proton neutron and proton pairs, these nuclear forces can be both attracting and positive.
Both types of nuclear forces come under the four fundamental forces of nature. There are mainly four fundamental forces of nature electromagnetic force, gravitational force, strong nuclear force, and weak nuclear force.
Thus, Option C is the appropriate response since, when compared to the force of gravity, the strong nuclear force is the greater force because it is tied to mass.
Learn more about nuclear forces here
brainly.com/question/4346079
#SPJ2
Assume that the speed of light in a vacuum has the hypothetical value of 18.0 m/s. A car is moving at a constant speed of 14.0 m/s along a straight road. A home owner sitting on his porch sees the car pass between two telephone poles in 6.76 s. How much time does the driver of the car measure for his trip between the poles
Answer:
4.245s
Explanation:
Given that,
Hypothetical value of speed of light in a vacuum is 18 m/s
Speed of the car, 14 m/s
Time given is 6.76 s, and we're asked to find the observed time, T
The relationship between the two times can be given as
T = t / √[1 - (v²/c²)]
The missing variable were looking for is t, and we can find it if we rearrange the formula and make t the subject
t = T / √[1 - (v²/c²)]
And now, we substitute the values and insert into the equation
t = 6.76 * √[1 - (14²/18²)]
t = 6.76 * √[1 - (196/324)]
t = 6.76 * √(1 - 0.605)
t = 6.76 * √0.395
t = 6.76 * 0.628
t = 4.245 s
Therefore, the time the driver measures for the trip is 4.245s
Describe and name the different types of collision. In which are the linear momentum and kinetic energy conserved
Answer:
1. Elastic collision
2. Inelastic collision
Explanation:
Elastic collision: collision is said to be elastic if total kinetic energy is not conserved and if there is a rebound after collision
the collision is described by the equation bellow
[tex]m1U1+ m2U2= m1V1+m2V2[/tex]
Inelastic collision: this type of collision occurs when the total kinetic energy of a body is conserved or when the bodies sticks together and move with a common velocity
the collision is described by the equation bellow
[tex]m1U1+ m2U2= V(m1+m2)[/tex]
A thick wire with a radius of 4.0 mm carries a uniform electric current of 1.0 A, distributed uniformly over its cross-section. At what distance from the axis of the wire, and greater than the radius of the wire, is the magnetic field strength equal to that at a distance 2.0 mm from the axis. distance
Answer:
8 mm
Explanation:
From the information given:
The Ampere circuital law can be used to estimate the magnetic field strength at two points when the distance is less than the radius and when the distance is greater than the radius.
when the distance is less than the radius ; we have:
[tex]B_1 = \dfrac{ \mu_o \times I \times r}{2 \times \pi \times \ d^2}[/tex]
when the distance is greater than the radius; we have:
[tex]B_2 = \dfrac{\mu_o \ \times I}{2 \ \times \pi \times R}[/tex]
Equating both equations together ; we have :
[tex]\dfrac{ \mu_o \times I \times r}{2 \times \pi \times \ d^2} = \dfrac{\mu_o \ \times I}{2 \ \times \pi \times R}[/tex]
[tex]\dfrac{1}{R}= \dfrac{r}{d^2}[/tex]
[tex]R= \dfrac{d^2}{r}[/tex]
where; d = radius of the wire and r = distance;
[tex]R =\dfrac{4^2}{2}[/tex]
[tex]R =\dfrac{16}{2}[/tex]
R = 8 mm
A car starts from rest and accelerates with a constant acceleration of 5 m/s2 for 4 s. The car continues for 18 s at constant velocity. How far has the car traveled from its starting point
The linear density rho in a rod 3 m long is 8/ x + 1 kg/m, where x is measured in meters from one end of the rod. Find the average density rhoave of the rod.
Answer:
The average density of the rod is 1.605 kg/m.
Explanation:
The average density of the rod is given by:
[tex] \rho = \frac{m}{l} [/tex]
To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 3, as follows:
[tex] \int_{0}^{3} \frac{8}{3(x + 1)}dx [/tex]
[tex] \rho = \frac{8}{3} \int_{0}^{3} \frac{1}{(x + 1)}dx [/tex] (1)
Using u = x+1 → du = dx → u₁= x₁+1 = 0+1 = 1 and u₂ = x₂+1 = 3+1 = 4
By entering the values above into (1), we have:
[tex] \rho = \frac{8}{3} \int_{0}^{3} \frac{1}{u}du [/tex]
[tex]\rho = \frac{8}{3}*log(u)|_{1}^{4} = \frac{8}{3}[log(4) - log(1)] = 1.605 kg/m[/tex]
Therefore, the average density of the rod is 1.605 kg/m.
