Answer:
Therefore the number of proton in the given system is 450.
Explanation:
Given that, a system has 1223 particles.
Let x number of proton be present in the system.
Then the number of electron is =(1223-x)
The charge of a proton is = 1.602×10⁻¹⁹ C
The charge of an electron = - 1.602×10⁻¹⁹ C
The charge of x protons is =( 1.6×10⁻¹⁹×x) C
The charge of (1223-x) electrons is = - 1.6×10⁻¹⁹ (1223-x) C
According to the problem,
(1.6×10⁻¹⁹×x) +{ - 1.6×10⁻¹⁹ (1223-x)}= -5.328×10⁻¹⁷
⇒1.6×10⁻¹⁹(x-1223+x)=-5.328×10⁻¹⁷
[tex]\Rightarrow (2x-1223)=\frac{-5.328\times 10^{-17}}{1.6\times 10^{-19}}[/tex]
⇒2x-1223= -333
⇒2x= -333+1223
⇒2x=900
[tex]\Rightarrow x=\frac{900}{2}[/tex]
⇒x=450
Therefore the number of proton is 450.
Consider a spring that does not obey Hooke’s law very faithfully. One end of the spring is fixed. To keep the spring stretched or compressed an amount x, a force along the x-axis with x-component Fx=kx−bx2+cx3 must be applied to the free end. Here k = 100 N/m, b=700N/m2, and c=12,000N/m3. Note that x < 0 when the spring is stretched and x > 0 when it is compressed. How much work must be done
(a) to stretch this spring by 0.050 m from its unstretched length?
(b) To compress this spring by 0.050 m from its unstretched length?
(c) Is it easier to stretch or compress this spring? Explain why in terms of the dependence of Fx on x. (Many real springs behave qualitatively in the same way.)
Final answer:
To find the work done when stretching or compressing a non-Hookean spring by 0.050 m, we calculate the definite integral of the force function over the respective intervals. The work done for stretching is the integral from 0 to -0.050 m, and for compression, it is the integral from 0 to 0.050 m. The ease of stretching versus compressing depends on the non-linear force constants.
Explanation:
Work Done to Stretch or Compress a Non-Hookean Spring
To compute the work (W) required to stretch or compress the spring an amount x, we need to integrate the force function Fx = kx - bx2 + cx3 from the unstretched length (0) to the final position (x).
For stretching (x < 0), substitute the given constants (k = 100 N/m, b = 700 N/m2, c = 12,000 N/m3) and integrate from 0 to -0.050 m.
For compression (x > 0), use the same constants and integrate from 0 to 0.050 m.
Analyze the dependence of Fx on x to determine which process (stretching or compressing) requires less work.
The work done for both stretching and compressing can be calculated by evaluating the definite integral of Fx over the interval [0, x].
(a) Work done to stretch:
W = ∫0-0.050 (100x - 700x² + 12,000x3) dx
(b) Work done to compress:
W = ∫00.050 (100x - 700x² + 12,000x3) dx
(c) Comparison of ease:
The comparison is based on the shape of the force-displacement curve. With the given non-linear characteristics, it might be easier or harder to stretch or compress the spring depending on the values of b and c which affect the quadratic and cubic terms, respectively.
Suppose a person whose mass is m is being held up against the wall with a constant tangential velocity v greater than the minimum necessary. Find the magnitude of the frictional force between the person and the wall
The magnitude of the frictional force between the person and the wall is
[tex]\[ f = \frac{mv^2}{r} \][/tex]
To find the magnitude of the frictional force between the person and the wall, we can use the concept of centripetal force.
First, let's define the variables:
m = mass of the person
v = tangential velocity of the person
R = radius of the circular path (distance from the person to the axis of rotation, in this case, the distance from the person to the wall)
The centripetal force [tex](\( F_c \))[/tex] required to keep the person moving in a circular path is given by:
[tex]\[ F_c = \frac{{mv^2}}{R} \][/tex]
Since the person is held against the wall, the frictional force [tex](\( F_f \))[/tex] provides the centripetal force. Therefore,
[tex]\[ F_f = F_c \][/tex]
[tex]\[ F_f = \frac{{mv^2}}{R} \][/tex]
Therefore, the magnitude of the frictional force between the person and the wall is [tex]\( \frac{{mv^2}}{R} \)[/tex].
A 5.93 5.93 kg ball is attached to the top of a vertical pole with a 2.03 2.03 m length of massless string. The ball is struck, causing it to revolve around the pole at a speed of 4.25 4.25 m/s in a horizontal circle with the string remaining taut. Calculate the angle, between 0° and 90°, that the string makes with the pole. Take g = 9.81 g=9.81 m/s2.
Final answer:
To find the angle that the string makes with the pole, we can use the concept of centripetal force. The tension in the string provides the centripetal force that keeps the ball in circular motion.Therefore angle is approximately 45.6°.
Explanation:
To find the angle that the string makes with the pole, we can use the concept of centripetal force. The tension in the string provides the centripetal force that keeps the ball in circular motion.
At the top of the circle, the tension in the string is equal to the weight of the ball, which is given by T = mg. We can use this equation to find the angle the string makes with the pole.
tan(theta) = T/ (m*v²/ R), where m is the mass of the ball, v is the speed of the ball, and R is the length of the string. Plugging in the values given, we get tan(theta) = (5.93 kg * 4.25 m/s²) / (5.93 kg * 9.81 m/s²). Solving for theta, we find that the angle is approximately 45.6°.
A quarter-wave monopole radio antenna (also called a Marconi antenna) consists of a long conductor of one quarter the length of the transmitted wavelength. The lower end of the antenna is connected to a conducting ground plane (often simply the Earth) which reflects the transmitted signal, creating an effective antenna length of a half wavelength
(a) What must be the height of the antenna (in m) for a radio station broadcasting at 604 kHz?
