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
wo plastic bowling balls, 1 and 2, are rubbed with cloth until they each carry a uniformly distributed charge of magnitude 0.70 nC . Ball 1 is negatively charged, and ball 2 is positively charged. The balls are held apart by a 500-mm stick stuck through the holes so that it runs from the center of one ball to the center of the other.
What is the magnitude of the dipole moment of the arrangement?
Answer:
the magnitude of the dipole moment is 3.5*10^-11Cm
Explanation:
The dipole moment is given by the following formula:
[tex]\mu=qr[/tex]
r: distance between the centers of the charges = 500mm
q: charges of the bowling balls = 0.70nC
By replacing you obtain:
[tex]\mu=(0.70*10^{-9}C)(500*10^{-4}m)=3.5*10^{-11}Cm[/tex]
hence, the magnitude of the dipole moment is 3.5*10^-11Cm
Interference in 2D. Coherent red light of wavelength λ = 700 nm is incident on two very narrow slits. The light has the same phase at both slits. (a) What is the angular separation in radians between the central maximum in intensity and an adjacent maximum if the slits are 0.025 mm apart? (b) What would be the separation between maxima in intensity on a screen located 1 m from the slits? (c) What would the angular separation be if the slits were 2.5 mm apart? (d) What would the separation between maxima be on a screen 25 mm from the slits? The answer is relevant to the maximum resolution along your retina that would be useful given the size of your pupil. The spacing between cones on the retina is about 10 µm. (Of course, you would never want to look directly into a laser beam like this....) (e) How would the location of the maxima change if one slit was covered by a thin film with higher index of refraction that shifted the phase of light leaving the slit by π? Would the separation change? 3. Combined two source interference and
Answer:
Explanation:
λ
given λ = 700 nm
a) for first maxima, d*sin(θ) = λ
sin(theta) = λ/d
= 700*10^-9/(0.025*10^-3)
= 0.028
theta θ = sin^-1(0.028)
= 1.60 degrees
b) given R = 1 m,
delta_y = λ*R/d
= 700*10^-9*1/(0.025*10^-3)
= 0.028 m or 2.8 cm
c) for first maxima, d*sin(θ) = λ
sin(θ) = λ/d
= 700*10^-9/(2.5*10^-3)
= 0.00028
theta = sin^-1(0.00028)
= 0.0160 degrees
d) R = 25 mm = 0.025 m
δ_y = λ*R/d
= 700*10^-9*0.025/(2.5*10^-3)
= 7*10^-6 m or 7 micro m
e) No. But the position of maxima and minima will be shifted.
Write a function to model the volume of a rectangular prism if the length is 24cm and the sum of the width and height is 28cm
[tex]w + h = 28[/tex]
[tex]l \times w \times h = v[/tex]
[tex]v = w \times h \times 24[/tex]
[tex]w = 28 - h[/tex]
[tex]h(28 - h) \times 24 = v[/tex]
[tex]24( - {h}^{2} + 28h) = v[/tex]
When you hear a noise, you usually know the direction from which it came even if you cannot see the source. This ability is partly because you have hearing in two ears. Imagine a noise from a source that is directly to your right. The sound reaches your right ear before it reaches your left ear. Your brain interprets this extra travel time (Δt) to your left ear and identifies the source as being directly to your right. In this simple model, the extra travel time is maximal for a source located directly to your right or left (Δt = Δtmax). A source directly behind or in front of you has equal travel time to each ear, so Δt = 0. Sources at other locations have intermediate extra travel times (0 ≤ Δt ≤ Δtmax). Assume a source is directly to your right.(a) Estimate the distance between a person's ears. (they gave us the answer of .2... apparently the program is messed up and we have to use .2(b) If the speed of sound in air at room temperature is vs = 338 m/s,find Δtmax. (Use your estimate.)
(c) Find Δtmax if instead you and the source are in seawater at the same temperature, where vs = 1534 m/s. (Use your estimate.)
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
The average distance calculated between a person's ears is 0.2 meters. The interaural time difference (ITD) is maximal when the sound source is directly to the side. For air (vs=338 m/s), Δ[tex]t_{max}[/tex] is 591.72 microseconds, and for seawater (vs= 1534 m/s), it is 130.38 microseconds. These calculations illustrate sound localization and the impact of medium on sound propagation.
Understanding Sound Localization and Interaural Time Difference
When a noise occurs directly to the right of a person, the sound waves reach the right ear before the left ear. We can calculate the maximum interaural time difference ( Δ[tex]t_{max}[/tex]) for sound reaching the ears using the given distance between the ears.
(a) The average distance between a person's ears was estimated as 0.2 meters (20 cm).