I hope it helps you!
The average density of the rod is [tex]1.605 \;\rm kg/m^{3}[/tex].
Given data:
The length of rod is, L = 3 m.
The linear density of rod is, [tex]\rho=\dfrac{8}{x+1} \;\rm kg/m[/tex].
To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 3, The expression for the average density is given as,
[tex]\rho' = \int\limits^3_0 { \rho} \, dx\\\\\\\rho' = \int\limits^3_0 { \dfrac{m}{L}} \, dx\\\\\\\rho' = \int\limits^3_0 {\dfrac{8}{3(x+1)}} \, dx[/tex]............................................................(1)
Using u = x+1
du = dx
u₁= x₁+1 = 0+1 = 1
and
u₂ = x₂+1 = 3+1 = 4
By entering the values above into (1), we have:
[tex]\rho' =\dfrac{8}{3} \int\limits^3_0 {\dfrac{1}{u}} \, du\\\\\\\rho' =\dfrac{8}{3} \times [log(u)]^{4}_{1}\\\\\\\rho' =\dfrac{8}{3} \times [log(4)-log(1)]\\\\\\\rho' =1.605 \;\rm kg/m^{3}[/tex]
Thus, we can conclude that the average density of the rod is [tex]1.605 \;\rm kg/m^{3}[/tex].
Learn more about the average density here:
https://brainly.com/question/1371999
A rock has mass 1.80 kg. When the rock is suspended from the lower end of a string and totally immersed in water, the tension in the string is 10.8 N . What is the smallest density of a liquid in which the rock will float?
Answer:
The density is [tex]\rho_z = 2544 \ kg /m^3[/tex]
Explanation:
From the question we are told that
The mass of the rock is [tex]m_r = 1.80 \ kg[/tex]
The tension on the string is [tex]T = 10.8 \ N[/tex]
Generally the weight of the rock is
[tex]W = m * g[/tex]
=> [tex]W = 1.80 * 9.8[/tex]
=> [tex]W = 17.64 \ N[/tex]
Now the upward force(buoyant force) acting on the rock is mathematically evaluated as
[tex]F_f = W - T[/tex]
substituting values
[tex]F_f = 17.64 - 10.8[/tex]
[tex]F_f = 6.84 \ N[/tex]
This buoyant force is mathematically represented as
[tex]F_f = \rho * g * V[/tex]
Here [tex]\rho[/tex] is the density of water and it value is [tex]\rho = 1000\ kg/m^3[/tex]
So
[tex]V = \frac{F_f}{ \rho * g }[/tex]
[tex]V = \frac{6.84}{ 1000 * 9.8 }[/tex]
[tex]V = 0.000698 \ m^3[/tex]
Now for this rock to flow the upward force (buoyant force) must be equal to the length
[tex]F_f = W[/tex]
[tex]\rho_z * g * V = W[/tex]
Here z is smallest density of a liquid in which the rock will float
=> [tex]\rho_z = \frac{W}{ g * V}[/tex]
=> [tex]\rho_z = \frac{17.64}{ 0.000698 * 9.8}[/tex]
=> [tex]\rho_z = 2544 \ kg /m^3[/tex]
There are 5510 lines per centimeter in a grating that is used with light whose wavelegth is 467 nm. A flat observation screen is located 1.03 m from the grating. What is the minimum width that the screen must have so the centers of all the principal maxima formed on either side of the central maximum fall on the screen
Answer:
1.696 nm
Explanation:
For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,
dsinθ = mλ = (1)λ = λ
dsinθ = λ
sinθ = λ/d.
Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m
From trig ratios 1 + cot²θ = cosec²θ
1 + (1/tan²θ) = 1/(sin²θ)
substituting the values of sinθ and tanθ we have
1 + (D/w)² = (d/λ)²
(D/w)² = (d/λ)² - 1
(w/D)² = 1/[(d/λ)² - 1]
(w/D) = 1/√[(d/λ)² - 1]
w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹ = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.
w is also the distance from the center to the other principal maximum on the other side.
So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm
So, the minimum width of the screen must be 1.696 nm
Coherent light with wavelength 601 nm passes through two very narrow slits, and the interference pattern is observed on a screen a distance of 3.00 m from the slits. The first-order bright fringe is a distance of 4.84 mm from the center of the central bright fringe. For what wavelength of light will thefirst-order dark fringe be observed at this same point on the screen?
