Answer:
speed when it reaches y = 4.00cm is
v = 14.9 g.m/s
Explanation:
given
q₁=q₂ =2.00 ×10⁻⁶
distance along x = 3.00cm= 3×10⁻²
q₃= 4×10⁻⁶C
mass= 10×10 ⁻³g
distance along y = 4×10⁻²m
r₁ = [tex]\sqrt{3^{2} +2^{2} }[/tex] = [tex]\sqrt{13}[/tex] = 3.61cm = 0.036m
r₂ = [tex]\sqrt{4^{2} + 3^{2} }[/tex] = [tex]\sqrt{25}[/tex] = 5cm = 0.05m
electric potential V = [tex]\frac{kq}{r}[/tex]
change in potential ΔV = [tex]V_{1} - V_{2}[/tex]
ΔV = [tex]\frac{2kq_{1} }{r_{1}} - \frac{2kq_{2} }{r_{2} }[/tex] , where [tex]q_{1} = q_{2}=[/tex]2.00μC
ΔV = [tex]2kq(\frac{1}{r_{1}} - \frac{1}{r_{2} })[/tex]
ΔV = 2 × 9×10⁹ × 2×10⁻⁶ × [tex](\frac{1}{0.036} - \frac{1}{0.05} )[/tex]
ΔV= 2.789×10⁵
[tex]\frac{1}{2}mv^{2}[/tex] = ΔV × q₃
[tex]\frac{1}{2}[/tex] ˣ 10×10⁻³ ×v² = 2.789×10⁵× 4 ×10⁻⁶
v² = 223.12 g.m/s
v = 14.9 g.m/s
The speed of the charge q₃ when it starts from rest at y = 2 cm and reaches y = 4 cm is; v = 14.89 m/s
We are given;
Charge 1; q₁ = 2.00 μC = 2 × 10⁻⁶ C
Charge 2; q₂ = 2.00 μC = 2 × 10⁻⁶ C
Distance of charge 1 along x = 3 cm = 3 × 10⁻² m
Distance of charge 2 along x = -3 cm = -3 × 10⁻² m
Charge 3; q₃ = +4.00 μC = 4 × 10⁻⁶ C
mass; m = 0.01 g
distance of charge 3 along y = 4 cm = 4 × 10⁻² m
q₃ starts from rest at y = 2 × 10⁻² m and reaches y = 4 × 10⁻² m.
Thus;
Distance of charge 1 from the initial position of q₃;
r₁ = √((3 × 10⁻²)² + ((2 × 10⁻²)²)
r₁ = 0.0361 m
Distance of charge 2 from the final position of q₃;
r₂ = √((3 × 10⁻²)² + ((4 × 10⁻²)²)
r₂ = 0.05 m
Now, formula for electric potential is;
V = kq/r
Where k = 9 × 10⁹ N.m²/s²
Thus,change in potential is;
ΔV = V₁ - V₂
Now, Net V₁ = 2kq₁/r₁
Net V₂ = 2kq₂/r₂
Thus;
ΔV = 2kq₁/r₁ - 2kq₂/r₂
ΔV = (2 × 9 × 10⁹)[(2 × 10⁻⁶/0.0361) - (2 × 10⁻⁶/0.05)]
ΔV = 277229.92 V
Now, from conservation of energy;
½mv² = q₃ΔV
Thus;
½ × 0.01 × v² = 4 × 10⁻⁶ × 277229.92
v² = 2 × 4 × 10⁻⁶ × 277229.92/0.01
v = √(221.783936)
v = 14.89 m/s
Read more about point charges at;https://brainly.com/question/13914561
An alternating current is supplied to an electronic component with a warning that the voltage across it should never exceed 12 V. What is the highest rms voltage that can be supplied to this component while staying below the voltage limit in the warning?
Answer:
The highest rms voltage will be 8.485 V
Explanation:
For alternating electric current, rms (root means square) is equal to the value of the direct current that would produce the same average power dissipation in a resistive load
If the peak or maximum voltage should not exceed 12 V, then from the relationship
[tex]V_{rms} = \frac{V_{p} }{\sqrt{2} }[/tex]
where [tex]V_{rms}[/tex] is the rms voltage
[tex]V_{p}[/tex] is the peak or maximum voltage
substituting values into the equation, we'll have
[tex]V_{rms} = \frac{12}{\sqrt{2} }[/tex] = 8.485 V
The voltage across the terminals of an ac power supply varies with time according to V=V0cos(t). The voltage amplitude is V0 = 41.0V .