(b) To calculate Δ[tex]t_{max}[/tex] with the speed of sound in air (vs = 338 m/s), we can use the formula Δ[tex]t_{max}[/tex] = d / vs, where d is the distance between ears. Substituting the values, we get:
Δ[tex]t_{max}[/tex] = 0.2 m / 338 m/s = 0.0005917159763 seconds, or approximately 591.72 microseconds.
(c) Lastly, for sound traveling in seawater at room temperature where vs = 1534 m/s, we similarly get:
Δ[tex]t_{max}[/tex] = 0.2 m / 1534 m/s = 0.0001303763441 seconds, or approximately 130.38 microseconds.
This demonstrates the role of the medium in sound propagation and how it affects the interaural time difference.
En una competición de tiro con arco, la diana de 80 cm de diámetro se encuentra a 50 m de distancia y a 1,5 metros del suelo. En uno de los tiros la flecha sale a 230 km hora con una de 3,5 grados desde una altura de 1,60 metros,despreciando el rozamiento con el aire.¿ Impactará la flecha en la diana? En caso afirmativo¿ con qué velocidad y en qué dirección?
Answer:
a .
15.68
m/s
(b).
21.72
m
Explanation:
To determine if the arrow will hit the target, we can analyze the projectile motion. By calculating the time of flight and horizontal displacement, we can determine if the arrow will hit the target. The velocity and direction of the arrow at impact can also be determined using the horizontal and vertical components of velocity.
To determine if the arrow will hit the target, we need to analyze the projectile motion of the arrow. First, we need to calculate the time of flight using the equation t = 2 * (v * sin(theta))/g, where v is the initial velocity of the arrow and theta is the launch angle. Next, we can calculate the horizontal displacement using the equation x = v * cos(theta) * t, where x is the distance traveled horizontally. Comparing the calculated horizontal displacement with the distance to the target, we can determine if the arrow will hit the target.
Given the initial velocity of the arrow (230 km/hr), launch angle (3.5 degrees), and distance to the target (50 m), we can convert the initial velocity to m/s (230 km/hr * (1000 m/km) * (1 hr/3600 s)) and use the formulas to calculate time of flight and horizontal displacement. If the horizontal displacement is less than or equal to the distance to the target, the arrow will hit the target.
To find the velocity and direction of the arrow at impact, we can use the equations v_x = v * cos(theta) and v_y = v * sin(theta), where v_x is the horizontal component of velocity and v_y is the vertical component of velocity. Since the angle of impact is not specified, the direction of the arrow at impact will depend on the particular situation.
Follow these steps to solve this problem: Two identical loudspeakers, speaker 1 and speaker 2, are 2.0 m apart and are emitting 1700-Hz sound waves into a room where the speed of sound is 340 m/s. Consider a point 4.0 m in front of speaker 1, which lies along a line from speaker 1, that is perpendicular to a line between the two speakers. Is this a point of maximum constructive interference, a point of perfect destructive interference, or something in between?
Answer:
It is somewhere in between
Explanation:
Wave length of sound from each of the speakers = 340 / 1700 = 0.2 m = 20 cm
Distance between first speaker and the given point = 4 m.
Distance between second speaker and the given sound
D = √(4² + 2²)
D = √(16 + 4)
D = √20
D = 4.472 m
Path difference = 4.472 - 4 =
0.4722 m.
Path difference / wave length = 0.4772 / 0.2 = 2.386
This is a fractional integer which is neither an odd nor an even multiple of half wavelength. Hence this point is of neither a perfect constructive nor a perfect destructive interference.
To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following general equation y(x,t)=Acos(kx−ωt). A transverse wave on a string is traveling in the +x direction with a wave speed of 8.50 m/s , an amplitude of 5.50×10−2 m , and a wavelength of 0.500 m . At time t=0, the x=0 end of the string has its maximum upward displacement. Find the transverse displacement y of a particle at x = 1.52 m and t = 0.150 s .
Answer:
0.0549 m
Explanation:
Given that
equation y(x,t)=Acos(kx−ωt)
speed v = 8.5 m/s
amplitude A = 5.5*10^−2 m
wavelength λ = 0.5 m
transverse displacement = ?
