Answer: 4.821m/s
Explanation:
initial momentum = final momentum
(80 x 4.1) + (85 x 5.5) = (80 + 85)u
328 + 467.5 = 165u = 795.5 / 165 = 4.821m/s
A baseball of mass m1 = 0.26 kg is thrown at another ball hanging from the ceiling by a length of string L = 1.45 m. The second ball m2 = 0.64 kg is initially at rest while the baseball has an initial horizontal velocity of V1 = 2.5 m/s. After the collision the first baseball falls straight down (no horizontal velocity).
The angle is approximately 50°
Explanation:
We know,
h = L – L * cos θ
= 1.55 – 1.55 (cos θ )
To determine the maximum height, we use conservation potential and kinetic energy. As 2nd ball rises to its maximum height, the increase of its potential energy is equal to the decrease of its kinetic energy. Below is the equation for determining the value of h from the 2nd ball’s initial velocity.
Potential energy, PE = mgh = m X 9.8 X h
Kinetic energy, KE = [tex]\frac{1}{2} mv^2[/tex]
Set PE equal to KE and solve for h.
h = v²/19.6
To determine the initial kinetic energy of the second ball, we need to the velocity of the 2nd ball immediately after the collision. To do this we need to use conservation of momentum. Below is the equation for determining the value of h from the 2nd ball’s initial velocity.
For the 1st ball, horizontal momentum = 0.26 X 2.5 = 0.65
Since this ball falls straight down after the collision, its final horizontal momentum is 0.
For the 2nd ball, horizontal momentum = 0.61 X vf
0.64 * vf = 0.65
vf = 0.65/.0.64
vf = 0.02m/s
h = (0.65/0.64)²/19.6
h = 0.00002m
0.00002 = 1.45 – 1.45 * cos θ
Subtract 1.55 from both sides.
0.00002 – 1.45 = -1.45 * cos θ
θ = 50°
wo cars start moving from the same point. One travels south at 60 mi/h and the other travels west at 25 mi/h. At what rate is the distance between the cars increasing three hours later?
Answer:
[tex]\frac{dz}{dt} = 65 mi/h[/tex]
Explanation:
let distance between two cars is = z mi
we have to find =[tex]\frac{dz}{dt}[/tex]
One travels south at = 60 mi/h = [tex]\frac{dx}{dt}[/tex] (given)
the other travels west at =25 mi/h.= [tex]\frac{dy}{dt}\\[/tex] (given)
since both car have constant speed
at t = 3 hrs
x = 3× 60 = 180 mi/h
y = 3 × 25 = 75 mi/h
from the figure (i) we get
[tex]z = \sqrt{( x^2+ y^2)}[/tex] ...............(i)
put x and y values
we get
[tex]z = \sqrt{(180)^2 + 75^2}[/tex]
[tex]z = \sqrt{32400 + 5625} \\z = \sqrt{38025} \\z = 195 mi/h[/tex]
differentiate the equation (i) w r to t
[tex]z^2 = x^2 +y^2[/tex]
[tex]2z\frac{dz}{dt} = 2x\frac{dx}{dt}+ 2y\frac{dy}{dt}\\[/tex]
put each values
[tex]2 \times195\frac{dz}{dt} = 2 \times 180\frac{dx}{dt}+2 \times75\frac{dy}{dt}\\[/tex]
[tex]2 \times195\frac{dz}{dt} = 2 \times 180\times 60}+2 \times75\times25\\\frac{dz}{dt} = \frac{{2 \times 180\times 60+2 \times75\times25}}{ 2 \times195}\\\frac{dz}{dt} = 65 mi/h[/tex]
Answer: Both cars have equal kinetic energy
Explanation:
A 210-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.600 rev/s in 2.00 s
Answer:
The value of the constant force is [tex]\bf{296.88~N}[/tex].