A. What is the root-mean-square potential difference Vrms?
B. What is the average potential difference Vav between the two terminals of the power supply?
Answer:
A) V_rms = 29 V
B) Vav = 0 V
Explanation:
A) We are told that;
V = V_o cos ωt
voltage amplitude; V = V_o = 41.0V
Now, the formula for the root-mean-square potential difference Vrms is given as;
V_rms = V/√2
Thus plugging in relevant values, we have;
V_rms = 41/√2
V_rms = 29 V
B) Due to the fact that the voltage is sinusoidal from the given V = V_o cos ωt, we can say that the average potential difference Vav between the two terminals of the power supply would be zero.
Thus; Vav = 0 V
A. The root-mean-square potential difference ([tex]V_{rms}[/tex]) is equal to 28.99 Volts.
B. For this voltage with a sinusoidal waveform (sine wave), the average potential difference ([tex]V_{ave}[/tex]) between the two terminals of the power supply is equal to zero (0).
Given the following data:
Voltage amplitude = 41.0 Volts.The voltage across the terminals of an alternating current (AC) power supply varies directly with time according to the equation:
[tex]V_0 = V_0cos(t)[/tex]
A. To find the root-mean-square potential difference ([tex]V_{rms}[/tex]):
Mathematically, root-mean-square for voltage in an alternating current (AC) power supply (circuit) is given by the formula:
[tex]V_{rms} = \frac{V}{\sqrt{2} }[/tex]
Substituting the given parameter into the formula, we have;
[tex]V_{rms} = \frac{41}{\sqrt{2} }\\\\V_{rms} = \frac{41}{1.4142 }\\\\V_{rms} = 28.99\; Volts[/tex]
B. To find the average potential difference ([tex]V_{ave}[/tex]) between the two terminals of the power supply:
For this voltage with a sinusoidal waveform (sine wave), the average potential difference ([tex]V_{ave}[/tex]) between the two terminals of the power supply is equal to zero (0).
Read more: https://brainly.com/question/15121836
A small barge is being used to transport trucks across a river. If the barge is 10.00 m long by 8.00 m wide and sinks an additional 3.75 cm into the river when a loaded truck pulls onto it, determine the weight of the truck and load.
Answer: Weight truck+load = 29.4×[tex]10^{3}[/tex] N
Explanation: When an object floats in a fluid, there is an upward force, caused by the liquid, acting on the object that opposes the weight of the immersed object. This force is called Buyoyant Force and is determined by:
B = d*V*g
where
d is density of the fluid;
V is volume of liquid displaced due to the immersed object;
g is acceleration due to gravity;
For the truck, the system is in equilibrium, which means buyoyant force is equal weight. Then:
Volume displaced is
V = 10*8*0.0375
V = 3 [tex]m^{3}[/tex]
Density of water: 1000kg/[tex]m^{3}[/tex]
[tex]F_{P} = F_{B}[/tex]
[tex]F_{P}[/tex] = 1000*3*9.8
[tex]F_{P}[/tex] = 29.4×[tex]10^{3}[/tex] N
The weight of the truck and the load is 29.4×[tex]10^{3}[/tex] Newtons
A soap bubble is 115 nm thick and illuminated by white light incident perpendicular to its surface. What wavelength (in nm) and color of visible light is most constructively reflected, assuming the same index of refraction as water (nw = 1.33)?
Answer:
The wavelength is [tex]\lambda = 612nm[/tex] and the color is Orange
Explanation:
from the question we are told that
The thickness is [tex]D = 115 nm = 115 *10^{-9} \ m[/tex]
The refractive index of water is [tex]n_w = 1.33[/tex]
Generally the condition for constrictive interference is
[tex]2 * D = \frac{\lambda _n}{2}[/tex]
Where [tex]\lambda _n[/tex] is the wavelength of light in a particular medium
Now considering the medium of water(soap bubble )
The wavelength of light in this medium is mathematically represented as
[tex]\lambda = \frac{\lambda }{n }[/tex]
So
[tex]2 * D = \frac{\frac{\lambda }{n} }{2}[/tex]
[tex]2 * D = \frac{\lambda }{2 * n }[/tex]
=> [tex]\lambda = 4 *n * D[/tex]
substituting values
[tex]\lambda = 4 *1.33 * 115*10^{-9}[/tex]
[tex]\lambda = 6.118 *10^{-7} \ m[/tex]
[tex]\lambda = 612nm[/tex]
The color is orange because the wavelength range of yellow is
590–625 nm
a football is kicked toward a goal keeper with an initial speed of 20m/s at an angle of 45 degrees with the horizontal .at the moment the ball is kicked the goal keeper is 50m from the player .at what speed and in what direction must the goalkeeper run in order to catch the ball at the same height at which it was kicked
Answer:
3.18 m/s
Explanation:
Given that
Initial speed of the ball, u = 20 m/s
Angle of inclination, θ = 45°
Distance from the ball, h = 50 m
Using equations of projectile to solve this, we have
We start by finding the time of flight, T
T = 2Usinθ/g
T = (2 * 20 * sin45)/9.8
T = (40 * 0.7071) / 9.8
T = 28.284/9.8
T = 2.89 s
Next we find the Range, R
R = u²sin2θ/g
R = (20² * sin 90) / 9.8
R = (400 * 1) / 9.8
R = 400/9.8 = 40.82 m
Distance the gk must cover
40.82 - 50 m
-9.18 m or 9.18 m in the opposite direction.
