Answer:
Magnitude of magnetic field is 1.5 x 10^(-8) T in the positive y-direction
Explanation:
From maxwell's equations;
B = E/v
Where;
B is maximum magnitude of magnetic field
E is maximum electric field
v is speed of light which has a constant value of 3 x 10^(8) m/s
We are given, E = 4.5 V/m
Thus; B = 4.5/(3 x 10^(8))
B = 1.5 x 10^(-8) T
Now, for Electric field, vector E to be in the positive x-direction, the product of vector E and vector B will have to be in the positive z-direction when vector B is in the positive y-direction
Thus,
Magnitude of magnetic field is 1.5 x 10^(-8) T in the positive y-direction
Magnitude of magnetic field in the space at given instant in time is [tex]\bold { 1.5 x 10^{-8}\ T}[/tex] in the positive y-direction.
From Maxwell's equations,
[tex]\bold {B = \dfrac Ev}[/tex]
Where;
B - maximum magnitude of magnetic field = ?
E- maximum electric field = 4.5 V/m
v- speed of light = 3 x 10^(8) m/s
Put the values in the formula,
[tex]\bold {B = \dfrac {4.5}{3 x 10^8}}\\\\\bold {B = 1.5 x 10^{-8}\ T}[/tex]
When Electric field, is in the positive x-direction, vector B is in the positive y-direction and the product of vector E and vector B will have to be in the positive z-direction.
Therefore, magnitude of magnetic field in the space at given instant in time is [tex]\bold { 1.5 x 10^{-8}\ T}[/tex] in the positive y-direction.
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Ehren is trying to increase his mile run from eight to six minutes. He has decided to run some sandy hills near his home. Which principle of overload is at work? progression time frequency intensity
Answer:
The correct answer is intensity.
Explanation:
The principle of overload is a basic sports fitness training concept which means that in order to improve, athletes must continually work harder as they their bodies adjust to existing workouts. Thus, overloading also plays a role in skill learning.
Now, since erhen wants to increase his mile run from eight to six minutes by run some sandy hills near his home, it means the principle at work is making his training runs more intense for better speed results. Thus the principle at work is intensity because he is trying to do something that makes him run faster.
Final answer:
Ehren is employing the Overload Principle, focusing on the component of intensity to improve his mile run time by running up sandy hills. This method increases resistance and is aligned with the Progression Principle to safely enhance his fitness and performance.
Explanation:
Ehren is working on improving his mile run time and has decided to include running up sandy hills as part of his training. This implementation of intensity in his workouts is a component of the Overload Principle. The principle of overload necessitates a "greater than normal workload or exertion" for an individual to improve in aspects such as aerobic endurance, muscular strength, endurance, and flexibility. By running on sandy hills, Ehren increases the resistance and difficulty compared to running on a flat surface, thus intensifying his training sessions to drive physiological adaptations.
The Progression Principle is also at play here, which entails the gradual increase in stress placed on the body to safely enhance fitness without risking overuse or injury. This principle aligns with the Overload Principle, as it supports the idea of incrementally adding stress to the body through exercises to foster continual improvements. Overall, Ehren's choice to run sandy hills is applying both the intensity and progression components of the overload to achieve his goal of increasing his running pace.
A point charge q1 is at the center of a sphere of radius 20 cm. Another point charge q2 = 10 nC is located at a distance r = 10 cm from the center of the sphere. If the net flux through the surface of the sphere is 800 N.m2/C, find q1.
Answer:
q₁ = -2.92 nC
Explanation:
Given;
first point charge, q₁ = ?
second point charge, q₂ = 10 nC
net flux through the surface of the sphere, Φ = 800 N.m²/C
According to Gauss’s law, the flux through any closed surface (Gaussian surface), is equal to the net charge enclosed divided by the permittivity of free space.
[tex]\phi = \frac{q_{enc.}}{\epsilon_o}[/tex]
where;
Φ is net flux
[tex]q_{enc.}[/tex] net charge enclosed
ε₀ is permittivity of free space.
[tex]q_{enc.}[/tex] = Φε₀
= 800 x 8.85 x 10⁻¹²
= 7.08 x 10⁻⁹ C
[tex]q_{enc.}[/tex] = 7.08 nC
q₁ + q₂ = [tex]q_{enc.}[/tex]
q₁ = [tex]q_{enc.}[/tex] - q₂
q₁ = 7.08nC - 10 nC
q₁ = -2.92 nC
A piece of glass has a temperature of 72.0 degrees Celsius. The specific heat capacity of the glass is 840 J/kg/deg C. A liquid that has a temperature of 40.0 degrees Celsius is poured over the glass, completely covering it, and the temperature at equilibrium is 57.0 degrees Celsius. The mass of the glass and the liquid is the same. Determine the specific heat capacity of the liquid
Answer:
741 J/kg°C
Explanation:
Given that
Initial temperature of glass, T(g) = 72° C
Specific heat capacity of glass, c(g) = 840 J/kg°C
Temperature of liquid, T(l)= 40° C
Final temperature, T(2) = 57° C
Specific heat capacity of the liquid, c(l) = ?
Using the relation
Heat gained by the liquid = Heat lost by the glass
m(l).C(l).ΔT(l) = m(g).C(g).ΔT(g)
Since their mass are the same, then
C(l)ΔT(l) = C(g)ΔT(g)
C(l) = C(g)ΔT(g) / ΔT(l)
C(l) = 840 * (72 - 57) / (57 - 40)
C(l) = 12600 / 17
C(l) = 741 J/kg°C
106 grams of liquid water are in a cylinder with a piston maintaining 1 atm (101325 Pa) of pressure. It is exactly at the boiling point of water, 373.15 K. We then add heat to boil the water, converting it all to vapor. The molecular weight of water is 18 g/mol and the latent heat of vaporization is 2260 J/g. 1) How much heat is required to boil the water?