Explanation:
Given:
Mass of the merry-go-round, [tex]m = 210~Kg[/tex]
Radius of the horizontal disk, [tex]r = 1.5~m[/tex]
Time required, [tex]t = 2.00~s[/tex]
Angular speed, [tex]\omega = 0.600~rev/s[/tex]
Torque on an object is given by
[tex]\tau = F.r = I.\alpha~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)[/tex]
where [tex]I[/tex] is the moment of inertia of the object, [tex]\alpha[/tex] is the angular acceleration and [tex]F[/tex] is the force on the disk.
The moment of inertia of the horizontal disk is given by
[tex]I = \dfrac{1}{2}mr^{2}[/tex]
and the angular acceleration is given by
[tex]\alpha = \dfrac{2\pi \omega}{t}[/tex]
Substituting all these values in equation (1), we have
[tex]F &=& \dfrac{I\alpha}{r}\\&=& \dfrac{\pi m r \omega}{t}\\&=& \dfarc{\pi(210~Kg)(1.5~m)(0.600~rev/s)}{2.00~s}\\&=& 296.88~N[/tex]
Does a 0.14 kg baseball moving at 41 m/s or a 0.058-kg tennis ball moving at 67 m/s
have more kinetic energy?
Answer:
The tennis ball has more kinetic energy
Explanation:
Recall the formula for kinetic energy: [tex]K=\frac{1}{2} \,m\,v^2[/tex] , so we can estimate it for each case and compare the results:
For the baseball:
[tex]K=\frac{1}{2} \,m\,v^2\\K=\frac{1}{2} \,0.14\,(41)^2\,\,J\\K=117.67\,\,J[/tex]
For the tennis ball:
[tex]K=\frac{1}{2} \,m\,v^2\\K=\frac{1}{2} \,0.058\,(67)^2\,\,J\\K=130.181\,\,J[/tex]
Therefore, the tennis ball has more kinetic energy
You know very well that the classical trajectory of a charged particle (charge q, mass m) in the uniform magnetic field B (directed, say, along the z-axis) is helical: the particle moves with constant velocity along the field lines and executes a circular motion in the perpendicular (xy) plane with the cyclotron frequency
Answer:
Check the explanation
Explanation:
Stationary states might as well be illustrated in a simpler form of the Schrödinger equation, It utilizes the theory of energy conservation (Kinetic Energy + Potential Energy = Total Energy) to acquire information about an electron’s behavior that is been bound to a nucleus.
Kindly check the attached image below to get the step by step explanation to the above question.
Two taut strings of identical mass and length are stretched with their ends fixed, but the tension in one string is 1.10 times greater than in the other. Waves on the string with the lower tension propagate at 35.2 m/s. The fundamental frequency of that string is 258 Hz. What is the beat frequency when each string is vibrating at its fundamental frequency?
Answer:
The beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz
Explanation:
Given;
velocity of wave on the string with lower tension, v₁ = 35.2 m/s
the fundamental frequency of the string, F₁ = 258 Hz
velocity of wave on the string with greater tension;
[tex]v_1 = \sqrt{\frac{T_1}{\mu }[/tex]
where;
v₁ is the velocity of wave on the string with lower tension
T₁ is tension on the string
μ is mass per unit length
[tex]v_1 = \sqrt{\frac{T_1}{\mu} } \\\\v_1^2 = \frac{T_1}{\mu} \\\\\mu = \frac{T_1}{v_1^2} \\\\ \frac{T_1}{v_1^2} = \frac{T_2}{v_2^2}\\\\v_2^2 = \frac{T_2v_1^2}{T_1}[/tex]
Where;
T₁ lower tension
T₂ greater tension
v₁ velocity of wave in string with lower tension
v₂ velocity of wave in string with greater tension
From the given question;
T₂ = 1.1 T₁
[tex]v_2^2 = \frac{T_2v_1^2}{T_1} \\\\v_2 = \sqrt{\frac{T_2v_1^2}{T_1}} \\\\v_2 = \sqrt{\frac{1.1T_1*(35.2)^2}{T_1}}\\\\v_2 = \sqrt{1.1(35.2)^2} = 36.92 \ m/s[/tex]
Fundamental frequency of wave on the string with greater tension;
[tex]f = \frac{v}{2l} \\\\2l = \frac{v}{f} \\\\thus, \frac{v_1}{f_1} =\frac{v_2}{f_2} \\\\f_2 = \frac{f_1v_2}{v_1} \\\\f_2 =\frac{258*36.92}{35.2} \\\\f_2 = 270.6 \ Hz[/tex]
Beat frequency = F₂ - F₁
= 270.6 - 258
= 12.6 Hz
Therefore, the beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz
The fundamental frequency for the second string is found to be 272 Hz, resulting in a beat frequency of 14 Hz.