(b) What must be the height of the antenna (in m) for radio stations broadcasting at 1,710 kHz?
Answer:
a) Height of the antenna (in m) for a radio station broadcasting at 604 kHz = 124.17 m
b)Height of the antenna (in m) for radio stations broadcasting at 1,710 kHz =43.86 m
Explanation:
(a) Radiowave wavelength= λ = c/f
As we know, Radiowave speed in the air = c = 3 x 10^8 m/s
f = frequency = 604 kHz = 604 x 10^3 Hz
Hence, wavelength = (3x10^8/604x10^3) m
λ = 496.69 m
So the height of the antenna BROADCASTING AT 604 kHz = λ /4 = (496.69/4) m
= 124.17 m
(b) As we know , f = 1710 kHz = 1710 x 10^3 Hz (1kHZ = 1000 Hz)
Hence, wavelength = λ = (3 x 10^8/1710 x 10^3) m
λ= 175.44 m
So, height of the antenna = λ /4 = (175.44/4) m
= 43.86 m
In the lab, you did not include friction in your calculations for the acceleration. Explain why it was not necessary. What would your equation for acceleration look like if you were to include friction
Final answer:
Friction was not included in the lab calculations because it opposes motion between surfaces, resulting in a smaller acceleration. The equation for acceleration without friction is a = g sinθ, while the equation for acceleration with friction is a = (g sinθ - μk cosθ) / (1 + μk sinθ), where μk is the coefficient of kinetic friction.
Explanation:
The reason friction was not included in the calculations for acceleration in the lab is because friction always opposes motion between surfaces, resulting in a smaller acceleration when it is present. In the absence of friction, all objects slide down a frictionless incline with the same acceleration, regardless of mass. The equation for acceleration without friction is a = g sinθ, where g is the acceleration due to gravity and θ is the angle of the incline.
If friction were to be included in the calculations, the equation for acceleration would be different. It would depend on the coefficient of friction (μ) and the normal force (N) of the object. The equation for acceleration with friction is a = (g sinθ - μk cosθ) / (1 + μk sinθ), where μk is the coefficient of kinetic friction. This equation takes into account the opposing force of friction and provides a more accurate representation of the object's acceleration.
You have a device that takes temperature measurements and runs off of solar power. How often it is programmed to take a measurement will affect how much power it uses--more frequent measurements, more power. You have one installed at the equator and one installed in the Antarctic. During which season of the year can you set each device to take more frequent measurements
Answer:
In the Equator:
As far as the temperature is concerned equator is more or less the same throughout the year, however, there are some fluctuations also, I will set this device in March and September because, in this month, the Sun is exactly over the equator and I would be able to get the more results in this month.
In Antarctic:
As far as the climatic conditions of Antarctic are concerned, it is all the same while it fluctuates in December because the Sun is very much close to Antarctic that's why I will choose this month.
Explanation:
An ice cube and a rubber ball are both placed at one end of a warm cookie sheet, and the sheet is then tipped up. The ice cube slides down with virtually no friction, and the ball rolls down without slipping. The ball and the ice cube have the same inertia. Which one reaches the bottom first?
The ice ball reaches the bottom first, when the sheet is then tipped up. In the given condition, the ball and the ice cube have the same inertia.
What is the friction force?It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).
Mathematically, it is defined as the product of the coefficient of friction and normal reaction.
The ice ball reaches the bottom first, when the sheet is then tipped up. In the given condition, the ball and the ice cube have the same inertia.
Because the ice cube slides down with virtually no friction. Without friction, an object can easily slide with the more speed.
The ball is spherical in nature and a spherical surface is in more contact with the surface during the inclined motion.
Hence, the ice ball reaches the bottom first, when the sheet is then tipped up.
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A thin-walled cylindrical steel water storage tank 30 ft in diameter and 62 ft long is oriented with its longitudinal axis vertical. The tank is topped with a hemispherical steel dome. The wall thickness of the tank and dome is 0.68 in. If the tank is pressurized to 55 psig and contains water 55 ft above its base, and considering the weight of the tank, determine the maximum state of stress in the tank and the corresponding principal stresses (normal and shear). The weight density of water is 62.4 lbf/ft3.
Answer:
ρ
=
55.0
π
⋅
15.0
2
⋅
62.0
Explanation:
in a mass spectrometer ionized molecules are accelerated from rest through a potential difference V into a uniform magnetic field. only those molecules with the correct mass will have the travel radius that allows them to go through the hole and into the detector. what potential differeence is needed so that singly ionized carbon dioxide molecules
Answer:
Potential difference = 245 V
Explanation:
First, we have to find velocity at which ionized molecules are accelerated into a magnetic field. Here, magnetic force acts as a centripetal force due to circular path of molecules towards detector slot.
B= 0.25 T,d= 12 cmmass = (12+16+16)×1.67×10-27kgNow, the magnetic force to the centripetal force equates as:
qvB = [tex]\frac{mv2}{r}[/tex] = [tex]\frac{mv2}{d/2}[/tex]
For Velocity (v) :
v = [tex]\frac{qBd}{2m}[/tex]
According to law of conservation of energy, we know that the initial potential energy of the molecules is converted into kinetic energy as they enter into the magnetic field. So,
[tex]qV=\frac{1}{2}mv2[/tex]
Substitute the value of v in above equation:
qV = [tex]\frac{1}{2}[/tex] m [tex]\frac{q2B2d2}{4m2}[/tex]
V = [tex]\frac{qB2d2}{8m}[/tex]
V= 245 V
A cube of mass m = 0.49 kg is set against a spring with a spring constant of k1 = 606 N/m which has been compressed by a distance of 0.1 m. Some distance in front of it, along a frictionless surface, is another spring with a spring constant of k2 = 233 N/m. How far d2, in meters, will the second spring compress when the block runs into it?