Answer:
239.55 KJ
Explanation:
Given:
Mass 'm' = 106 g
Latent heat of vaporization'L'= 2260 J/g.
Molecular weight of water'M' = 18 g/mol
Pressure 'P' = 101325 Pa
Temperature 'T' = 373.15 K
Using the formula of phase change, in order to determine the amount of heat required, we have
Q = mL
Q = 106 x 2260
Q = 239560J = 239.55 KJ
Acceleration is measured in m/s/s (meters per second squared).
True or False
Answer:
true
Explanation:
Answer:true
Explanation:
Acceleration is measured in meters per second squared
An imaginary cubical surface of side L has its edges parallel to the x-, y- and z-axes, one corner at the point x = 0, y = 0, z = 0 and the opposite corner at the point x=L, y=L, z=L. The cube is in a region of uniform electric field E⃗ =E1i^+E2j^, where E1 and E2 are positive constants. Calculate the electric flux through the cube face in the plane x = 0 and the cube face in the plane x=L. For each face the normal points out of the cube.
Find the given attachment for solution
The electric flux trough x = 0 plane is = - EL² and the electric flux trough x = l plane is = EL².
What is electric flux?Although an electric field cannot flow by itself, electric flux in electromagnetism is a measure of the electric field passing through a specific surface.
An electric field surrounds an electric charge, such as a solitary electron in space. Field lines have no physical significance and are merely a graphic representation of field strength and direction.
The number of "lines" per unit area, also known as the electric flux density, is inversely proportional to the electric field strength. The total number of electric field lines passing through a surface determines the amount of electric flux.
Hence, electric flux trough x = 0 plane is = - EL² and the electric flux trough x = l plane is = EL².
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In a long distance race, the athletes were expected to take four rounds of the track such that the line of finish was the same as the track was 200 m. (i) What is the total distance to be covered by the athletes? (ii) What is the displacement of the athletes when they touch the finish line? (iii) Is the motion of the athletes uniform or non uniform? (iv) Is the distance moved by and displacement of athletes at the end of the race equal?
Answer:
1) The track is 200 meters long, the athletes do 4 rounds, so the total distance ran is:
4*200 meters = 800 meters
2) The displacement is defined as the difference between the final position and the initial position.
When you are in a track, and you run a full round, you end in the same position where you started, so the total displacement is 0 meters.
3) As they are running in a closed track (for example, a circular track), they must change the direction of motion at some point, so the motion is not uniform (uniform motion happens when the movement is always in the same direction and always at the same speed)
4) no, is not equal because in the end, the total distance is 800 meters and the displacement is 0 meters.
Vectors A and B lie in the x-y plane. Vector A has a magnitude of 17.6 and is at an angle of 120.5° counter-clockwise from the x-axis. Vector B has a magnitude of 21.7 and is 240.3° from the x-axis. Resolve A and B into components, and express in unit vector form below.
The unit vector forms of the given vectors is required.
The required vectors are [tex]A=-8.93\hat{i}+15.16\hat{j}[/tex] and [tex]B=-10.75\hat{i}-18.84\hat{j}[/tex]
VectorsMagnitude of vector A = [tex]|A|=17.6[/tex]
Angle vector A makes with positive x axis counter clockwise = [tex]\theta_1=120.5^{\circ}[/tex]
Magnitude of vector B = [tex]|B|=21.7[/tex]
Angle vector B makes with positive x axis counter clockwise = [tex]\theta_2=240.3^{\circ}[/tex]
The vectors need to be resolved in order to write in the unit vector forms.
The vectors are
[tex]A=|A|(\cos\theta_1\hat{i}+\sin\theta_1\hat{j})\\\Rightarrow A=17.6(\cos120.5\hat{i}+\sin120.5\hat{j})\\\Rightarrow A=-8.93\hat{i}+15.16\hat{j}[/tex]
[tex]B=|B|(\cos\theta_2\hat{i}+\sin\theta_2\hat{j})\\\Rightarrow B=21.7(\cos240.3\hat{i}+\sin240.3\hat{j})\\\Rightarrow B=-10.75\hat{i}-18.84\hat{j}[/tex]
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Final answer:
To resolve the vectors into components, apply trigonometric functions -- cosine for x components and sine for y components -- to their magnitudes and angles, then express them in unit vector form with i and j.
Explanation:
When resolving vectors A and B into their components in the x-y plane, the general method involves using trigonometry, specifically, the cosine and sine functions for the x and y components, respectively. Given that vector A has a magnitude of 17.6 and an angle of 120.5° from the x-axis, its components can be calculated as follows:
Ax = A * cos(θ) = 17.6 * cos(120.5°)
Ay = A * sin(θ) = 17.6 * sin(120.5°)
Similarly, vector B with a magnitude of 21.7 and an angle of 240.3° from the x-axis has components:
Bx = B * cos(θ) = 21.7 * cos(240.3°)
By = B * sin(θ) = 21.7 * sin(240.3°)
The resulting components should then be written in unit vector form by attaching the unit vectors i (for the x-axis) and j (for the y-axis) to the respective components.
A mass is hung from a spring and set in motion so that it oscillates continually up and down. The velocity v of the weight at time t is given by the equation v=−2 cos(3πt) with v measured in feet per second and t measured in seconds. Determine the maximum velocity of the mass and the amount of time it takes for the mass to move from its lowest position to its highest position.