Speed of the GK = d/t
9.18 / 2.89 = 3.18 m/s
The angle between the axes of two polarizing filters is 41.0°. By how much does the second filter reduce the intensity of the light coming through the first?
Answer:
The amount by which the second filter reduces the intensity of light emerging from the first filter is
z = 0.60
Explanation:
From the question we are told that
The angle between the axes is [tex]\theta = 41^o[/tex]
The intensity of polarized light that emerges from the second filter is mathematically represented as
[tex]I= I_o cos^2 \theta[/tex]
Where [tex]I_o[/tex] is the intensity of light emerging from the first filter
[tex]I = I_o [cos(41.0)]^2[/tex]
[tex]I =0.60 I_o[/tex]
This means that the second filter reduced the intensity by z = 0.60
•• A metal sphere carrying an evenly distributed charge will have spherical equipotential surfaces surrounding it. Suppose the sphere’s radius is 50.0 cm and it carries a total charge of (a) Calculate the potential of the sphere’s surface. (b)You want to draw equipotential surfaces at intervals of 500 V outside the sphere’s surface. Calculate the distance between the first and the second equipotential surfaces, and between the 20th and 21st equipotential surfaces. (c) What does the changing spacing of the surfaces tell you about the electric field?
Answer:
Explanation:
For this exercise we will use that the potential is created by the charge inside the equinoctial surface and just like in Gauss's law we can consider all the charge concentrated in the center.
Therefore the potential on the ferric surface is
V = k Q / r
where k is the Coulomb constant, Q the charge of the sphere and r the distance from the center to the point of interest
a) On the surface the potential
V = 9 10⁹ Q / 0.5
V = 18 10⁹ Q
Unfortunately you did not write the value of the load, suppose a value to complete the calculations Q = 1 10⁻⁷ C, with this value the potential on the surfaces V = 1800 V
b) The equipotential surfaces are concentric spheres, let's look for the radii for some potentials
for V = 1300V let's find the radius
r = k Q / V
r = 9 109 1 10-7 / 1300
r = 0.69 m
other values are shown in the following table
V (V) r (m)
1800 0.5
1300 0.69
800 1,125
300 3.0
In other words, we draw concentric spheres with these radii and each one has a potential difference of 500V
C) The spacing of the spheres corresponds to lines of radii of the electric field that have the shape
E = k Q / r²
Now, let's see what happens when the cannon is high above the ground. Click on the cannon, and drag it upward as far as it goes (15 m above the ground). Set the initial velocity to 14 m/s, and fire several pumpkins while varying the angle. For what angle is the range the greatest?
choices:
A. 45∘
B. 20∘
C. 30∘
D. 40∘
E. 50∘
Answer:
B. 20°Explanation:
Range in projectile is defined as the distance covered in the horizontal direction. It is expressed as R = U²sin2Ф/g
U is the initial velocity of the body (in m/s)
Ф is the angle of projection
g is the acceleration due to gravity.