Answer:0.161 m
Explanation:
Given
mass of cube [tex]m=0.49\ kg[/tex]
Spring constant [tex]k_1=606\ N/m[/tex]
compression in the spring [tex]x_1=0.1\ m[/tex]
When this cube is released then it will compress another spring of spring constant [tex]k_2=233\ N/m[/tex]
Conserving energy
[tex]\frac{1}{2}k_1x^2=\frac{1}{2}mv^2=\frac{1}{2}k_2x'^2[/tex]
[tex]\frac{x'}{x}=\sqrt{\frac{k_1}{k_2}}[/tex]
[tex]x'=0.1\times \sqrt{\frac{606}{233}}[/tex]
[tex]x'=0.1\times 1.61[/tex]
[tex]x'=0.161\ m[/tex]
You throw a football straight up. Air resistance can be neglected. (a) When the football is 4.00 m above where it left your hand, it is moving upward at 0.500 m/s. What was the speed of the football when it left your hand
To find the initial speed of the football, you can use the kinematic equation. Given that the football is moving upward at 0.500 m/s and the displacement is 4.00 m, the equation can be used to find the initial velocity.
Explanation:To find the initial speed of the football, we can use the kinematic equation that relates the final velocity, initial velocity, acceleration, and displacement: vf = vi + at. Given that the football is moving upward at 0.500 m/s and the displacement is 4.00 m, we can plug in the values to solve for the initial velocity:
0.500 m/s = vi + (-9.8 m/s2) x t
Since the football is thrown straight up, the acceleration due to gravity is negative. Let's assume the time taken for the football to reach a height of 4.00 m is t. Since the football goes up and then comes back down, the time taken to reach the height of 4.00 m in the upward direction will be half of the total time. Let's call this time tu. Since the motion is symmetric, the time taken to reach the height of 4.00 m on the way down will also be tu. Therefore, the total time taken for the football's motion is 2tu.
Given that the total time is 2tu, we can substitute it into the equation:
0.500 m/s = vi + (-9.8 m/s2) x 2tu
Now, we can solve for vi by isolating it:
vi = 0.500 m/s - (-9.8 m/s2) x 2tu
Since the time taken for the motion is unknown, we cannot determine the precise initial velocity without more information.
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The slit-to-screen distance is D = 200 cm , and the laser wavelength is 633 nm, use the formula for single-slit diffraction minima to find the slit width a.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The slit width [tex]a = \frac{2L \lambda}{W}[/tex]
Explanation:
Assuming the unit on the graph is cm
Given that the slit to screen distance is D = 200 cm = 20 000 m
The wavelength [tex]\lambda[/tex] = 633 nm = [tex]633*10^{-9}m[/tex]
slit width a = ?
The width of the spot that is the width of the peak from the graph is
W = 1.6 × 2 = 3.2 cm
Where the 1.6 is the distance from 0 to the right end point of the peak
The change in y i.e [tex]\Delta y[/tex] has a formula
[tex]\Delta y[/tex] = Ltanθ
An the width of the spot is 2 × [tex]\Delta y[/tex]
W = 2Ltanθ
Applying this formula qsinθ = m[tex]\lambda[/tex]
where m = 1 because we a focused on the first zeros ,using small angle approximation we have y
[tex]a\theta = (1) \lambda[/tex]
[tex]\theta = \frac{\lambda}{a}[/tex]
Substituting this into W = 2ltanθ
Using small angle approximation
W = 2ltanθ = 2Lθ
[tex]W = 2L\frac{\lambda}{a}[/tex]
[tex]a = \frac{2L \lambda}{W}[/tex] and this is the slit width
An electric field can exist solely due to a source charge. An electric force requires at least two charges, the source charge to set up the field and the test charge to feel the field.
A. TRUE
B. FALSE
Answer:
TRUE.
Explanation:
The first part of the statement is understood as certain since the definition of the electric field warns that its existence is dependent on the charge of the source. At the same time, we know that for there to be an electric force it is necessary to consider two charges. Either for the generation of the force of repulsion or for the force of attraction.
The statement is true. An electric field can exist due to a single source charge, while an electric force requires a test charge to respond to the field generated by the source charge.
Explanation:The statement is
TRUE
. The phenomenon takes place within the study of
electromagnetism
. An
electric field
indeed can exist solely due to a source charge. The source charge creates an electric field in the space around it. However, to cause an electric force, there must be at least one other charge (test charge) that is capable of sensing the presence of this field. The greater the source charge, the stronger the electric field. The test charge responds to the field by experiencing a force. The direction of the electric force is the same as the direction of the electric field if the test charge is positive, but opposite if the test charge is negative.