Answer:
the maximum velocity of the mass v (max) = 2 ft/s
the amount of time it takes for the mass to move from its lowest position to its highest position∆t = 1/3 seconds = 0.33 seconds
Explanation:
Given the velocity equation;
v=−2 cos(3πt)
The maximum velocity would be at cos(3πt) = 1 or cos(3πt) = -1
v (max) = -2 × -1 = 2 ft/s
The time taken for the mass to move from lowest position to highest position
At Lowest position, vertical velocity equals zero.
At highest position, vertical velocity equals zero.
The time taken for the mass to move from one v = 0 to the next v = 0
Cos(π/2) = 0 and
Cos(3π/2) = 0
For the first;
Cos(3πt) = cos(π/2)
3πt1 = π/2
t1 = π/2(3π)
t1 = 1/6 second
For the second;
Cos(3πt) = cos(3π/2)
3πt2 = 3π/2
t2 = 3π/2(3π)
t2 = 1/2 second
∆t = t2 - t1 = 1/2 - 1/6 = 3/6 - 1/6 = 2/6 = 1/3 seconds
∆t = 1/3 seconds
A small, positively charged ball is moved close to a large, positively charged ball. Which describes how the small ball likely responds when it is released?
It will move toward the large ball because like charges repel.
It will move toward the large ball because like charges attract.
It will move away from the large ball because like charges repel.
It will move away from the large ball because like charges attract.
Answer:
C
Explanation:
i took the test
A small, positively charged ball is moved close to a large, positively charged ball. "It will move away from the large ball because like charges repel." The correct option is A.
Charge Interaction: The behavior of charged objects is governed by the fundamental principle that opposite charges attract each other, and like charges repel each other.
Coulomb's Law: Coulomb's Law describes the electrostatic force between two charged objects. It states that the force of attraction or repulsion between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Repulsion: Because both the small and large balls have a positive charge, they will exert a repulsive force on each other when they are in close proximity.
Movement Away: When the small ball is released near the large ball, it will experience this repulsive force, causing it to move away from the large ball.
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Problem (2) A 16 kg cylinder, initially at rest, is held by a cord connected to a grooved drum whose mass is 20 kg. The drum has an outer radius �% = 250 mm and an inner radius of �& = 160 mm. If the drum experiences a constant frictional moment of 3 N∙m at O, how far has the cylinder dropped when it has a downward velocity of 2 m/s? Neglect the mass of the cord and treat the drum as a thin disk. Use the Principle of Work and Energy.
Answer:
Explanation:
The solution to the problem is given in the pictures attached below; the three pictures explains the problem fully and I hope it helps you. Thank you
Two forces act on a block as shown in the picture.
Right
20 N
Left
10 N
What is the net force on the block?
Answer:
10n to the right
Explanation:
Which statements are true about the flow of blood in the body? Check all that apply.
Answer:
i need some explanation
Explanation:
The correct statements are D, C, and E. Blood flows from the heart to the lungs to pick up oxygen and then to the rest of the body to deliver oxygen, sugar, and nutrients while collecting carbon dioxide.
Understanding the flow of blood in the body is essential. Here are the correct statements regarding blood circulation:
D.) Blood flows from the heart to the lungs to pick up oxygen.C.) Blood picks up carbon dioxide from the cells of the body.E.) Blood delivers sugar and nutrients to cells in the body.The heart pumps oxygen-poor blood to the lungs through the pulmonary circuit where it releases carbon dioxide and picks up oxygen.
The oxygen-rich blood is then pumped through the systemic circuit to the rest of the body, delivering oxygen, sugar, and nutrients to the cells and collecting carbon dioxide to be expelled during the next circulation.
Therefore, the correct statements are D, C, and E.
Complete Question
Which statements are true about the flow of blood in the body? Check all that apply.
A. Blood picks up oxygen from the cells of the body.
B. Blood delivers carbon dioxide to cells in the body.
C. Blood picks up carbon dioxide from the cells of the body.
D. Blood flows from the heart to the lungs to pick up oxygen.
E. Blood delivers sugar and nutrients to cells in the body.
F. Blood flows from the lungs to the heart to pick up oxygen.
If an arrow is shot upward on the moon with a velocity of 58 m/s, its height (in meters) after t seconds is given by H = 58t − 0.83t2. (a) Find the velocity of the arrow after two seconds. m/s (b) Find the velocity of the arrow when t = a. m/s (c) When will the arrow hit the surface? (Round your answer to one decimal place.) t = s (d) With what velocity will the arrow hit the surface? m/s
Answer:
a) v = 54.7m/s
b) v = (58 - 1.66a) m/s
c) t = 69.9 s
d) v = -58.0 m/s
Explanation:
Given;
The height equation of the arrow;
H = 58t - 0.83t^2
(a) Find the velocity of the arrow after two seconds. m/s;
The velocity of the arrow v can be given as dH/dt, the change in height per unit time.
v = dH/dt = 58 - 2(0.83t) ......1
At t = 2 seconds
v = dH/dt = 58 - 2(0.83×2)
v = 54.7m/s
(b) Find the velocity of the arrow when t = a. m/s
Substituting t = a into equation 1
v = 58 - 2(0.83×a)
v = (58 - 1.66a) m/s
(c) When will the arrow hit the surface? (Round your answer to one decimal place.) t = s
the time when H = 0
Substituting H = 0, we have;
H = 58t - 0.83t^2 = 0
0.83t^2 = 58t
0.83t = 58
t = 58/0.83
t = 69.9 s
(d) With what velocity will the arrow hit the surface? m/s
from equation 1;
v = dH/dt = 58 - 2(0.83t)
Substituting t = 69.9s
v = 58 - 2(0.83×69.9)
v = -58.0 m/s
After landing on an unexplored Klingon planet, Spock tests for the direction of the magnetic field by firing a beam of electrons in various directions and by observing the following: Electrons moving upward feel a magnetic force in the NW direction; Electrons moving horizontally North are pushed down; Electrons moving horizontally South-East are pushed upward. He naturally concludes that the magnetic field at this landing site is in which direction?