Answer:
The wavelength is [tex]\lambda = 1805 nm[/tex]
Explanation:
From the question we are told that
The wavelength of the light is [tex]\lambda = 601 \ nm = 601 *10^{-9} \ m[/tex]
The distance of the screen is D = 3.0 m
The fringe width is [tex]y = 4.84 \ mm = 4.84 *10^{-3} \ m[/tex]
Generally the fringe width for a bright fringe is mathematically represented as
[tex]y = \frac{ \lambda * D }{d }[/tex]
=> [tex]d = \frac{ \lambda * D }{ y }[/tex]
=> [tex]d = \frac{ 601 *10^{-9} * 3}{ 4.84 *10^{-3 }}[/tex]
=> [tex]d = 0.000373 \ m[/tex]
Generally the fringe width for a dark fringe is mathematically represented as
[tex]y_d = [m + \frac{1}{2} ] * \frac{\lambda D }{d }[/tex]
Here m = 0 for first order dark fringe
So
[tex]y_d = [0 + \frac{1}{2} ] * \frac{\lambda D }{d }[/tex]
looking at which we see that [tex]y_d = y[/tex]
[tex]4.84 *10^{-3} = [0 + \frac{1}{2} ] * \frac{\lambda * 3 }{ 0.000373 }[/tex]
=> [tex]\lambda = 1805 *10^{-9} \ m[/tex]
=> [tex]\lambda = 1805 nm[/tex]
a transformer changes 95 v acorss the primary to 875 V acorss the secondary. If the primmary coil has 450 turns how many turns does the seconday have g
Answer:
The number of turns in the secondary coil is 4145 turns
Explanation:
Given;
the induced emf on the primary coil, [tex]E_p[/tex] = 95 V
the induced emf on the secondary coil, [tex]E_s[/tex] = 875 V
the number of turns in the primary coil, [tex]N_p[/tex] = 450 turns
the number of turns in the secondary coil, [tex]N_s[/tex] = ?
The number of turns in the secondary coil is calculated as;
[tex]\frac{N_p}{N_s} = \frac{E_p}{E_s}[/tex]
[tex]N_s = \frac{N_pE_s}{E_p} \\\\N_s = \frac{450*875}{95} \\\\N_s = 4145 \ turns[/tex]
Therefore, the number of turns in the secondary coil is 4145 turns.
A charged particle enters a magnetic field with an angle theta If theta equals 90 degrees what bath it will follow - If theta larger than zero and less than 90 degrees what path will it follow?
Given that,
A charged particle enters a magnetic field with an angle theta .
If [tex]\theta=90^{\circ}[/tex]
We know that,
If the angle is 90° then the charged particle enters perpendicular to the B.
B is magnetic field.
The charged particle will be follow of the circular path.
If the angle is greater than 0 and less than 90° then the charged particle will be show the helical path.
Hence, This is required answer.
wo 10-cm-diameter charged rings face each other, 25.0 cm apart. Both rings are charged to + 20.0 nC . What is the electric field strength
Complete question:
Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field strength at:
a) the midpoint between the two rings?
b) the center of the left ring?
Answer:
a) the electric field strength at the midpoint between the two rings is 0
b) the electric field strength at the center of the left ring is 2712.44 N/C
Explanation:
Given;
distance between the two rings, d = 25 cm = 0.25 m
diameter of each ring, d = 10 cm = 0.1 m
radius of each ring, r = [tex]\frac{0.1}{2} = 0.05 \ m[/tex]
the charge on each ring, q = 20 nC
Electric field strength for a ring with radius r and distance x from the center of the ring is given as;
[tex]E = \frac{kxQ}{(x^2 +r^2)^{3/2}}[/tex]
The electric field strength at the midpoint;
the distance from the left ring to the mid point , x = 0.25 m / 2 = 0.125 m
[tex]E = \frac{kxQ}{(x^2 +r^2)^{3/2}} \\\\E = \frac{8.99*10^{9}*0.125*20*10^{-9}}{(0.125^2 + 0.05^2)^{3/2}} \\\\E = 9210.5 \ N/C[/tex]
[tex]E_{left} = 9210.5 \ N/C[/tex]
The electric field strength due to right ring is equal in magnitude to left ring but opposite in direction;
[tex]E_{right} = -9210.5 \ N/C[/tex]
The electric field strength at the midpoint;
[tex]E_{mid} = E_{left} + E_{right}\\\\E_{mid} = 9210.5 \ N/C - 9210.5 \ N/C\\\\E_{mid} = 0[/tex]
(b)
The distance from the right ring to center of the left ring, x = 0.25 m.
[tex]E = \frac{KxQ}{(x^2 +r^2)^{3/2}} \\\\E = \frac{8.99*10^{9} *0.25*20*10^{-9}}{(0.25^2 + 0.05^2)^{3/2}} \\\\E = 2712.44 \ N/C[/tex]