v = angular frequency / wave number
and
wave number = 2π/ λ
wave number = 2 * 3.142 / 0.5
wave number = 12.568
angular frequency = v k
angular frequency = 8.5 * 12.568
angular frequency = 106.828 rad/sec ~= 107 rad/sec
so
equation y(x,t)=Acos(kx−ωt)
y(x,t)= 5.5*10^−2 cos(12.568 x−107t)
when x =0 and and t = 0
maximum y(x,t)= 5.5*10^−2 cos(12.568 (0) − 107 (0))
maximum y(x,t)= 5.5*10^−2 m
and when x = x = 1.52 m and t = 0.150 s
y(x,t)= 5.5*10^−2 cos(12.568 (1.52) −107(0.150) )
y(x,t)= 5.5*10^−2 × (0.9986)
y(x,t) = 0.0549 m
so the transverse displacement is 0.0549 m
A girl launches a toy rocket from the ground. The engine experiences an average thrust of 5.26 N. The mass of the engine plus fuel before liftoff is 25.0 g, which includes fuel mass of 12.7 g. The engine fires for a total of 1.90 s. (Assume all the fuel is consumed.) (a) Calculate the average exhaust speed of the engine (in m/s). (b) This engine is situated in a rocket case of mass 63.0 g. What is the magnitude of the final velocity of the rocket (in m/s) if it were to be fired from rest in outer space with the same amount of fuel
Answer:
a) v = 786.93 m/s
b) v = 122.40 m/s
Explanation:
a) To find the average exhaust speed (v) of the engine we can use the following equation:
[tex] F = \frac{v\Delta m}{\Delta t} [/tex]
Where:
F: is the thrust by the engine = 5.26 N
Δm: is the mass of the fuel = 12.7 g
Δt: is the time of the burning of fuel = 1.90 s
[tex]v = \frac{F*\Delta t}{\Delta m} = \frac{5.26 N*1.90 s}{12.7 \cdot 10^{-3} kg} = 786.93 m/s[/tex]
b) To calculate the final velocity of the rocket we need to find the acceleration.
The acceleration (a) can be calculated as follows:
[tex] a = \frac{F}{m} [/tex]
In the above equation, m is an average between the mass of the engine plus the rocket case mass and the mass of the engine plus the rocket case minus the fuel mass:
[tex]m = \frac{(m_{engine} + m_{rocket}) + (m_{engine} + m_{rocket} - m_{fuel})}{2} = \frac{2*m_{engine} + 2*m_{rocket} - m_{fuel}}{2} = \frac{2*25.0 g + 2*63.0 g - 12.7 g}{2} = 81.65 g[/tex]
Now, the acceleration is:
[tex] a = \frac{5.26 N}{81.65 \cdot 10^{-3} kg} = 64.42 m*s^{-2} [/tex]
Finally, the final velocity of the rocket can be calculated using the following kinematic equation:
[tex]v_{f} = v_{0} + at = 0 + 64.42 m*s^{-2}*1.90 s = 122.40 m/s[/tex]
I hope it helps you!
Final answer:
The average exhaust speed of the engine is calculated to be approximately 786.929 m/s, and the magnitude of the final velocity of the rocket if fired in outer space is roughly 94.916 m/s.
Explanation:
To calculate the average exhaust speed (v_ex), we can use the impulse-momentum theorem, which states that impulse is equal to the change in momentum of the system. Impulse is given by the product of the average force exerted by the engine (thrust) and the time interval during which the thrust is applied. If all the fuel is consumed, the change in mass (Δm) is the mass of the fuel.
Impulse = Thrust × Time = 5.26 N × 1.90 s = 9.994 N·s
The momentum change is equal to the mass of the fuel expelled times the average exhaust speed.
Δ(momentum) = Δm × v_ex
Substituting the impulse and solving for v_ex, we get:
v_ex = Impulse / Δm
v_ex = 9.994 N·s / 0.0127 kg = 786.929 m/s
Part B: Final Velocity of the Rocket in Space
The final velocity (V_final) of the rocket in space can be determined using the rocket equation, also known as Tsiolkovsky's rocket equation:
V_final = v_ex × ln(m_initial / m_final)
Where:
m_initial = mass of the rocket with fuel = 25.0 g + 63.0 g = 88.0 g = 0.088 kgm_final = mass of the rocket without fuel = 25.0 g - 12.7 g + 63.0 g = 75.3 g = 0.0753 kgCalculating the final velocity:
V_final = 786.929 m/s × ln(0.088 kg / 0.0753 kg) ≈ 94.916 m/s
Use the drop-down menus to complete the scenarios.
A patient has an ongoing history of cancer. She has a tumor in the abdominal region, and has been undergoing
treatment for it. There may be other tumors and a potential blockage in the surrounding area that need to be
investigated. The imaging technique that might provide the most information in this case is
Joe has ongoing issues with his throat and feels some sort of blockage or abnormality as he swallows. The doctor
decides to use X-ray imaging to visualize Joe's internal anatomy as he swallows to help determine the nature of the
problem.
will be used for this procedure.
First: CT Scan
Second: Fluoroscopy
Explanation:
Both correct
A CT scan can provide detailed images of a tumor and its surrounding area, which can be beneficial for a cancer patient. For ongoing throat issues, a Barium swallow study can be used, where the patient swallows a barium solution that is then visualized with an X-ray to identify abnormalities.