Answer:
Magnetic field is in south west direction .
Explanation:
Let us represent various direction by i , j, k . i representing east , j representing north and k representing vertically upward direction .
magnetic field is represented vectorially as follows
B = B₀ ( - i - j )
In the first case velocity of electron
v = v k
Force = q ( v x B )
= -e [ vk x B₀ ( - i - j ) ]
= evB₀ ( j -i )
Direction of force is north -west .
In the second case velocity of electron
v = vj
Force = -e [ vj x B₀ ( - i - j ) ]
= - evB₀ k
force is downward
In the third case, velocity of electron
v = v( -j +i )
Force = -e [ v( -j +i ) x B₀ ( - i - j ) ]
= 2 evB₀ k
Force is upward.
Consider an incident normal shock wave that reflects from the end wall of a shock tube. The air in the driven section of the shock tube (ahead of the incident wave) is at p, = 0.01 atm and TI = 300 K. The pressure ratio across the incident shock is 1050. With the use of Eq. (7.23), calculate a. The reflected shock wave velocity relative to the tube b. The pressure and temperature behind the reflected shock
Answer:
Find the given attachments for complete solution
Kyle lays a mirror flat on the floor and aims a laser at the mirror. The laser beam reflects off the mirror and strikes an adjacent wall. The plane of the incident and reflected beams is perpendicular to the wall. The beam from the laser strikes the mirror at a distance a = 16.3 cm a=16.3 cm from the wall. The reflected beam strikes the wall at a height b = 32.5 cm b=32.5 cm above the surface of the mirror. Find the angle of incidence θ i θi at which the laser beam strikes the mirror.
Answer:
26.64°
Explanation:
Given:
a = 16.3 cm
b = 32.5 cm
Angle when laser beam reflects off the mirror and strike the wall =
θ [tex] = tan^-^1(\frac{b}{a}) [/tex]
[tex] = tan^-^1(\frac{32.5}{16.3}) [/tex]
= 63. 36°
For angle of reflection, we have:
θr = 90° - 63.36°
θr = 26.64°
Since angle of incidence, θi is equal to angle of reflection θr, the angle of incidence θi at which the laser beam strikes the mirror is =
θi = θr = 26.64°
A plane flies toward a stationary siren at 1/4 the speed of sound. Then the plane stands still on the ground and the siren is driven toward it at 1/4 the speed of sound. In both cases, a person sitting in the plane will hear the same frequency of sound from the siren. A plane flies toward a stationary siren at 1/4 the speed of sound. Then the plane stands still on the ground and the siren is driven toward it at 1/4 the speed of sound. In both cases, a person sitting in the plane will hear the same frequency of sound from the siren. True False
Answer:
The question above is repeated twice.
Removing the repetition, we have: A plane flies toward a stationary siren at 1/4 the speed of sound. Then the plane stands still on the ground and the siren is driven toward it at 1/4 the speed of sound. In both cases, a person sitting in the plane will hear the same frequency of sound from the siren. True or False?
The correct answer to the question is "False"
Explanation:
The question above, illustrates a phenomenon referred to as "Doppler effect"
The Doppler effect only changes the frequency of the sound which explains how wavelength changes when a wave source is moving toward or away from an object. The Doppler effect occurs when a source of waves and/or observer move relative to each other.
When a sound source is moving toward the observer (a person sitting in the plane) in the case above, the observer will hear a higher pitch as the source approaches. That is, the plane stands still on the ground and the siren is driven toward it.This is due to a decrease in the amplitude of the sound wave.
However, If the observer moves toward the stationary source, the observed frequency is higher than the source frequency. In this case, A plane flies toward a stationary siren.
λ = v/f = vT,
where T is the period,
The relationship between frequency, speed, and wavelength is:
f = v/λ
v represents the speed of sound through the medium.
Doppler effect depends on things moving, as the observer moves, the frequency becomes higher as the distance decreases. If the observer moves and the distance becomes larger, it means that the sound frequency becomes lower.
Final answer:
The Doppler effect explains why a person sitting in a plane moving toward or away from a stationary siren at 1/4 the speed of sound perceives the same frequency of sound. This phenomenon is true due to the relative motion between the source of sound and the observer.
Explanation:
True
The phenomenon described in the question is related to the Doppler effect in physics. When a source of sound and an observer are in motion relative to each other, the frequency of the sound waves changes due to this motion. In this case, when the plane is moving toward or away from the siren at 1/4 the speed of sound, the observer perceives the same frequency of sound from the siren.
A 300 MHz electromagnetic wave in air (medium 1) is normally incident on the planar boundary of a lossless dielectric medium with ϵr = 9 (medium 2). What is the wavelength of the incident wave and the wave in medium 2? What are the intrinsic impedances of media 1 and 2? What are the reflection coefficient and the transmission coefficient at the boundary? If the amplitude of the incident electric field is 10 V/m, what are the maximum amplitudes of the total fields in media 1 and 2? A standing wave pattern appears in medium 1. What are the locations of the first minimum and maximum?