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Basidiomycete fungi ballistically eject millions of spores into the air by releasing the surface tension energy of a water droplet condensing on the spore. The spores are ejected with typical speeds of 1.11 m/s, allowing them to clear the "boundary layer" of still air near the ground to be carried away and dispersed by winds, (a) If a given spore is accelerated from rest to 1.11 m/s in 7.40 us, what is the magnitude of the constant acceleration of the spore (in m/s) while being ejected? m/s(b) Find the maximum height of the spore (in cm) if it is ejected vertically. Ignore air resistance and assume that the spore is ejected at ground level. cm(c) Find the maximum horizontal range of the spore (in cm) if it is ejected at an angle to the ground. Ignore air resistance and assume that the spore is ejected at ground level CM
a) [tex]1.5\cdot 10^5 m/s^2[/tex]
b) 6.3 cm
c) 12.6 cm
Explanation:
a)
The acceleration of an object is the rate of change of its velocity; it is given by:
[tex]a=\frac{v-u}{t}[/tex]
where
u is the initial velocity
v is the final velocity
t is the time interval taken for the velocity to change from u to t
In this problem for the spore, we have:
u = 0 (the spore starts from rest)
v = 1.11 m/s (final velocity of the spore)
[tex]t=7.40\mu s = 7.40\cdot 10^{-6}s[/tex] (time interval in which the spore accelerates from zero to 1.11 m/s)
Substituting, we find the acceleration:
[tex]a=\frac{1.11-0}{7.40\cdot 10^{-6}}=1.5\cdot 10^5 m/s^2[/tex]
b)
Since the upward motion of the spore is a free fall motion (it is subjected to the force of gravity only), it is a uniformly accelerated motion (=constant acceleration, equal to the acceleration due to gravity: [tex]g=9.8 m/s^2[/tex]). Therefore, we can apply the following suvat equation:
[tex]v^2-u^2=2as[/tex]
where:
v = 0 is the final velocity of the spore (when it reaches the maximum height, its velocity is zero)
u = 1.11 m/s is the initial velocity (the velocity at which it is ejected)
[tex]a=-g=-9.8 m/s^2[/tex] is the acceleration (negative because it is downward)
s is the vertical displacement of the spore, which corresponds to the maximum height reached by the spore
Solving for s, we find:
[tex]s=\frac{v^2-u^2}{2a}=\frac{0^2-(1.11)^2}{2(-9.8)}=0.063 m = 6.3 cm[/tex]
c)
If the spore is ejected at a certain angle [tex]\theta[/tex] from the ground, then its motion is a projectile motion, which consists of two independent motions:
- A uniform horizontal motion, with constant horizontal velocity
- A uniformly accelerated motion along the vertical direction (free fall motion)
The horizontal range of a projectile, which can be derived from the equations of motion, is given by:
[tex]d=\frac{v^2 sin(2\theta)}{g}[/tex]
where
v is the initial velocity
[tex]\theta[/tex] is the angle or projection
g is the acceleration of gravity
From the equation, we observe that the maximum range is achevied when
[tex]\theta=45^{\circ}[/tex]
For this angle, the range is
[tex]d=\frac{v^2}{g}[/tex]
For the spore in this problem, the initial velocity is
v = 1.11 m/s
Therefore, the maximum range is
[tex]d=\frac{(1.11)^2}{9.8}=0.126 m = 12.6 cm[/tex]
The rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3 × 10-11 e-250/T and 2 × 10-12 e-940/T, where T is the Kelvin temperature. Calculate the ratio of the rates of ozone destruction by these catalysts at 20 km, given that at this altitude the average concentration of OH is about 100 times that of Cl and that the temperature is about -50 °C. Calculate the rate constant for ozone destruction by chlorine under conditions in the Antarctic ozone hole, when the temperature is about -80 °C and the concentration of atomic chlorine increases by a factor of one hundred to about 4 × 105 molecules cm-3.
Answer:
Calculate the ratio of the rates of ozone destruction by these catalysts at 20 km, given that at this altitude the average concentration of OH is about 100 times that of Cl and that the temperature is about -50 °C
Knowing
Rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3x[tex]10^{-11} e^{-255/T}[/tex] and 2x[tex]10^{-12} e^{-940/T}[/tex]
T = -50 °C = 223 K
The reaction rate will be given by [Cl] [O3] 3x[tex]10^{-11} e^{-255/223} = 9.78^{-12} [Cl] [O3][/tex]
Than, the reaction rate of OH with O3 is
Rate = [OH] [O3] 2x[tex]10^{-12} e^{-940/223} = 2.95^{-14} [OH] [O3][/tex]
Considering these 2 rates we can realize the ratio of the reaction with Cl to the reaction with OH is 330 * [Cl] / [OH]
Than, the concentration of OH is approximately 100 times of Cl, and the result will be that the reaction with Cl is 3.3 times faster than the reaction with OH
Calculate the rate constant for ozone destruction by chlorine under conditions in the Antarctic ozone hole, when the temperature is about -80 °C and the concentration of atomic chlorine increases by a factor of one hundred to about 4 × 105 molecules cm-3
Knowing
Rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3x[tex]10^{-11} e^{-255/T}[/tex] and 2x[tex]10^{-12} e^{-940/T}[/tex]
T = -80 °C = 193 K
The reaction rate will be given by [Cl] [O3] 3x[tex]10^{-11} e^{-255/193} = 8.21^{-12} [Cl] [O3][/tex]
Than, the reaction rate of OH with O3 is
Rate = [OH] [O3] 2x[tex]10^{-12} e^{-940/193} = 1.53^{-14} [OH] [O3][/tex]
Considering these 2 rates we can realize the ratio of the reaction with Cl to the reaction with OH is 535 * [Cl] / [OH]
Than, considering the concentration of Cl increases by a factor of 100 to about 4 × [tex]10^{5}[/tex] molecules [tex]cm^{-3}[/tex], the result will be that the reaction with OH will be 535 + (100 to about 4 × [tex]10^{5}[/tex] molecules [tex]cm^{-3}[/tex]) times faster than the reaction with Cl
Explanation:
A 60-Hz single-phase transformer with capacity of 150 kVA has the following parameters: RP = 0.35 Ω RS = 0.002 Ω Rc = 5.2 kΩ XP = 0.5 Ω XS = 0.008 Ω Xm = 1.1 kΩ The primary transformer voltage is 2.8 kV and the secondary is 230 V. The transformer is connected to a variable load (0 to 300 kW) with a lagging power factor of 0.83 and a load voltage equal to the rated transformer secondary. Determine: (a) the total input impedance of the transformer when the secondary is shorted; and (b) the input current, voltage, power and power factor at full load (150 kW). (c) Plot the voltage regulation versus load, and determine the load that produces 5% regulation.