Given U = 14m/s, g = 9.8m/s and range R = 15 m
we will substitute this value into the formula to get the projection angle Ф as shown;
15 = 15²sin2Ф/9.8
15*9.8 = 15²sin2Ф
147 = 225sin2Ф
sin2Ф = 147/225
sin2Ф = 0.6533
2Ф = sin⁻¹0.6533
2Ф = 40.79°
Ф = 40.79°/2
Ф = 20.39° ≈ 20°
Hence, the range is greatest at angle 20°
The nonreflective coating on a camera lens with an index of refraction of 1.21 is designed to minimize the reflection of 570-nm light. If the lens glass has an index of refraction of 1.52, what is the minimum thickness of the coating that will accomplish this ta
Answer: 117.8 nm
Explanation:
Given,
Nonreflective coating refractive index : n = 1.21
Index of refraction: [tex]n_0[/tex] = 1.52
Wave length of light = λ = 570 nm = [tex]570\times10^{-9}\ m[/tex]
[tex]\text{ Thickness}=\dfrac{\lambda}{4n}[/tex]
[tex]=\dfrac{570\times10^{-9}\ m}{4\times1.21}\\\\\approx\dfrac{117.8\times 10^{-9}\ m}{1}\\\\=117.8\text{ nm}[/tex]
Hence, the minimum thickness of the coating that will accomplish= 117.8 nm
If an astronomer wants to find and identify as many stars as possible in a star cluster that has recently formed near the surface of a giant molecular cloud (such as the Trapezium cluster in the Orion Nebula), what instrument would be best for her to use
Answer:
Infrared telescope and camera
Explanation:
An infrared telescope uses infrared light to detect celestial bodies. The infrared radiation is one of the known forms of electromagnetic radiation. Infrared radiation is given off by a body possessing some form of heat. All bodies above the absolute zero temperature in the universe radiates some form of heat, which can then be detected by an infrared telescope, and infrared radiation can be used to study or look into a system that is void of detectable visible light.
Stars are celestial bodies that are constantly radiating heat. In order to see a clearer picture of the these bodies, Infrared images is better used, since they are able to penetrate the surrounding clouds of dust, and have located many more stellar components than any other types of telescope, especially in dusty regions of star clusters like the Trapezium cluster.
Exercise 1 - Questions 1. Hold the grating several inches from your face, at an angle. Look at the grating that you will be using. Record what details you see at the grating surface. 0 Words 2. Hold the diffraction grating up to your eye and look through it. Record what you see. Be specific. 0 Words 3. Before mounting the diffraction grating, look through the opening that you made for your grating. Record what you see across the back of your spectroscope.
Answer:
1) on the surface you can see the slits with equal spacing, on the one hand and on the other hand it is smooth.
2)If the angle is zero we see a bright light called undispersed light
For different angles we see the colors of the spectrum
3) must be able to see the well-collimated light emission source
Explanation:
1) A diffraction grating (diffraction grating) is a surface on which a series of indentations are drawn evenly spaced.
These crevices or lines are formed by copying a standard metal net when the plastic is melted and after hardening is carefully removed, or if the nets used are a copy of the master net.
The network can be of two types of transmission or reflection, in teaching work the most common is the transmission network, on the surface you can see the slits with equal spacing, on the one hand and on the other hand it is smooth.
The number of lines per linear mm determines which range of the spectrum a common value can be observed to observe the range of viable light is 600 and 1200 lines per mm.
2) when looking through the diffraction grating what we can observe depends on the relative angle between the eye and the normal to the network.
If the angle is zero we see a bright light called undispersed light
For different angles we see the colors of the spectrum, if it is an incandescent lamp we see a continuum with all the colors in the visible range and if it is a gas lamp we see the characteristic emission lines of the gas.
3) Before mounting the grid on the spectrometer, we must be able to see the well-collimated light emission source, this means that it is clearly observed.
The spectrometers have several screws to be able to see the lamp clearly, this is of fundamental importance in optical experiments.
A crate is given a big push, and after it is released, it slides up an inclined plane which makes an angle 0.44 radians with the horizontal. The frictional coefficients between the crate and plane are (\muμs = 0.61, \muμk = 0.23 ). What is the magnitude of the acceleration (in meters/second2) of this crate as it slides up the incline?
Answer:
The acceleration is [tex]a = 6.2 m/s^2[/tex]
Explanation:
From the question we are told that
The angle which the inclined plane make with horizontal is [tex]\theta = 0.44 \ rad[/tex]
The frictional coefficients are [tex]\mu_{\mu s} = 0.61[/tex] and [tex]\mu_{\mu k} = 0.23[/tex]
The force acting on the crate is mathematically represented as
[tex]f = F_w + F_N[/tex]
Here f is the net force at which the crate is sliding down the plane which is mathematically represented as
[tex]f = ma[/tex]
[tex]F_w[/tex] is the force due to weight which is mathematically represented as
[tex]F_w = mg sin (\theta)[/tex]
and [tex]F_N[/tex] the force due to friction which is mathematically represented as
[tex]F_N = \mu_{\mu k } * mg cos(\theta )[/tex]
So
[tex]ma = mgsin(\theta ) + \mu_{\mu k} mg cos(\theta )[/tex]
[tex]a = gsin(\theta ) + \mu_{\mu k } * g cos(\theta)[/tex]
substituting values
[tex]a = 9.8 sin(0.44 ) + 0.23 * 9.8* cos(0.44)[/tex]
[tex]a = 6.2 m/s^2[/tex]
A 3 kg rock is swung in a circular path and in a vertical plane on a 0.25 m length string. At the top of the path, the angular velocity is 11 rad/s. What is the tension in the string at that point
Answer:
The tension in the string at that point is 90.75 N
Explanation:
Given;
mass of the object, m = 3 kg
length of string, r = 0.25 m
the angular velocity, ω = 11 rad/s
The tension on string can be equated to the centrifugal force on the object;
T = mω²r
Where;
T is the tension in the string
m is mass of the object
ω is the angular velocity
r is the radius of the circular path
T = 3 x (11)² x 0.25
T = 90.75 N
Therefore, the tension in the string at that point is 90.75 N
what is the mass of an oil drop having two extra electrons that is suspended motionless by the field between the plates
Answer:
m = 3,265 10⁻²⁰ E
Explanation:
For this exercise we can use Newton's second law applied to our system, which consists of a capacitor that creates the uniform electric field and the drop of oil with two extra electrons.