Answer:
Wavelength of the incident wave in air = 1 m
Wavelength of the incident wave in medium 2 = 0.33 m
Intrinsic impedance of media 1 = 377 ohms
Intrinsic impedance of media 2 = 125.68 ohms
Check the explanation section for a better understanding
Explanation:
a) Wavelength of the incident wave in air
The frequency of the electromagnetic wave in air, f = 300 MHz = 3 * 10⁸ Hz
Speed of light in air, c = 3 * 10⁸ Hz
Wavelength of the incident wave in air:
[tex]\lambda_{air} = \frac{c}{f} \\\lambda_{air} = \frac{3 * 10^{8} }{3 * 10^{8}} \\\lambda_{air} = 1 m[/tex]
Wavelength of the incident wave in medium 2
The refractive index of air in the lossless dielectric medium:
[tex]n = \sqrt{\epsilon_{r} } \\n = \sqrt{9 }\\n =3[/tex]
[tex]\lambda_{2} = \frac{c}{nf}\\\lambda_{2} = \frac{3 * 10^{6} }{3 * 3 * 10^{6}}\\\lambda_{2} = 1/3\\\lambda_{2} = 0.33 m[/tex]
b) Intrinsic impedances of media 1 and media 2
The intrinsic impedance of media 1 is given as:
[tex]n_1 = \sqrt{\frac{\mu_0}{\epsilon_{0} } }[/tex]
Permeability of free space, [tex]\mu_{0} = 4 \pi * 10^{-7} H/m[/tex]
Permittivity for air, [tex]\epsilon_{0} = 8.84 * 10^{-12} F/m[/tex]
[tex]n_1 = \sqrt{\frac{4\pi * 10^{-7} }{8.84 * 10^{-12} } }[/tex]
[tex]n_1 = 377 \Omega[/tex]
The intrinsic impedance of media 2 is given as:
[tex]n_2 = \sqrt{\frac{\mu_r \mu_0}{\epsilon_r \epsilon_{0} } }[/tex]
Permeability of free space, [tex]\mu_{0} = 4 \pi * 10^{-7} H/m[/tex]
Permittivity for air, [tex]\epsilon_{0} = 8.84 * 10^{-12} F/m[/tex]
ϵr = 9
[tex]n_2 = \sqrt{\frac{4\pi * 10^{-7} *1 }{8.84 * 10^{-12} *9 } }[/tex]
[tex]n_2 = 125.68 \Omega[/tex]
c) The reflection coefficient,r and the transmission coefficient,t at the boundary.
Reflection coefficient, [tex]r = \frac{n - n_{0} }{n + n_{0} }[/tex]
You didn't put the refractive index at the boundary in the question, you can substitute it into the formula above to find it.
[tex]r = \frac{3 - n_{0} }{3 + n_{0} }[/tex]
Transmission coefficient at the boundary, t = r -1
d) The amplitude of the incident electric field is [tex]E_{0} = 10 V/m[/tex]
Maximum amplitudes in the total field is given by:
[tex]E = tE_{0}[/tex] and [tex]E = r E_{0}[/tex]
E = 10r, E = 10t
g Two cars, car 1 and car 2 are traveling in opposite directions, car 1 with a magnitude of velocity v1=13.0 m/s and car 2 v2= 7.22 m/s. If car 1’s exhaust system is loud enough to be heard by car 2 and the frequency fe produced from the exhaust is 2.10 kHz. What frequencies would be heard by car 2 when the cars are approaching, passing, and retreating from one another?
Answer:
When they are approaching each other
[tex]f_a = 2228.7 \ Hz[/tex]
When they are passing each other
[tex]f_a = 2100Hz[/tex]
When they are retreating from each other
[tex]f_a = 1980.7 Hz[/tex]
Explanation:
From the question we are told that
The velocity of car one is [tex]v_1 = 13.0 m/s[/tex]
The velocity of car two is [tex]v_2 = 7.22 m/s[/tex]
The frequency of sound from car one is [tex]f_e = 2.10 kHz[/tex]
Generally the speed of sound at normal temperature is [tex]v = 343 m/s[/tex]
Now as the cars move relative to each other doppler effect is created and this can be represented mathematically as
[tex]f_a = f_o [\frac{v \pm v_o}{v \pm v_s} ][/tex]
Where [tex]v_s[/tex] is the velocity of the source of sound
[tex]v_o[/tex] is the velocity of the observer of the sound
[tex]f_o[/tex] is the actual frequence
[tex]f_a[/tex] is the apparent frequency
Considering the case when they are approaching each other
[tex]f_a = f_o [\frac{v + v_o}{v - v_s} ][/tex]
[tex]v_o = v_2[/tex]
[tex]v_s = v_1[/tex]
[tex]f_o = f_e[/tex]
Substituting value
[tex]f_a = 2100 [\frac{343 + 7.22}{ 343 - 13} ][/tex]
[tex]f_a = 2228.7 \ Hz[/tex]
Considering the case when they are passing each other
At that instant
[tex]v_o = v_s = 0m/s[/tex]
[tex]f_o = f_e[/tex]
[tex]f_a = f_o [\frac{v }{v } ][/tex]
[tex]f_a = f_o[/tex]
Substituting value
[tex]f_a = 2100Hz[/tex]
Considering the case when they are retreating from each other
[tex]f_a = f_o [\frac{v - v_o}{v + v_s} ][/tex]
[tex]v_o = v_2[/tex]
[tex]v_s = v_1[/tex]
[tex]f_o = f_e[/tex]
Substituting value
[tex]f_a = 2100 [\frac{343 - 7.22}{343 + 13} ][/tex]
[tex]f_a = 1980.7 Hz[/tex]
Blood flow rates in the umbilical cord can be found by measuring the Doppler shift of the ultrasound signal reflected by the red blood cells. If the source emits a frequency f, what is the measured reflected frequency fR? Assume that all of the red blood cells move directly toward the source. Let c be the speed of sound in blood and v be the speed of the red blood cells.
Answer:
fR = f(c + v)/c
Explanation:
The speed of a wave is its frequency x wavelenght. Therefore,
Frequency is speed of wave over the wavelength.