Explanation:In the first scenario, the patient with an ongoing history of cancer might benefit most from a Computed Tomography (CT) scan. CT scans are capable of creating detailed pictures of organs, bones, and other tissues, making it an excellent tool for capturing the size and position of a tumor and surrounding blockages in the abdominal area. Furthermore, it can reveal whether the cancer has spread to other parts of the body.
In the second scenario, Joe's doctor chooses to use X-ray imaging to detect any abnormalities in his throat. The specific procedure used for this is known as a Barium swallow study. This involves swallowing a barium solution that coats the esophagus, enabling the X-ray to capture clear images of the region as the patient swallows.
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Describe the process of nuclear fission. How does it work and when is it used? What is meant by a chain reaction, and how do you control it?
A 0.5 Kg pinball is initially at rest against a 120 N/m spring. The shooter is pulled back and has the spring compressed a distance of 0.2 m. The spring is released and the ball is shot up the ramp. It hits nothing and eventually comes to rest before it begins to roll down. We can ignore friction. The game board ramp is at an angle of 30o . How far did the ball travel on the board from the place of maximal compression to the first stop
Answer:
[tex]\Delta s = 0.978\,m[/tex]
Explanation:
The pinball-spring system is modelled after the Principle of Energy Conservation:
[tex]U_{g,1} + U_{k,1} + K_{1} = U_{g,2} + U_{k,2} + K_{2}[/tex]
[tex]-(0.5\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right) \cdot \Delta h + \frac{1}{2}\cdot \left(120\,\frac{N}{m}\right)\cdot (-0.2\,m)^{2} = 0[/tex]
The height reached by the pinball is:
[tex]\Delta h = 0.489\,m[/tex]
The distance travelled by the pinball is:
[tex]\Delta s =\frac{0.489\,m}{\sin 30^{\circ}}[/tex]
[tex]\Delta s = 0.978\,m[/tex]
When you run around a track at 5 km/h, your velocity is constant. Please select the best answer from the choices provided T F
Answer:
T
Explanation:
Answer:
true
Explanation:
n optician is performing Young's double-slit experiment for her clients. She directs a beam of monochromatic light to a pair of parallel slits, which are separated by 0.134 mm from each other. The portion of this light that passes through the slits goes on to form an interference pattern upon a screen, which is 4.50 meters distant. The light is characterized by a wavelength of 553 nm. (a) What is the optical path-length difference (in µm) that corresponds to the fourth-order bright fringe on the screen? (This is the fourth fringe, not counting the central bright band, that one encounters moving from the center out to one side.)
Answer:
Explanation:
This is a problem based on interference pf light waves.
wavelength of light λ = 553 nm
slit separation d = .134 x 10⁻³ m
screen distance D = 4.5 m
for fourth order bright fringe, path- length difference = 4 x λ
= 4 x 553
= 2212 nm .
= 2.212 μm
A cart of mass 300 g is placed on a frictionless horizontal air track. A spring having a spring constant of 9.0 N/m is attached between the cart and the left end of the track. The cart is displaced 3.8 cm from its equilibrium position. (a) Find the period at which it oscillates. Correct: Your answer is correct. s (b) Find its maximum speed. Incorrect: Your answer is incorrect. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m/s (c) Find its speed when it is located 2.0 cm from its equilibrium position. m/s
Answer:
Explanation:
the angular frequency ω of the pendulum is given by the formula
ω = [tex]\sqrt{\frac{k}{m} }[/tex] , k is spring constant , m is mass attached .
= [tex]\sqrt{\frac{9}{.3} }[/tex]
= 5.48 rad /s
time period = 2π / ω
= 2 x 3.14 / 5.48
= 1.146 s
b ) formula for speed
v = ω[tex]\sqrt{(a^2-\ x^2)}[/tex] , a is amplitude , x is displacement from equilibrium point.
for maximum speed x = 0
max speed = ωa
= 5.48 x 3.8 x 10⁻² ( initial displacement becomes amplitude that is 3.8 cm )
= .208 m /s
20.8 cm / s
c )
when x = .02 m , velocity = ?
v = ω[tex]\sqrt{(a^2-\ x^2)}[/tex]
= 5.48 [tex]\sqrt{(.038^2-\ .02^2)}[/tex]
= 5.48 x .0323109
= .177 m /s
17.7 cm /s .