∑ F = 0
[tex]F_{e}[/tex] - W = 0
the electric force is
F_{e} = q E
as they indicate that the charge is two electrons
F_{e} = 2e E
The weight is given by the relationship
W = mg
we substitute in the first equation
2e E = m g
m = 2e E / g
let's put the value of the constants
m = (2 1.6 10⁻¹⁹ / 9.80) E
m = 3,265 10⁻²⁰ E
The value of the electric field if it is a theoretical problem must be given and if it is an experiment it can be calculated with measures of the spacing between plates and the applied voltage, so that the system is in equilibrium
Consider a house at sea level that has 2500 square feet of floor area. What is the total force that the air inside the house exerts upward on the ceiling
Answer:
Total Force that the air inside the house exerts upward on the ceiling is 5.25 × 10⁶ lb
Explanation:
Force = Atmospheric Pressure × Floor Area
Where; Standard Atmospheric Pressure = 2100 lb/ft²
Floor Area = 2500 ft²
Substitute the data
∴ Total Force = 2100 lb/ft² × 2500 ft²
Total Force = 5.25 × 10⁶ lb
A string on the violin has a length of 24.20 cm and a mass of 0.0992 g. The fundamental frequency of the string is 659.3 Hz.
Required:
a. What is the speed of the wave on the string?
b. What is the tension in the string?
Answer:
a. The speed of the wave is 319.1m
b. The tension in the string is 41.74N
Explanation:
Please see the attachments below
Which of the following statements about Masters programs is not correct?
A. Most Masters athletes did not compete when they were in school.
B. The social life is as important as the athletics on most Masters
teams.
C. The level of competition is not very high in most Masters
programs.
D. Masters programs allow adults to work out and socialize with
people who share their love of a sport.
SUBMIT
The correct answer is C. The level of competition is not very high in most Masters programs.
Explanation:
In sports, the word "master" is used to define athletes older than 30 and that usually are professional or have trained for many years, although novates are also allowed. This means in most cases in Master programs and teams a high level of competition can be expected due to the experience and extensive training of Master athletes. Indeed, many records in the field of sport belong to Master athletes rather than younger athletes. According to this, the incorrect statement is "The level of competition is not very high in most Masters programs".
Which equations are used to calculate the velocity of a wave? velocity = distance × time velocity = wavelength × frequency velocity = distance/time velocity = wavelength/frequency velocity = distance/time velocity = wavelength × frequency velocity = distance × time velocity = wavelength/frequency
Answer:
velocity = distance/time
velocity = wavelength × frequency
Both of these are commonly known equations to calculate velocity with different variables.
Scattered light in the atmosphere is often partially polarized. The best way to determine whether or not light from a particular direction in the sky shows polarization is to
Answer:
Rotate a piece of polaroid film about an axis perpendicular to the ray while looking through it in that sky direction.
Explanation:
Polarization involves constraining a transverse wave e.g light waves to vibrate in one phase only. Since unpolarized light vibrates in all direction during propagation. Polarization can be achieved by a polaroid.
A polaroid is a material the make transverse waves to vibrate in one direction after passing through it. It has various applications in sun glasses, wind shield of a car etc.
If the slit of the polaroid is perpendicular to the polarized light from a particular direction in the sky, there would be no propagation of the light. But when it is parallel to the polarized light from the direction, the light would propagate through the polaroid.
a small bar magnet is suspended horizontally by a string. When placed in a uniform horizontal magnetic field, it will
Answer:
It will neither translate in the opposite direction nor .rotate so as to be at right angles, it will also neither rotate so as to be vertical direction
Which of the following options is correct and why?