This answer explains the calculation of turns in a transformer's secondary winding based on given voltage outputs and primary turns, along with determining maximum output currents for different voltage levels.
Explanation:Input Voltage: 240 V
Primary Coil Turns: 280 turns
Secondary Parts Turns: 5.60 V - 77 turns, 12.0 V - 56 turns, 480 V - 7 turns
Maximum Output Currents: 5.60 V - 5 A, 12.0 V - 2.1 A, 480 V - 0.042 A
Final answer:
The question delves into advanced transformer analysis, focusing on impedance during a short circuit, input parameters at full load, and how voltage regulation varies with load. However, specific calculations for these scenarios are complex and require more detailed analysis beyond the provided data.
Explanation:
The student's question involves a complex analysis of a 60-Hz single-phase transformer under different loading conditions, focusing on electrical parameters such as impedance, current, power, and voltage regulation. It covers several fundamental concepts in electrical engineering, particularly those relating to transformers' performance and efficiency under load.
Total Input Impedance
To calculate the total input impedance when the secondary is shorted, we must consider the impedances in parallel and series as per the equivalent circuit of the transformer. However, given the specific data provided, a direct calculation isn't presented here.
Input Current, Voltage, Power, and Power Factor at Full Load
At full load (150 kW) with a power factor of 0.83 and a secondary voltage of 230 V, the input current, voltage, power, and power factor can be derived from the transformer's ratings and load characteristics. However, to accurately calculate these values, one would typically apply the transformer's equivalent circuit model, considering real and reactive power components.
Voltage Regulation vs. Load
The voltage regulation versus load plot and the determination of the load that produces 5% regulation require an analysis of how the secondary voltage varies with load, against rated conditions. This involves calculations and assumptions not fully detailed in the question.
Which statement below is NOT true about electric field lines?
A) They can start on + charges
B) They are perpendicular to the electric field at every point
C) They get closer together where the field is stronger
D) Field lines cannot cross
E) They can end on - charges
a wave is 8 meters long and has a frequency of 3 Hz. Find speed
Answer:
The speed is 24 [tex]\frac{meter}{s}[/tex]
Explanation:
A wave is a disturbance that propagates through a certain medium or in a vacuum, with transport of energy but without transport of matter.
The wavelength is the minimum distance between two successive points of the wave that are in the same state of vibration. It is expressed in units of length (m).
Frequency is the number of vibrations that occur in a unit of time. Its unit is s⁻¹ or hertz (Hz).
The speed of propagation is the speed with which the wave propagates in the middle, that is, the magnitude that measures the speed at which the wave disturbance propagates along its displacement. Relate wavelength (λ) and frequency (f) inversely proportionally using the following equation:
v = f * λ.
In this case, λ= 8 meter and f= 3 Hz
Then:
v= 3 Hz* 8 meter
So:
v= 24 [tex]\frac{meter}{s}[/tex]
The speed is 24 [tex]\frac{meter}{s}[/tex]
The speed of the wave is 24 m/s.
To find the speed of a wave, you can use the relationship between speed, frequency, and wavelength.
The formula is:
Speed (v) = Wavelength (λ) × Frequency (f)Given in this problem:
Wavelength (λ) = 8 metersFrequency (f) = 3 HzWe substitute these values into the formula:
Speed (v) = 8 meters × 3 HzThus, the speed of the wave is:
v = 24 m/sThe width of a particular microwave oven is exactly right to support a standing-wave mode. Measurements of the temperature across the oven show that there are cold spots at each edge of the oven and at three spots in between. The wavelength of the microwaves is 12 cm . How wide is the oven?
Answer:
Explanation:
In standing wave pattern we find region of nodes where vibration is minimum or cold spots . The distance between any two consecutive node is half the wave length . There are 5 cold spot or node in between which is equal to 4 half wave length .
width of oven = 4 x half wave length
= 4 x (12 / 2 )
= 24 cm
An isolated charged conducting sphere of radius 11.0 cm creates an electric field of 4.90 ✕ 104 N/C at a distance 20.0 cm from its center. (a) What is its surface charge density?
Answer:
Surface charge density = 1.43 × 10⁶μC/m²
Explanation:
surface charge density = Q /A_____(1)
where charge Q is the uniformly distributed on surface area A and d surface charge density σ
The electric field due to uniformly charge sphere of charge Q a distance r from the center of the sphere
[tex]E = k\frac{Q}{r^2}[/tex]______(2)
where k is 8.988 × 10⁹N.m²C²
The surface area of the sphere is 4πR² ______(3)
The Capacitance is 4πε₀R
where ε₀ = 8.8542 × 10⁻¹²C/Nm² is a constant
Given that,
R = 11cm = 0.11m
E = 4.90 ✕ 10⁴ N/C
r = 20.0cm = 0.20m
substitute for Q in eqn(2) and for A in eqn(3)
surface charge density = [tex]\frac{Er^2 }{k(4\pi R^2)} \\\frac{(4.9 * 10^4)(0.20)^2}{4\pi (8.988 * 10^9)(0.11)^2 }[/tex]
= 1.43 * 10⁶C/m²
Surface charge density = 1.43 μC/m²
An isolated, charged conducting sphere of radius 11.0 cm creates an electric field of 4.90×10⁴ N/C at a distance 20.0 cm from its center. (a) What is its surface charge density? (b) What is its capacitance?