To measure specific heat, the student flows air with a velocity of 20 m/s and a temperature of 25C perpendicular to the length of the tube, as illustrated below. In this time she heats the tube electrically, passing a current of 50 mA through the tube wall. The initial temperature of the tube is 25 C. After 10 min of heating, a thermocouple buried inside the tube wall measures a temperature of 31.2 C. The tube core is sealed and the heat lost from the internal surface of the tube is negligible during measurements. Furthermore, the tube is suspended on two thermally insulating supports. What is the specific heat of the tube?
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 58.1 cm ( 0.581 m) and the flow speed of the petroleum is 10.1 m/s. At the refinery, the petroleum flows at 5.85 m/s. What is the volume flow rate of the petroleum along the pipe and what is the pipe's diameter at the refinery?
volume flow rate:________m^3/sdiameter________cm
Answer:
volume flow rate:2.68m³/s
diameter :58.27cm
Explanation:
flow rate = Q = πr² v = amount per second that flows through the pipe
Given:
pipe's diameter= 0.581
r= 0.581/2=>0.2905m
speed 'v'= 10.1 m/s
Q= (3.142)(0.2905)²(10.1)
so
volume flow rate= 2.68m³/s
->if no oil has been added or subtracted or compressed then Q is the same everywhere
therefore,
Q= πr²v
2.68 = π(d/2)²(5.85)
d= (2.68x4) /(3.142 x 5.85)
d= 0.5827m =>58.27cm
One particle has a mass of 3.12 x 10-3 kg and a charge of +8.8 C. A second particle has a mass of 7.1 x 10-3 kg and the same charge. The two particles are initially held in place and then released. The particles fly apart, and when the separation between them is 0.15 m, the speed of the 3.12 x 10-3 kg particle is 131 m/s. Find the initial separation between the particles.
Answer:[tex]r_i=0.016119\ m\approx 16.119\ mm[/tex]
Explanation:
Given
mass of first particle is [tex]m_1=3.12\times 10^{-3}\ kg[/tex]
mass of second particle is [tex]m_2=7.1\times 10^{-3}\ kg[/tex]
Charge on both the particle [tex]q=8.8\times 10^{-6}\ C[/tex]
Now final speed of first particle is [tex]v_1=131\ m/s[/tex]
Final separation between particles is [tex]r=0.15\ m[/tex]
As there is no external force therefore linear momentum is conserved
[tex]0+0=m_1v_1+m_2v_2[/tex]
[tex]0=3.12\times 10^{-3}\times 131+7.1\times 10^{-3}\times v_2[/tex]
[tex]v_2=-\dfrac{3.12\times 10^{-3}}{7.1\times 10^{-3}}\times 131[/tex]
[tex]v_2=-57.56\ m/s[/tex]
Conserving total energy
Initial Kinetic energy +Initial Potential energy=Final Kinetic energy +Final Potential energy
[tex]\Rightarrow 0+\frac{kq^2}{r_i}=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2+\frac{kq^2}{r_f}[/tex]
[tex]\Rightarrow \frac{9\times 10^9\times 8.8^2\times 10^{-12}}{r_i}=\frac{1}{2}\times 3.12\times 10^{-3}\times 131^2+\frac{1}{2}7.1\times 10^{-3}\times (57.56)^2+\frac{9\times 10^9\times 8.8^2\times 10^{-12}}{0.15}[/tex]
[tex]\Rightarrow \frac{0.696}{r_i}=26.771+11.76+4.646[/tex]
[tex]\Rightarrow \frac{0.696}{r_i}=43.177[/tex]
[tex]r_i=0.016119\ m\approx 16.119\ mm[/tex]
Using the law of conservation of linear momentum and conservation of total energy, the obtained initial separation between the particles is 0.01612 m.
Conservation of Linear MomentumGiven that the masses of the two particles are;
[tex]m_1 = 3.12\times 10^{-3}\,kg\\m_2 = 7.1\times 10^{-3}\,kg[/tex]
Also, the charges of both the particles are equal.
[tex]q_1 = q_2 = +8.8\times 10^{-6} \,C[/tex]
The final separation between the particles is;
[tex]r_f = 0.15\,m[/tex]
Also, the final speed of the first particle is;
[tex](v_1)_{final} = 131\,m/s[/tex]
There is no external force applied here; so by the law of conservation of linear momentum, we have;
[tex](m_1v_1)_{initial} +(m_2v_2)_{initial}=(m_1v_1)_{final} +(m_2v_2)_{final}[/tex]
But initially, the particles are at rest, so the initial velocities are zero.
[tex]0+0=(3.12\times 10^{-3}\,kg \, \times 131\,m/s ) +(7.1\times 10^{-3}\,kg\,)\,(v_2)_{final}\\[/tex]
[tex]\implies (v_2)_{final}=-\frac{3.12\times 10^{-3}\,kg \, \times 131\,m/s}{7.1\times 10^{-3}\,kg} =-57.57\,m/s[/tex]
Conservation of EnergyNow, applying the law of conservation of total energy, we get;
[tex](KE)_{\,initial}\, + \,(PE)_{\,initial} = (KE)_{\,final}\, + \,(PE)_{\,final}[/tex]
But initially, the particles are at rest; so they have no initial kinetic energy.