Consider a spherical Gaussian surface of radius R centered at the origin. A charge Q is placed inside the sphere. To maximize the magnitude of the flux of the electric field through the Gaussian surface, the charge should be located
(a) at x = R/2, y = 0, z = 0.
(b) at the origin.
(c) at x = 0, y = 0, z = R/2.
(d) at x = 0, y = R/2, z = 0.
(e) The charge can be located anywhere since flux does not depend on the position of the charge as long as it is inside the sphere.
Answer:
Option (e) = The charge can be located anywhere since flux does not depend on the position of the charge as long as it is inside the sphere.
Explanation:
So, we are given the following set of infomation in the question given above;
=> "spherical Gaussian surface of radius R centered at the origin."
=> " A charge Q is placed inside the sphere."
So, the question is that if we are to maximize the magnitude of the flux of the electric field through the Gaussian surface, the charge should be located where?
The CORRECT option (e) that is " The charge can be located anywhere since flux does not depend on the position of the charge as long as it is inside the sphere." Is correct because of the reason given below;
REASON: because the charge is "covered" and the position is unknown, the flux will continue to be constant.
Also, the Equation that defines Gauss' law does not specify the position that the charge needs to be located, therefore it can be anywhere.
The velocity of an object is given by the following function defined on a specified interval. Approximate the displacement of the object on this interval by sub-dividing the interval into the indicated number of sub-intervals. Use the left endpoint of each sub-interval to compute the height of the rectangles.
v= 4t + 5(m/s) for 3 < t < 7; n = 4
The approximate displacement of the object is______m.
Answer:
The approximate displacement of the object is 23 m.
Explanation:
Given that:
v = 4t + 5 (m/s) for 3< t< 7; n= 4
The approximate displacement of the object can be calculated as follows:
The velocities at the intervals of t are :
3
4
5
6
the velocity at the intervals of t = 7 will be left out due the fact that we are calculating the left endpoint Reimann sum
n = 4 since there are 4 values for t, Then there is no need to divide the velocity values
v(3) = 4(3)+5
v(3) = 12+5
v(3) = 17
v(4)= 4(4)+5
v(4) = 16 + 5
v(4) = 21
v(5)= 4(5)+5
v(5) = 20 + 5
v(5) = 25
v(6) = 4(6)+5
v(6) = 24 + 5
v(6) = 29
Using Left end point;
[tex]= \dfrac{1}{4}(17+21+25+29)[/tex]
= 23 m
Which two types of electromagnetic waves have higher frequencies than the waves that make up ultraviolet light?
radio waves and infrared light
visible light and X-rays
microwaves and gamma rays
gamma rays and X-rays
The two types of electromagnetic waves that have higher frequencies than the waves that make up ultraviolet light are gamma rays and X-rays.
WHAT ARE ELECTROMAGNETIC WAVES?Electromagnetic waves are components of the electromagnetic spectrum, which is made up of the following:
Radio wavesInfraredUltravioletVisible lightX-raysGamma raysmicrowaveEach electromagnetic wave have a specific frequency and wavelength.
However, the two types of electromagnetic waves that have higher frequencies than the waves that make up ultraviolet light are gamma rays and X-rays.
Learn more about electromagnetic waves at: https://brainly.com/question/8553652
Answer:
gamma rays and X-rays
Explanation:
d on edge I got 100%
Find the rms (a) electric and (b)magnetic fields at a point 2.00 m from a lightbulb that radiates 90.0 W of light uniformly in all directions.
Answer:
a) rms of electric field =
[tex]E_{rms}[/tex]= 25.97 V/m
b) rms of magnetic field
[tex]B_{rms}[/tex] = 8.655 × 10⁻⁸
[tex]B_{rms}[/tex] = 86.55nT
Explanation:
given
power p = 90.0W
distance d = 2.00m
Intensity = [tex]\frac{power}{area}[/tex]
I = [tex]\frac{p}{A}[/tex]
A = [tex]4\pi d^{2}[/tex]
I = [tex]\frac{p}{4\pi d^{2} }[/tex]
I = [tex]\frac{90}{4\pi(2^{2}) }[/tex]
I = 1.79 W/m²
a) [tex]I_{ave}[/tex] = ε₀ × [tex]E^{2} _{rms}[/tex] × c
where ε₀ is permittivity of free space = 8.85×10⁻¹², [tex]E^{2} _{rms}[/tex] is the root mean value and c is speed of light = 3×10⁸m/s
1.79 = 8.85×10⁻¹² × [tex]E^{2} _{rms}[/tex] × 3×10⁸
[tex]E^{2} _{rms}[/tex] = [tex]\frac{1.79}{8.85x10^{-12} x 3x10^{8} }[/tex]
[tex]E^{2} _{rms}[/tex]= 674.1996
[tex]E_{rms}[/tex]= 25.97 V/m
b)for rems magnetic field
[tex]E_{rms}[/tex]= c [tex]B_{rms}[/tex]
[tex]B_{rms}[/tex] = [tex]\frac{E_{rms} }{c}[/tex]
[tex]B_{rms}[/tex] = [tex]\frac{25.97 V/m}{3x10^{8} }[/tex]
[tex]B_{rms}[/tex] = 8.655 × 10⁻⁸
[tex]B_{rms}[/tex] = 86.55nT
An FM radio station transmits a signal with a frequency of 89.1 MHz. Give the wavelength in meters. (use at least three significant digits)
Answer:
3m
Explanation:
89.1 MHz means
89.1×10^6 cycles/second.