Since the source (ultrasound machine) is stationary, and the receiver red blood cell is moving towards it. The wavelenght of the wave sent out towards the observer is c/f
The speed of the reflected sound wave is (c + v), so that the reflected frequency fR is given by
fR = f(c + v)/c
The measured reflected frequency ( fR) in Doppler-shifted ultrasound, when blood cells move towards the source, is calculated using the formula fR = f * (c + v) / c, where f is the original frequency, c is the speed of sound in blood, and v is the velocity of blood cells.
The question relates to the principle of the Doppler effect and its application in calculating the velocity of blood flow using Doppler-shifted ultrasound. The Doppler effect occurs when a wave source and an observer are in relative motion, resulting in a change in the observed frequency. Specifically, when the blood cells move towards the ultrasound source, the frequency of the reflected ultrasound increases. This change in frequency can be measured for diagnostic purposes.
To find the measured reflected frequency fR when blood cells are moving towards the source, you can use the formula:
fR = f * (c + v) / c
Where,
f = original frequency of the ultrasound
c = speed of sound in blood (or human tissue)
v = velocity of blood cells relative to the ultrasound source
The simple pendulum above consists of a bob hanging from a light string. You wish to experimentally determine the frequency of the swinging pendulum. (a) By checking the line next to each appropriate item on the list below, select the equipment that you would need to do the experiment. ____ Meter Stick ____ Protractor ____ Additional string ____ Stopwatch ____ Photogate ____ Additional masses (b) Describe the experimental procedure that you would use. In your description, state the measurements you would make, how you would use the equipment to make them, and how you would determine the frequency from those measurements. (c) You next wish to discover which parameters of a pendulum affect its frequency. State one parameter that could be varied, describe how you would conduct the experiment, and indicate how you would analyze the data to show whether there is a dependence. (d) After swinging for a long time, the pendulum eventually comes to rest. Assume that the room is perfectly thermally insulated. How will the temperature of the room change while the pendulum comes to rest
A) the item required for the experiment is a stopwatch.
B) You are looking for time savings. To get this, divide the time by 10. The result is the period. The formula is given as
F = [tex]\frac{1}{T}[/tex] Where F is the frequency i.e. the number of cycles per second and T is the number of seconds per cycle.
C) One parameter of the pendulum that can be altered in order to affect the frequency of the pendulum is its length. A pendulum with a longer string will have a lower frequency.
The one with a shorter length will have a higher frequency
D) In an environment that is thermally insulated perfectly, it means that any heat generated within the room is trapped within it. As the pendulum comes to rest, the room will experience a slight increase in temperature due to the conversion of mechanical energy to thermal energy.
Uses of the pendulum experimentsUnderstanding the physics of pendulums helps one to get a better grasp of gravity, inertia, and centripetal force.
Pendulums are used for the construction or engineering of clocks, metronomes, sismometers, amusement park rides.
See more exercises with pendulums in the link below:
https://brainly.com/question/6951404
The simple pendulum above consists of a bob hanging from a light string the experiments is :
A. Stopwatch
B. You are looking for time savings. To get this, divide the time by 10.
C. The one parameter of the pendulum that can be modified in arrange to influence the recurrence of the pendulum is its length.
D. In an environment that's thermally protects superbly
"Simple Pendulum"Answer A:
The item that is required for the experiment is a stopwatch.Answer B:
The experimental procedure that you would use is :
You are looking for time savings. To get this, divide the time by 10. The result is the period. Formula is given as :F = 1/10Where :
F is the frequency and T is the number of seconds per cycle.Answer C:
The one parameter of the pendulum that can be modified in arrange to influence the recurrence of the pendulum is its length. A pendulum with a longer string will have a lower frequency. The one with a shorter length will have a better frequency.Answer D:
In an environment that's thermally protects superbly, it implies that any warm created inside the room is caught inside it. As the pendulum comes to rest, the room will encounter a slight increment in temperature due to the change of mechanical vitality to warm energy.Uses of the pendulum experiments :
Understanding the material science of pendulums helps one to induce distant better a much better, a higher, a stronger ,an improved an improved get a handle on of gravity, inactivity, and centripetal force. Pendulums are utilized for the development or building of clocks, metronomes, sismometers, entertainment stop rides.Learn more about pendulum experiments :
https://brainly.com/question/14480660?referrer=searchResults
What is the distance from the center of the Moon to the point between Earth and the Moon where the gravitational pulls of Earth and Moon are equal? The mass of Earth is 5.97 × 1024 kg, the mass of the Moon is 7.35 × 1022 kg, the center-to-center distance between Earth and the Moon is 3.84 × 108 m, and G = 6.67 × 10-11 N ∙ m2/kg2. Group of answer choices 4.69 × 107 m 4.69 × 106 m 3.83 × 106 m 3.45 × 108 m
Answer:
3.83*10^7 m
Explanation:
Assume that the distance at which gravitational force due to both Earth and moon is zero and it l is given by force force balance
F(moon) = F(earth)
F(moon) = GM(moon) / r²
F(earth) = GM(earth) / (d - r)²
If F(moon) = F(earth), then
GM(moon) / r² = GM(earth) / (d - r)²
7.35*10^22 / r² = 5.97*10^24 / (3.84*10^8 - r)²
now, we take the square root of both sides, we have
2.71*10^11 / r² = 24.4*10^11 / (3.84*10^8 - r) =>
2.71 / r² = 24.4 / (3.84*10^8 - r)
if we cross multiply, we have
24.4r = 1.04064*10^9 - 2.71r
24.4r + 2.71r = 1.04064*10^9
27.11r = 1.04064*10^9
r = 1.04064*10^9 / 27.11
r = 3.83*10^7 m
A 969-kg satellite orbits the Earth at a constant altitude of 99-km. (a) How much energy must be added to the system to move the satellite into a circular orbit with altitude 195 km? 469 Incorrect: Your answer is incorrect. How is the total energy of an object in circular orbit related to the potential energy? MJ (b) What is the change in the system's kinetic energy? MJ (c) What is the change in the system's potential energy?