They have electrostatic potential alone initially.
[tex]0+k\frac{q^2}{r_i} = \frac{1}{2}(m_1 v_1)_{initial} + \frac{1}{2}(m_2 v_2)_{final} + k\frac{q^2}{r_f}[/tex]
Substituting the known values, we get;
[tex](9\times 10^9 \, Nm^2/C^2)\frac{(8.8\times 10^{-6} \,C)^2}{r_i} =[ \frac{1}{2}(3.12\times 10^{-3}\,kg )\times (131\,m/s)^2] + [\frac{1}{2}(7.1\times 10^{-3}\,kg) \times (-57.57\,m/s)^2 + (9\times 10^9 \, Nm^2/C^2)\frac{(8.8\times 10^{-6} \,C)^2}{r_f}[/tex]
[tex]\implies\frac{0.696\,Nm^2}{r_i} =26.77\,J+11.76\,J+ 4.64\,J[/tex]
[tex]\implies r_i =\frac{0.696\,Nm^2}{43.17\,J}=0.01612\,m[/tex]
Learn more about conservation of momentum and energy in a collision here;
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Does this chemical equation support the Law of Conservation of Mass? Why or why not?
N₂ + H₂ → NH₃
Answer:No it does not the chemical equation is unbalanced,this is because according to the law of conservation it states that mass cannot be created or destroyed ,the chemical equation N₂ + H₂ → NH₃ shows that energy has been created which is not possibly so the correct balanced chemical is equation is 1N₂ +3H₂ → 3NH₃
Explanation:
Characteristics help scientists ________ objects.
Answer:Identify
Explanation:
Answer:
b
Explanation:
The air in a car tire la compressed when the car rolls over a rock. If the air
outside the tire becomes slightly warmer, what are the correct signs of heat
and work for this change?
Answer:
the signs of heat and work are; -Q and -W
Explanation:
The first law of thermodynamics is given by; ΔU = Q − W
where;
ΔU is the change in internal energy of a system,
Q is the net heat transfer (the sum of all heat transfer into and out of the system)
W is the net work done (the sum of all work done on or by the system).
Now, The system in this case is the tire and since the air gets warmer, heat must have left the system. Therefore Q is negative (-Q).
Since work is done by the system, W remains negative.
Thus, the signs of heat and work are; -Q and - W
When a rocket is traveling toward a mountain at 100 m/s, the sound waves from this rocket's engine approach the mountain at speed V.
If the rocket doubles its speed to 200 m/s, the sound waves from the engine will now approach the mountain at speed
A. 2–√V
B. 2V
C. V
D. 4V
Answer:
The correct option is C
Explanation:
From the question we are told that
The initial speed of the rocket is [tex]v_i = 100 m/s[/tex]
The speed of the rocket engine sound is [tex]V[/tex]
The final speed of the rocket is [tex]v_f = 200 \ m/s[/tex]
The speed of the sound at [tex]v_f[/tex] would still remain V this because the speed of sound wave is constant and is not dependent on the speed of the observer(The mountain ) or the speed of the source (The rocket ).
A clear example when lightning strikes you will first see (that is because it travels at the speed of light which is greater than the speed of sound) but it would take some time before you hear the sound of the lightning
Here we see that the speed of the lightning(speed of sound) does not affect the speed of the sound it generates
Which pair shows the law of reflection?
Answer:
The answer is A and C .
Explanation:
Reflection of object is reflected through the Normal (mirror) .
*incident angle = refracted angle
A sinusoidal electromagnetic wave is propagating in a vacuum in the z-direction. If at a particular instant and at a certain point in space the electric field is in the x-direction and has a magnitude of 4.50 V/m, what is the magnitude of the magnetic field of the wave at this same point in space and instant in time
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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A uniform-density 8 kg disk of radius 0.25 m is mounted on a nearly frictionless axle. Initially it is not spinning. A string is wrapped tightly around the disk, and you pull on the string with a constant force of 41 N through a distance of 0.9 m. Now what is the angular speed
Answer:
Explanation:
moment of inertia of the disc I = 1/2 m r²
= .5 x 8 x .25²
= .25 kg m²
The wok done by force will be converted into rotational kinetic energy
F x d = 1/2 I ω²
F is force applied , d is displacement , I is moment of inertia of disc and ω
is angular velocity of disc
41 x .9 = 1/2 x .25 ω²
ω² = .25
ω = 17.18 rad / s
The angular speed should be 17.18 rad / s
Calculation of the angular speed:Since
moment of inertia of the disc I = 1/2 m r²
= .5 x 8 x .25²
= .25 kg m²
Now the work done by force should be converted into the rotational kinetic energy
F x d = 1/2 I ω²
here,
F is the force applied,
d is displacement,
I is moment of inertia of disc
and ω is angular velocity of disc
So,
41 x .9 = 1/2 x .25 ω²
ω² = .25
ω = 17.18 rad / s
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When an earthquake occurs, two types of sound waves are generated and travel through the earth. The primary, or P, wave has a speed of about 8.0 km/s and the secondary, or S, wave has a speed of about 4.5 km/s. A seismograph, located some distance away, records the arrival of the P wave and then, 77.2 s later, records the arrival of the S wave. Assuming that the waves travel in a straight line, how far (in terms of m) is the seismograph from the earthquake?