Electromagnetic radiation (including radio waves) travel at
3.0×10^8meters/second
Wavelength = Speed/Frequency
The wavelength of a
89.1MHz radio signal is
3.0×10^8/89.1x10^6
= 0.03x10^2
= 3meters
An electron is accelerated through 2.35 103 V from rest and then enters a uniform 2.30-T magnetic field. (a) What is the maximum magnitude of the magnetic force this particle can experience
the maximum magnitude is 5.5
If you were in a smooth-riding train with no windows, could you sense the difference between uniform motion and rest or between accelerated motion and rest?
1. Both acclerated and uniform motion can be sensed.
2. Only uniform motion can be sensed.
3. Only accelerated motion can be sensed.
4. No motion can be sensed.
Answer:
3. Only accelerated motion can be sensed
Explanation:
Without windows on such a train, you'd have no frame of reference for your speed. By that I mean, without being able to see how fast you are moving past other things, it's almost as if you aren't moving at all... almost.
At rest you obviously aren't moving and in uniform motion, with a constant speed, it would feel as though you aren't moving. But during periods of acceleration you'll feel the force on your body (F=ma) and would be able to tell if you were moving in a particular direction.
You've probably felt this before. Maybe not on a windowless train but perhaps in a car or on a roller coaster. Speeding up makes you go back into your seat a bit and slowing down makes you lean forward a bit. Both speeding up and slowing down are examples of acceleration (just in different directions) and how fast you accelerate will affect how much force you experience.
So the answer would be option 3.
Side note: If the train wasn't smooth riding then there would be some amount of friction going on and you could probably tell if you were in motion by the products of that friction (like sound and vibrations) even at a constant speed.
Suppose a 185 kg motorcycle is heading toward a hill at a speed of 29 m/s. The two wheels weigh 12 kg each and are each annular rings with an inner radius of 0.280 m and an outer radius of 0.330 m.
Randomized Variables
m = 185 kg
v = 29 m/s
h = 32 m
A. How high can it coast up the hill. if you neglect friction in m?
B. How much energy is lost to friction if the motorcycle only gains an altitude of 33 m before coming to rest?
Answer:
a) Height reached before coming to rest is 42.86 m
b) Energy lost to friction is 17902.45 J
Explanation:
mass of the motorcycle = 185 kg
speed of the towards the hill = 29 m/s
The wheels weigh 12 kg each
Wheels are annular rings with an inner radius of 0.280 m and outer radius of 0.330 m
a) To go up the hill, the kinetic energy of motion of the motorcycle will be converted to the potential energy it will gain in going up a given height
the kinetic energy of the motorcycle is given as
[tex]KE[/tex] = [tex]\frac{1}{2}mv^{2}[/tex]
where m is the mass of the motorcycle
v is the velocity of the motorcycle
[tex]KE[/tex] = [tex]\frac{1}{2}*185*29^{2}[/tex] = 77792.5 J
This will be converted to potential energy
The potential energy up the hill will be
[tex]PE[/tex] = mgh
where m is the mass
g is acceleration due to gravity 9.81 m/s^2
h is the height reached before coming to rest
[tex]PE[/tex] = 185 x 9.81 x m = 1814.85h
equating the kinetic energy to the potential energy for energy conservation, we'll have
77792.5 = 1814.85h
height reached before coming to rest = 77792.5/1814.85 = 42.86 m
b) if an altitude of 33 m was reached before coming to rest, then the potential energy at this height is
[tex]PE[/tex] = mgh
[tex]PE[/tex] = 185 x 9.81 x 33 = 59890.05 J
The energy lost to friction will be the kinetic energy minus this potential energy.
energy lost = 77792.5 - 59890.05 = 17902.45 J
A) The motorcycle can coast up the hill by ; 42.86m
B) The amount of energy lost to friction : 17902.45 J
A) Determine how high the motorcycle can coast up the hill when friction is neglected
apply the formula for kinetic and potential energies
K.E = 1/2 mv² ---- ( 1 )
P.E = mgH ---- ( 2 )
As the motorcycle goes uphiLl the kinetic energy is converted to potential energy.