Answer:(a) 1.47 x 10⁻⁶ C/m² or 1.47 μC/m²
(b) 12.07 x 10⁻¹² F or 12.07 pF
Explanation:The surface charge density, σ, of a surface (sphere, in this case) of area A which has a charge Q uniformly distributed on it is given by;
σ = [tex]\frac{Q}{A}[/tex] -----------------(i)
Also, the electric field, E, due to the charge Q, at a distance r, from the center of the sphere to another point on the sphere is given as;
E = k x [tex]\frac{Q}{r^2}[/tex] --------------(ii)
Where;
k = Coulomb's constant = 8.99 x 10⁹Nm²/C²
(a) i. First calculate the charge on the sphere as follows;
From the question;
r = 20.0cm = 0.20m
E = 4.90 x 10⁴ N/C
Substitute these values into equation (ii) as follows;
4.90 x 10⁴ = 8.99 x 10⁹ x [tex]\frac{Q}{0.20^2}[/tex]
4.90 x 10⁴ = 8.99 x 10⁹ x [tex]\frac{Q}{0.04}[/tex]
4.90 x 10⁴ = 224.75 x 10⁹ x Q
Q = [tex]\frac{4.90*10^4}{224.75*10^9}[/tex]
Q = 0.022 x 10⁻⁵
Q = 0.22 x 10⁻⁶ C
(a) ii. Also calculate the area A, of the sphere as follows;
A = 4π R²
Where;
R = radius of the sphere = 11.0cm = 0.11m
Substitute this value into equation above;
A = 4π (0.11)² [Take π = 3.142]
A = 4(3.142)(0.0121)
A = 0.15m²
(a) iii. Now calculate the surface charge density, of the sphere as follows;
Substitute the values of A and Q into equation (i) as follows;
σ = [tex]\frac{0.22 * 10^{-6}}{0.15}[/tex]
σ = 1.47 x 10⁻⁶C/m²
Therefore the surface charge density is 1.47 x 10⁻⁶C/m²
==============================================================
(b) The capacitance C, of an isolated charged sphere with radius R, is given by;
C = Aε₀ / R ----------------(iii)
Where;
R = 11.0cm = 0.11m
A = area of the sphere = 0.15m² [as calculated above]
ε₀ = permittivity of free space = 8.85 x 10⁻¹² C/Nm² [a known constant]
Substitute these values into equation (iii) as follows;
C = 0.15 x 8.85 x 10⁻¹² / 0.11
C = 12.07 x 10⁻¹²F
Therefore, the capacitance of the charged sphere is 12.07 x 10⁻¹²F
An electron and a proton, moving side by side at the same speed, enter a 0.020-T magnetic field. The electron moves in a circular path of radius9.0 mm. What is the radius of the proton?
Answer:
16.5 m
Explanation:
Given,
Magnetic field = 0.02 T
radius of electron = 9 mm
speed of electron = speed of proton
radius of proton = ?
We know,
[tex]F_e = q_e vB[/tex].........(1)
[tex]F_p = q_p vB[/tex]
using newton second law
[tex]F = m a = m\dfrac{v^2}{r}[/tex]
equating Force due to electron and proton
[tex]F_e = F_p[/tex]
[tex]\dfrac{m_ev^2}{r_e}=\dfrac{m_pv^2}{r_p}[/tex]
[tex]\dfrac{m_e}{r_e} = \dfrac{m_p}{r_p}[/tex]
m_ e = 9.1 x 10⁻³¹ Kg and m_p = 1.67 x 10⁻²⁷ Kg
[tex]r_p = \dfrac{m_p}{m_e}\times r_e[/tex]
[tex]r_p = \dfrac{1.67\times 10^{-27}}{9.1 \times 10^{-31}}\times 9 \times 10^{-3}[/tex]
[tex]r_p = 16.5 m[/tex]
Hence, the radius of proton is equal to 16.5 m.
Two blocks, joined by a string, have masses of 6.0 kg and 9.0 kg. They rest on a frictionless horizontal surface. A 2nd string, attached only to the 9 kg block, has horizontal force = 30 N applied to it. Both blocks accelerate. Find the tension in the string between the blocks.
Answer:
Explanation:
30 N force is pulling total mass of 15 kg , so acceleration in the system of masses
= 30 / 15
= 2 m / s²
Let us now consider forces acting on 9 kg . 30 N is pulling it in forward direction . Tension T in the string attached to it is pulling it in reverse direction
so net force on it
30 - T
Applying Newton's law of motion on it
30 - T = mass x acceleration
30 - T = 9 x 2
30 - 18 = T
T = 12 N
Using Newton’s second law, the applied force is used to find the acceleration of the whole system. We then calculate the force (tension) required to move the 6 kg block using this acceleration which is 12N.
Explanation:In Physics, specifically Newtons’ second law of motion, the tension in the string between the two blocks can be calculated by using the equation F=ma, where F is the force, m is the mass, and a is the acceleration. The force applied on the 9 kg block can be considered to cause an acceleration in the entire system (both blocks) given the fact that they are connected by a string. First, find the acceleration of the whole system by using the formula a = F/total mass = 30N/(9kg+6kg) = 2 m/s². Then, to find the tension in the string between the blocks, we calculate the force required to move the 6 kg block with that acceleration: T = ma = 6kg * 2 m/s² = 12 N. Therefore, the tension in the string between the blocks is 12N.