Answer:[tex]d=7.94\times 10^5\ m[/tex]
Explanation:
Given
Speed of Primary wave [tex]v_1=8\ km/s[/tex]
Speed of secondary wave [tex]v_2=4.5\ km/s[/tex]
difference in timing of two waves are [tex]77.2\ s[/tex]
Suppose both travel a distance of d km then
[tex]t_1=\frac{d}{8}\quad \ldots (i)[/tex]
[tex]t_2=\frac{d}{4.5}\quad \ldots (ii)[/tex]
Subtract (ii) from (i)
[tex]\frac{d}{4.5}-\frac{d}{8}=77.2[/tex]
[tex]d[\frac{1}{4.5}-\frac{1}{8}]=77.2[/tex]
[tex]d[0.0972]=77.2[/tex]
[tex]d=794.23\ km[/tex]
[tex]d=7.94\times 10^5\ m[/tex]
Final answer:
To calculate the distance to the earthquake's epicenter, we use the time difference between the arrival of P-waves and S-waves and their speeds. By setting up an equation and solving for the distance, the seismograph is found to be approximately 618,940 meters from the epicenter.
Explanation:
When an earthquake occurs, two types of waves are generated: P-waves (primary waves) and S-waves (secondary waves), each with distinctive speeds. To determine the distance to the epicenter of the earthquake, we use the formula d = v × t, where d is distance, v is velocity, and t is time. Given that the P-wave has a speed of 8.0 km/s and the S-wave has a speed of 4.5 km/s, and the time difference of arrival between the two waves is 77.2 seconds, we can calculate the distance from the seismograph to the earthquake's epicenter.
Let the distance be d, then:
Time for P-wave to travel d: d / 8.0 km/sTime for S-wave to travel d: d / 4.5 km/sThe difference in travel time is 77.2 s, so: d / 4.5 km/s - d / 8.0 km/s = 77.2 sTo find the distance d, we solve the equation:
8.0×d - 4.5×d = 77.2 s × (8.0 km/s × 4.5 km/s)3.5×d = 77.2 s × 36 km²/s²d ≈ 618.94 kmNow, to convert kilometers to meters:
d ≈ 618.94 km × 1,000 m/kmd ≈ 618,940 mTherefore, the seismograph is approximately 618,940 meters from the earthquake's epicenter.
For an experiment, a student wants to maximize the power output for a circuit. The student has three resistors of resistances 512, 1012, and 20 S, and a 5 V battery. Which of the following best describes how the students should arrange the resistors so that when they are connected to the battery it will maximize the power output for the circuit and explains why? a) The student should place all resistors in parallel, because the battery supplies more current to a parallel combination than to a series combination b) The student should place all resistors in parallel, because the potential difference across all the resistors is less for a parallel combination than for a series combination c) The student should place all resistors in series, because this creates the largest equivalent resistance d) The student should place all resistors in series, because the battery supplies more current to a series combination than to a parallel combination e) The student should place all resistors in series, because the potential difference across all the resistors is greater in series than in parallel
Final answer:
To maximize the power output of a circuit with a 5V battery and three resistors, the resistors should be placed in parallel because this arrangement supplies more current than a series combination, thereby maximizing power.
Explanation:
To maximize the power output for a circuit with a given voltage source, the resistors should be arranged to draw the highest current possible. For resistors connected in parallel, the total resistance of the circuit decreases, as opposed to when they are in series, where it increases. Since power (P) is calculated using the formula P = V^2 / R, where V is the voltage and R is the resistance, minimizing R will maximize P.
Because a parallel configuration guarantees that each resistor gets the full voltage of the battery, the correct answer is: The student should place all resistors in parallel because the battery supplies more current to a parallel combination than to a series combination. This is because in a parallel circuit, the voltage across each resistor is equal to the voltage of the source, which maximizes the current through each resistor due to Ohm's law (I = V/R), and therefore maximizes the total current in the circuit.