Answer:
1.3*10^14 J
Explanation:
The energy of the satellite that orbits the earth is given by the second Newton law:
[tex]F=ma_c\\\\-G\frac{mM_s}{r^2}=m\frac{v^2}{r}\\\\v^2=\frac{GM}{r}\\\\E_T=K+U=G\frac{mM_s}{2r}-G\frac{mM}{r}=-G\frac{mM}{2r}[/tex]
where you have taken into account the centripetal acceleration of the satellite.
m: mass of the satellite
M_s: mass of the sun = 1.98*10^30 kg
G: Cavendish's constant = 6.67*10^-11 m^3/kg s^2
r: distance to the center of the Earth = Earth radius + distance satellite-Earth surface
To find the needed energy, you first compute the energy for a constant altitude of 99km:
r = 6.371*10^6m + 99*10^3m = 6.47*10^6 m
[tex]E_T=-(6.67*10^-11 m^3 /kg.s^2)\frac{(969kg)(1.98*10^{30}kg)}{2(6.47*10^6)}\\\\E_T=-9.88*10^{15} \ J[/tex]
Next, you calculate the energy for an altitude of 195km:
r = 6.371*10^6m + 195*10^{3}m = 6.56*10^6 m
[tex]E_T=-(6.67*10^-11 m^3 /kg.s^2)\frac{(969kg)(1.98*10^{30}kg)}{2(6.56*10^6)}\\\\E_T=-9.75*10^{15} \ J[/tex]
Finally, the energy required to put the satellite in the new orbit is:
-9.75*10^15 J - (-9.88*10^15 J) = 1.3*10^14 J
A sound is traveling through the air with a temperature of 35•C. The sound wave has a wavelength of 0.75 meters. What is the frequency of the sound
Answer:
f = 687.85 Hz
Explanation:
Given that,
The wavelength of sound wave, [tex]\lambda=0.75\ m[/tex]
We need to find the frequency of the sound wave. The relation between wavelength and frequency is given by :
[tex]v=f\lambda[/tex]
v is speed of sound at T = 35°C = 308.15 K
[tex]v=331+0.6T\\\\v=331+0.6\times 308.15 \\\\v=515.89\ m/s[/tex]
So,
[tex]f=\dfrac{v}{\lambda}\\\\f=\dfrac{515.89}{0.75}\\\\f=687.85\ Hz[/tex]
So, the frequency of the sound is 687.85 Hz.
Final answer:
To calculate the frequency of a sound wave, use the formula frequency = speed of sound / wavelength. For this specific scenario, the frequency of the sound wave is 3400 Hz.
Explanation:
The frequency of the sound wave can be calculated using the formula:
frequency = speed of sound / wavelength
Given that the speed of sound in air is 340 m/s and the wavelength is 0.10 m, we can plug in the values to find the frequency:
frequency = 340 m/s / 0.10 m = 3400 Hz
Therefore, the frequency of the sound wave is 3400 Hz.
A sphere of radius 5.00 cm carries charge 3.00 nC. Calculate the electric-field magnitude at a distance 4.00 cm from the center of the sphere and at a distance 6.00 cm from the center of the sphere if the sphere is a solid insulator with the charge spread uniformly throughout its volume. Express your answers in newtons per coulomb separated by a comma. Repeat part A if the sphere is a solid conductor.
Answer:
a) E = 8.63 10³ N /C, E = 7.49 10³ N/C
b) E= 0 N/C, E = 7.49 10³ N/C
Explanation:
a) For this exercise we can use Gauss's law
Ф = ∫ E. dA = [tex]q_{int}[/tex] /ε₀
We must take a Gaussian surface in a spherical shape. In this way the line of the electric field and the radi of the sphere are parallel by which the scalar product is reduced to the algebraic product
The area of a sphere is
A = 4π r²
if we use the concept of density
ρ = q_{int} / V
q_{int} = ρ V
the volume of the sphere is
V = 4/3 π r³
we substitute
E 4π r² = ρ (4/3 π r³) /ε₀
E = ρ r / 3ε₀
the density is
ρ = Q / V
V = 4/3 π a³
E = Q 3 / (4π a³) r / 3ε₀
k = 1 / 4π ε₀
E = k Q r / a³
let's calculate
for r = 4.00cm = 0.04m
E = 8.99 10⁹ 3.00 10⁻⁹ 0.04 / 0.05³
E = 8.63 10³ N / c
for r = 6.00 cm
in this case the gaussine surface is outside the sphere, so all the charge is inside
E (4π r²) = Q /ε₀
E = k q / r²
let's calculate
E = 8.99 10⁹ 3 10⁻⁹ / 0.06²
E = 7.49 10³ N/C
b) We repeat in calculation for a conducting sphere.
For r = 4 cm
In this case, all the charge eta on the surface of the sphere, due to the mutual repulsion between the mobile charges, so since there is no charge inside the Gaussian surface, therefore the field is zero.
E = 0
In the case of r = 0.06 m, in this case, all the load is inside the Gaussian surface, therefore the field is
E = k q / r²
E = 7.49 10³ N / C
Three balls with the same radius 21 cm are in water. Ball 1 floats, with
half of it exposed above the water level. Ball 2, with a density 893
kg/m3 is held below the surface by a cord anchored to the bottom of the
container, so that it is fully submerged. Ball 3, of density 1320 kg/m3, is
suspended from a rope so that it is fully submerged. Assume the
density of water is 1000 kg/m3 in this problem.