Learn more about Tension in a string here:https://brainly.com/question/34111688
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When its 75 kW (100 hp) engine is generating full power, a small single-engine airplane with mass 700 kg gains altitude at a rate of 2.5 m>s 1150 m>min, or 500 ft>min2. What fraction of the engine power is being used to make the airplane climb
Answer:
0.2289
Explanation:
Power required to climb= Fv where F is force and v is soeed. We know that F= mg hence Power, P= mgv and substituting 700 kg for m, 9.81 for g and 2.5 m/s for v then
P= 700*9.81*2.5=17167.5 W= 17.1675 kW
To express it as a fraction of 75 kw then 17.1675/75=0.2289 or 22.89%
The vertical displacement of an ocean wave is described by the function, y = A sin(ωt - kx). k is called the wave number (k = 2π/λ) and has a value of k = 18 rad/m. The remaining values are A = 9.5 m and ω = 14.5 rad/s.
a) Using y = A sin(B), input an expression for B where the wave would be traveling in the -x-direction. sig.gif?tid=7M79-31-9F-4E-8624-20536
b) What is the wave's velocity in m/s?
c) What is the wave's amplitude in m?
The expression for B for a wave traveling in the -x direction is B = ωt - kx + π. The wave velocity is 0.805 m/s and the wave amplitude is 9.5 m.
Explanation:To determine the expression for B where the wave would be traveling in the -x direction, we need to consider that the general equation for the wave function is y = A sin(B). In this case, the wave is traveling in the -x direction, which means the phase of the wave is shifted by π (180 degrees). So the expression for B would be B = ωt - kx + π.
The wave's velocity can be calculated using the formula v = ω/k. Substituting the given values, the wave's velocity is v = 14.5 rad/s / 18 rad/m = 0.805 m/s.
The wave's amplitude is given directly as A = 9.5 m.
A vice pushes on a system of three boards, each oriented vertically and held up by horizontal forces. The outer boards weigh 90 N and the inner board weighs 118 N. The coefficient of friction between the inner and the outer boards is 0.67. What is the magnitude of one of the compression forces acting on either side of the inner board?
Answer:
88.1 N
Explanation:
As shown in the free body diagram attached to the question, the only forces acting on the inner block include in the required vertical direction.
- Its Weight.
- Frictional forces as a result of the Two compression forces.
If the block is not to slip off, the weights must match the two frictional forces
Let the compressive forces be F and frictional force be Fr
Fr = μ F
Sum of force acting on the inner block in the y-direction
Fr + Fr - W = 0
μ F + μ F = mg
2 μF = 118
2(0.67) F = 118
F = 118/1.34
F = 88.1 N
Each of the compression forces is 88.1 N
Hope this Helps!!!
The magnitude of one of the compression forces acting on either side of the inner board in a vice setup is 88.1 N
Given that the coefficient of friction () between the inner and the outer boards is 0.67 and the normal force (weight of the inner board) is 118 N, we can use the relationship for static friction to find the frictional force. Since the inner board is at equilibrium and not moving, the magnitude of one of the compression forces that acts on either side of the inner board must be equal to the static frictional force.
Thus,
Let the compressive forces be F and the frictional force be Fr
Fr = μ F
The sum of force acting on the inner block in the y-direction
Fr + Fr - W = 0
μ F + μ F = mg
2 μF = 118
2(0.67) F = 118
F = 118/1.34
F = 88.1 N
Consider a magnetic force acting on an electric charge in a uniform magnetic field. Which of the following statements are true? Check all that apply. Check all that apply. An electric charge moving parallel to the magnetic field experiences a magnetic force. The direction of the magnetic force acting on a moving electric charge in the magnetic field is perpendicular to the direction of motion. An electric charge moving perpendicular to the magnetic field experiences a magnetic force. The magnetic force is exerted on a stationary electric charge in the uniform magnetic field. The direction of the magnetic force acting on a moving charge in the magnetic field is perpendicular to the direction of the magnetic field.
Answer: 5) "The direction of the magnetic force acting on a moving charge in the magnetic field is perpendicular to the direction of the magnetic field"
3) "An electric charge moving perpendicular to the magnetic field experiences a magnetic force"
Explanation:
When a charge of magnitude q, moving with a velocity v is placed in a magnetic field of strength B, it experiences a force, the magnitude of this force F is given mathematically as
F =qvB sinx
Where x is the angle between the magnetic field and the velocity of the charge.
From this equation, we can see that the force is zero when magnetic field strength B is parallel to velocity (x=0) or when velocity v is zero.
Also the force F is maximum when the angle between magnetic field strength B and velocity is 90 ( that's they are perpendicular).
By the right hand rule, the force, velocity and strength of magnetic field are perpendicular to each other.
These points have made statement 3 and 5 of the questions to be correct.
A parallel-plate capacitor in air has a plate separation of 1.51 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 260 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume the liquid is an insulator.
(a) What is the charge on the plates before immersion?
(b) What is the charge on the plates after immersion?
Answer:
(a) 380.96 pC
(b) 3.25V
Explanation:
(a) Before immersion,
[tex]C_{air}[/tex] = [tex]\frac{E_{0}A }{d}[/tex]
⇒(8.85E-12× 25E-4× 260) ÷(0.0151)
= 380.96 pC
(b) Charge on the plates after immersion can be calculated by,
Q = ΔV×C
= Δ[tex]V_{air}[/tex] ÷ K
where K is the constant for distilled water
= 260 ÷ 80
= 3.25V
Answer:
Explanation:
(a) Charge on the plates before immersion
as we know that,
C = εA/d
C = 8.85×[tex]10^{-12}[/tex]× 25×[tex]10^{-4}[/tex]/ 1.51×[tex]10^{-2}[/tex]
= 1.46×[tex]10^{-12}[/tex]
Q = CV
= (1.46×[tex]10^{-12}[/tex]) (260)
= 3.796×[tex]10^{-10}[/tex] C
(b) Charge on the plates after immersion
Q = 3.796×[tex]10^{-10}[/tex] C
The charge will remain the same, as the capacitor was disconnected before it was immersed.