Answer option (a) is thus correct: The student should place all resistors in parallel because the battery supplies more current to a parallel combination than to a series combination.
A 150 g pinball rolls towards a springloaded launching rod with a velocity of 2.0 m/s
to the west. The launching rod strikes the pinball and causes it to move in the
opposite direction with a velocity of 10.0 m/s. What impulse was delivered to the
pinball by the launcher?
A 0.75 kg•m/s to the east
B 1.2 kg•m/s to the east
C 1.8 kg•m/s to the east
D 3.0 kg•m/s to the east
The impulse delivered to the pinball by the launcher is 1.8 kg•m/s to the east
Impulse:Let us consider east as positive direction and west as negative direction.
Then, from the question, we get that the initial velocity of the pinball is
2m/s towards the west
or u = - 2 m/s
and the mass of the pinball is m = 150g = 0.15 kg
So, the initial momentum of the pinball is:
[tex]P_i=mu\\\\P_i=0.15\times(-2)\;kgm/s\\\\P_i=-0.3\;kgm/s[/tex]
Now, the final velocity of the pinball after being struck by the rod is 10 m/s towards the east,
or v = 10 m/s
So, the final momentum of the pinball is:
[tex]P_f=mu\\\\P_f=0.15\times(10)\;kgm/s\\\\P_f=1.5\;kgm/s[/tex]
Impulse is defined as the change in momentum, that is,
[tex]I=\Delta P\\\\I=P_f-P_i\\\\I=1.5-(-0.3)\\\\I=1.8\;kgm/s[/tex]
The impulse is 1.8 kgm/s towards the east.
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The potential energy of a 1.7 x 10-3 kg particle is described by U (x )space equals space minus (17 space J )space cos open square brackets fraction numerator x over denominator 0.35 space straight m end fraction close square brackets. What is the angular frequency of small oscillations around the point x = 0?
Answer:
Explanation:
Given a particle of mass
M = 1.7 × 10^-3 kg
Given a potential as a function of x
U(x) = -17 J Cos[x/0.35 m]
U(x) = -17 Cos(x/0.35)
Angular frequency at x = 0
Let find the force at x = 0
F = dU/dx
F = -17 × -Sin(x/0.35) / 0.35
F = 48.57 Sin(x/0.35)
At x = 0
Sin(0) =0
Then,
F = 0 N
So, from hooke's law
F = -kx
Then,
0 = -kx
This shows that k = 0
Then, angular frequency can be calculated using
ω = √(k/m)
So, since k = 0 at x = 0
Then,
ω = √0/m
ω = √0
ω = 0 rad/s
So, the angular frequency is 0 rad/s
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=33.7 cm from the wall. The reflected beam strikes the wall at a height b=36.7 cm above the surface of the mirror. Find the angle of incidence θi at which the laser beam strikes the mirror.
Answer:
The angle of incidence is [tex]\theta_i =42.6^o[/tex]
Explanation:
The diagram for this question is shown on the first uploaded image
From the question we are told that
The distance between the mirror and the wall is [tex]a = 33.7 cm[/tex]
The height of the above the mirror is [tex]b = 36.7 cm[/tex]
Generally the angle which the reflected ray make with the mirror is mathematically evaluated as
[tex]\alpha =tan ^{-1} (\frac{b}{a})[/tex]
substituting values
[tex]\alpha = tan ^{-1}( \frac{36.7}{33.7})[/tex]
[tex]\alpha =47.4^o[/tex]
From the diagram we can deduce that the angle of incidence is
[tex]\theta_i = 90 - \alpha[/tex]
So [tex]\theta_i = 90 - 47.4[/tex]
[tex]\theta_i =42.6^o[/tex]
A diffraction grating is illuminated with yellow light. The diffraction pattern seen on a viewing screen consists of three yellow bright spots, one at the central maximum (θ = 0°) and one on either side of it at θ = ±50°. Then the grating is simultaneously illuminated with red light. Where a red and a yellow spot overlap, an orange spot is produced. The new pattern consists of __________
Answer:
an orange fringe at 0°, yellow fringes at ±50° and red fringes farther out.
Explanation:
In the visible spectrum- red to violet, red has the highest wavelength.
The maximum internsity for diffraction grating is given by,
Sinθ[tex](_{max} )[/tex] = mλ/d
It is concluded that the angle of diffraction increases with increase in wavelength'λ' . So, red fringe will be farthest from the center, orange light will be at the center and yellow fringe will be at 50°.
Therefore, The new pattern consists of : an orange fringe at 0°, yellow fringes at ±50° and red fringes farther out.