A. Which is true for Ball 1?
B. What is the tension on the rope holding the second ball, in newtons?
C. What is the tension on the rope holding the third ball in N?
Ball 1 floats with half exposed above water level. Tension on rope holding Ball 2 is calculated using weight and buoyant force. Tension on rope holding Ball 3 is equal to buoyant force.
Explanation:A. Ball 1 is floating with half of it exposed above the water level.
This means that the buoyant force on the ball is equal to the weight of the ball.
Since the buoyant force is greater than the weight of the ball, the ball floats.
B. The tension on the rope holding Ball 2 can be found using the equation:
Tension = Weight - Buoyant force.
The weight of the ball is calculated by multiplying its volume by its density and acceleration due to gravity.
The buoyant force can be found by multiplying the volume of the ball submerged in water by the density of water and acceleration due to gravity.
C. The tension on the rope holding Ball 3 is the same as the buoyant force acting on it.
The buoyant force can be found by multiplying the volume of the ball submerged in water by the density of water and acceleration due to gravity.
Ball 1's density is 500 kg/m³. The tension on the rope holding Ball 2 is 41.15 N. The tension on the rope holding Ball 3 is 121.24 N.
Let's solve this problem step-by-step to better understand floating and submerged objects in water.
A.) Which is true for Ball 1?
Ball 1 floats with half of it exposed above the water level. This means the density of Ball 1 must be half the density of water. Since the density of water is 1000 kg/m³, the density of Ball 1 is:
500 kg/m³B.) What is the tension on the rope holding the second ball, in newtons?
Ball 2 has a density of 893 kg/m³ and is held below the surface of water. The buoyant force is equal to the weight of the volume of water displaced by Ball 2.
Calculate the volume of Ball 2: (Volume of a sphere = 4/3 π r³)
r = 21 cm r = 0.21 mVolume = (4/3) π (0.21)³ = 0.0388 m³Calculate the buoyant force:
Buoyant Force = Density of Water x Volume of Water x gBuoyant Force = 1000 kg/m³ x 0.0388 m³ x 9.8 m/s² Bouyant force = 380.24 NCalculate the weight of Ball 2:
Weight = Density of Ball 2 x Volume x gWeight = 893 kg/m³ x 0.0388 m³ x 9.8 m/s² = 339.09 NCalculate the tension in the rope:
Tension = Weight - Buoyant ForceTension = 339.09 N - 380.24 N = -41.15 N (negative indicates an upward force)Tension on the rope for Ball 2: 41.15 NC.) What is the tension on the rope holding the third ball in N?
Ball 3 has a density of 1320 kg/m³ and is also fully submerged, suspended by a rope.
Calculate the weight of Ball 3:
Weight = Density of Ball 3 x Volume x gWeight = 1320 kg/m³ x 0.0388 m³ x 9.8 m/s² = 501.48 NCalculate the tension in the rope:
Tension = Weight - Buoyant ForceTension = 501.48 N - 380.24 N = 121.24 NTension on the rope for Ball 3: 121.24 NThis example shows how to calculate buoyant forces and tensions for submerged and floating objects.
Tish scarf is 17.75 inches long and knits 2 3/8 per minute. Emma scarf is 4 inches knits 3 3/4 inches per minute. After how many minutes will Emma’s scarf be longer than tish scarf
Answer:
t = 10 minutes
Emma’s scarf will be longer than tish scarf after 10 minutes
Explanation:
Given;
The length of tish scarf after t minutes can be written as;
L1 = 17.75 + 2 3/8 ×t
L1 = 17.75 + 2.375t
The length of emma scarf after t minutes can be written as;
L2 = 4 + 3 3/4 × t
L2 = 4 + 3.75t
For emma scarf to be longer than tish;
L2 >/= L1
At L2 = L1
4+3.75t = 17.75 + 2.375t
Solving for t
3.75t - 2.375t = 17.75 - 4
1.375t = 13.75
t = 13.75/1.375
t = 10 minutes
Emma’s scarf will be longer than tish scarf after 10 minutes
Katie rubs a balloon against her hair. Electrons from her hair travel to the balloon, giving the balloon a negative charge and her hair a positive charge.
When the negatively charged balloon is brought near the strands of Katie's hair, they move to get closer to the balloon, without the balloon actually touching them. This shows that
A. electric attraction is a force that can only act on contact.
B. Katie's hair would move in this way with or without the balloon.
C. electric attraction is a force that can act at a distance.
D. particles in the air must be pulling Katie's hair toward the balloon.
Answer:
c. electric attraction is a force that can act at a distance.
Explanation:
stuisland
Final answer:
A negatively charged balloon attracting positively charged hair strands without contact illustrates that c. electric attraction can act at a distance.
Explanation:
When Katie rubs the balloon against her hair, electrons move from her hair to the balloon, resulting in the balloon having a negative charge and her hair having a positive charge. If the negatively charged balloon is brought near Katie's hair and the hair strands move toward it without direct contact, this demonstrates electric attraction as a force that can act at a distance.
Therefore, electric attraction is a force that can act at a distance. This example, similar to the effect observed when someone touches a Van de Graaff generator, shows charge separation and induction. It validates the scientific concept that electric forces can operate between charged objects even when they are not in physical contact.
how many grams of nitrogen gas are needed to produce 34 g of ammonia
Answer:
28 grams
Explanation:
The equation for the reaction is
3H(2) + N(2) -> 2NH(3)
Then we have.
The molar mass, M of ammonia is 17 g/mol.
34 grams of ammonia, NH3 then would be
34 g / 17 g/mol
= 2 moles
2 moles of ammonia will be obtained from
(2 * 1) / 2
= 1 mole of nitrogen
The molar masses of nitrogen is 28 g/mol
2 moles of nitrogen corresponds to 1 * 28 = 28 grams.