A 18.5-cm-diameter loop of wire is initially oriented perpendicular to a 1.3-T magnetic field. The loop is rotated so that its plane is parallel to the field direction in 0.18 s . What is the average induced emf in the loop? Express your answer to two significant figures and include the appropriate units.
Answer:
0.2v
Explanation:
Data given,
Diameter=18.5cm
Hence we can calculate the radius as D/2=18.5/2=9.25cm
radius=9.25cm/100=0.0925m
The area is calculated as
[tex]area=\pi r^{2}\\Area=0.0925^{2}*\pi \\Area=0.02688m^{2}\\[/tex]
magnetic field, B=1.3T
time,t=0.18s
The flux is expressed as
[tex]flux=BAcos\alpha \\[/tex]
since the loop is parallel, the angle is 0
Hence we can calculate the flux as
[tex]flux=1.3*0.02688cos(0)\\flux=0.0349Wb\\[/tex]
to determine the emf induced in the loop, we use Faraday law
[tex]E=-N\frac{d(flux)}{dt}\\ E=-0.0349/0.18\\E=0.19V\\E=-0.2v[/tex]
Note the voltage is not negative but the negative sign shows the current flows in other to oppose the flux
Two froghoppers sitting on the ground aim at the same leaf, located 35 cm above the ground. Froghopper A jumps straight up while froghopper B jumps at a takeoff angle of 58° above the horizontal.
Which froghopper experiences the greatest change in kinetic energy from the start of the jump to when it reaches the leaf?
Answer: A
Explanation: We can use the concept of conservation energy which implies that the kinetic energy of the froghoppers equals it potential energy from the ground level.
Where potential energy = mgh
Where m = mass of the object , g = acceleration due gravity and h = height from ground level.
The value of potential energy will reduce when the height is inclined at an angle.
Let us assume equal mass for both froghoppers, say m , g = 10 m/s^2 and a value of h.
For the first froghopper, potential energy = m×9.8×h = 9.8 mh
For the second froghopper, potential energy = m×9.8×hsin58 ( hsin58 is the vertical componet of height h inclined at angle 58),
potential energy = 8.3109 mh
As we can see , froghopper A has more potential energy than froghoppers B which implies that A has more kinetic energy than B
Final answer:
Both froghoppers experience the same change in kinetic energy, as the change in potential energy is the same for both due to reaching the same height.
Explanation:
To determine which froghopper experiences the greatest change in kinetic energy we can consider their jumps from an energy perspective. As both froghoppers are aiming for the same leaf at a height of 35 cm, the change in gravitational potential energy (UG) from the ground to the leaf will be the same for both, since potential energy depends only on the height and mass, which are constant for both froghoppers. Assuming that both froghoppers start at rest (kinetic energy Ki = 0), the change in kinetic energy during the jump (ΔK) will equal the change in potential energy (ΔUG) at the top of their trajectories where their velocity is zero.
If we follow the work-energy principle, the work done by the froghoppers' muscles will convert to potential energy at the peak of their jumps. Thus, the greatest change in kinetic energy will be equivalent to the potential energy gained at the leaf's height, which is same for both, meaning that neither froghopper experiences a greater change in kinetic energy than the other when reaching the leaf. The takeoff angle does not affect the change in kinetic energy in this case since both reached the same height.
A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 3.0 km/h. When a second tugboat applies an additional constant force of magnitude F2 in the same direction, the speed increases by 11 km/h during a 10 s interval. How do the magnitudes of F1 and F2 compare? (Neglect the effects of water resistance and air resistance.)
Answer:
The magnitude of F₁ is 3.7 times of F₂
Explanation:
Given that,
Time = 10 sec
Speed = 3.0 km/h
Speed of second tugboat = 11 km/h
We need to calculate the speed
[tex]v_{1}=\dfrac{3.0\times10^{3}}{3600}[/tex]
[tex]v_{1}=0.833\ m/s[/tex]
The force F₁is constant acceleration is also a constant.
[tex]F_{1}=ma_{1}[/tex]
We need to calculate the acceleration
Using formula of acceleration
[tex]a_{1}=\dfrac{v}{t}[/tex]
[tex]a_{1}=\dfrac{0.833}{10}[/tex]
[tex]a_{1}=0.083\ m/s^2[/tex]
Similarly,
[tex]F_{2}=ma_{2}[/tex]
For total force,
[tex]F_{3}=F_{2}+F_{1}[/tex]
[tex]ma_{3}=ma_{2}+ma_{1}[/tex]
The speed of second tugboat is
[tex]v=\dfrac{11\times10^{3}}{3600}[/tex]
[tex]v=3.05\ m/s[/tex]
We need to calculate total acceleration
[tex]a_{3}=\dfrac{v}{t}[/tex]
[tex]a_{3}=\dfrac{3.05}{10}[/tex]
[tex]a_{3}=0.305\ m/s^2[/tex]
We need to calculate the acceleration a₂
[tex]0.305=a_{2}+0.083[/tex]
[tex]a_{2}=0.305-0.083[/tex]
[tex]a_{2}=0.222\ m/s^2[/tex]
We need to calculate the factor of F₁ and F₂
Dividing force F₁ by F₂
[tex]\dfrac{F_{1}}{F_{2}}=\dfrac{m\times0.83}{m\times0.22}[/tex]
[tex]\dfrac{F_{1}}{F_{2}}=3.7[/tex]
[tex]F_{1}=3.7F_{2}[/tex]
Hence, The magnitude of F₁ is 3.7 times of F₂