∴ K.E = P.E
1/2 * mv² = mgH
∴ H = ( 1/2 * mv² ) / mg ---- ( 3 )
where ; m = 185 kg , v = 29 m/s , g = 9.81
Insert values into equation ( 3 )
H ( height travelled by motorcycle neglecting friction ) = 42.86m
B) Determine how much energy is lost to friction if the motorcycle attains 33m before coming to rest
P.E = mgh = 185 * 9.81 * 33 = 59890.05 J
where : h = 33 m , g = 9.81
K.E = 1/2 * mv² = 77792.5 J ( question A )
∴ Energy lost ( ΔE ) = ( 77792.5 - 59890.05 ) = 17902.45 J
Hence we can conclude that The motorcycle can coast up the hill by ; 42.86m , The amount of energy lost to friction : 17902.45 J.
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a wire of a certain material has resistance r and diameter d a second wire of the same material and length is found to have resistance r/9 what is the diameter of the second wire g
Answer:
d₂ = 3dThe diameter of the second wire is 3 times that of the initial wire.Explanation:
Using the formula for calculating the resistivity of an object to find the diameter.
Resistivity P = RA/L
R is the resistance of the material
A is the cross sectional area
L is the length of the material
Since A = πd²/4
P = R( πd²/4)/L
P = Rπd²/4L ... 1
If the second wire of the same material and length is found to have resistance R/9, the resistivity of the second material will be;
P₂ = (R/9)A₂/L₂
P₂ = (R/9)(πd₂²/4)/L₂
P₂ = (Rπd₂²/36)/L₂
P₂ = (Rπd₂²)/36L₂
Since the length and resistivity are the same;
P = P₂ and L =L₂
Equating 1 and 2;
Rπd²/4L = (Rπd₂²)/36L₂
Rπd²/4L = (Rπd₂²)/36L
d² = d₂²/9
d₂² = 9d²
Taking the square root of both sides;
√d₂² = √9d²
d₂ = 3d
Therefore the diameter of the second wire is 3 times that of the initial wire
Estimate the peak electric field inside a 1.2-kW microwave oven under the simplifying approximation that the microwaves propagate as a plane wave through the oven's 700-cm2 cross-sectional area.
Answer:
The peak electric field is [tex]E_o = 3593.6 V/m[/tex]
Explanation:
From the question we are told that
The power is [tex]P = 1.2 \ kW = 1.2 *10^{3} \ W[/tex]
The cross-sectional area is [tex]A = 700 \ cm^2 = 700 *10^{-4} \ m^2[/tex]
Generally the average intensity of the microwave is mathematically represented as
[tex]I = \frac{c * \epsilon _o * E_o^2 }{2}[/tex]
Where [tex]c[/tex] is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
and [tex]\epsilon_o[/tex] is the permitivity of free space with value [tex]\epsilon_o = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]
also [tex]E_o[/tex] is the peak electric field.
Now making [tex]E_o[/tex] the subject [tex]E_o = \sqrt{\frac{2 * I }{ c * \epsilon _o } }[/tex]
But this intensity of the microwave can also be represented mathematically as
[tex]I = \frac{ P }{A }[/tex]
substituting values
[tex]I = \frac{ 1.2 *10^{3} }{700 *10^{-4} }[/tex]]
[tex]I = 17142.85 \ W/m^2[/tex]
So
[tex]E_o = \sqrt{\frac{2 * 17142.85 }{ 3.0*10^{8}] * 8.85*10^{-12} } }[/tex]
[tex]E_o = 3593.6 V/m[/tex]
The peak electric field of the microwave is 3,593.1 V/m.
The given parameters;
power of the wave, P = 1.2 kW = 1,200 Warea of the plane, A = 700 cm²The intensity of the wave is calculated as follows;
[tex]I = \frac{P}{A} \\\\I = \frac{1,200}{700 \times 10^{-4}} \\\\I = 17,142.86 \ W/m^2[/tex]
The peak electric field is calculated as follows;
[tex]E_o = \sqrt{\frac{2I}{c \varepsilon _o} } \\\\E_o = \sqrt{\frac{2\times 17,142.86}{3\times 10^8 \times 8.85 \times 10^{-12}} } \\\\E_o = 3,593.1 \ V/m[/tex]
Thus, the peak electric field of the microwave is 3,593.1 V/m.
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