To find the beat frequency when two strings are vibrating at their fundamental frequencies, follow these steps:
Use the formula for wave speed in a string: v = √(T/μ), where T is the tension and μ is the mass per unit length.Given that the speed of waves in the string with lower tension is 35.2 m/s, and the fundamental frequency is 258 Hz, use the relationship: v = f × λ to calculate the wavelength: λ = v / f = 35.2 m/s / 258 Hz ≈ 0.136 m.The tension in the second string is 1.10 times that in the first string. So, if T1 is the tension in the first string, the tension in the second string T₂ is 1.10 × T₁The speed of waves in the second string is: v₂ = √(T₂/μ) = √(1.10 × T₁/μ). Since v₁ = √(T₁/μ) = 35.2 m/s, v₂ = 35.2 × √1.10 ≈ 37.0 m/s.Calculate the fundamental frequency of the second string: f₂ = v₂ / λ = 37.0 m/s / 0.136 m ≈ 272 Hz.The beat frequency is the absolute difference between the two fundamental frequencies: beat frequency = |f₂ - f₁| = |272 Hz - 258 Hz| = 14 Hz.Therefore, the beat frequency when both strings vibrate at their fundamental frequencies is 14 Hz.
A "home-made" solid propellant rocket has an initial mass of 9 kg; 6.8 kg of this is fuel. The rocket is directed vertically upward from rest, burns fuel at a constant rate of 0:225 kg=s, and ejects exhaust gas at a speed of 1980 m=s relative to the rocket. Assume that the pressure at the exit is atmospheric and that air resistance may be neglected. Calculate the rocket speed after 20 s and the distance traveled by the rocket in 20 s. Plot the rocket speed and the distance traveled as functions of time.
Answer:
v = 1176.23 m/s
y = 741192.997 m = 741.19 km
Explanation:
Given
M₀ = 9 Kg (Initial mass)
me = 0.225 Kg/s (Rate of fuel consumption)
ve = 1980 m/s (Exhaust velocity relative to rocket, leaving at atmospheric pressure)
v = ? if t = 20 s
y = ?
We use the equation
v = ∫((ve*me)/(M₀ - me*t)) dt - ∫g dt where t ∈ (0, t)
⇒ v = - ve*Ln ((M₀ - me*t)/M₀) - g*t
then we have
v = - 1980 m/s*Ln ((9 Kg - 0.225 Kg/s*20 s)/(9 Kg)) - (9.81 m/s²)(20 s)
v = 1176.23 m/s
then we apply the formula
y = ∫v dt = ∫(- ve*Ln ((M₀ - me*t)/M₀) - g*t) dt
⇒ y = - ve* ∫ Ln ((M₀ - me*t)/M₀) dt - g*∫t dt
⇒ y = - ve*(Ln((M₀ - me*t)/M₀)*t + (M₀/me)*(M₀ - me*t - M₀*Ln(M₀ - me*t))) - (g*t²/2)
For t = 20 s we have
y = Ln((9 Kg - 0.225 Kg/s*20 s)/9 Kg)*(20 s) + (9 Kg/0.225 Kg/s)*(9 Kg - 0.225 Kg/s*20 s - 9 Kg*Ln(9 Kg - 0.225 Kg/s*20 s)) - (9.81 m/s²*(20 s)²/2)
⇒ y = 741192.997 m = 741.19 km
The graphs are shown in the pics.
Following are the solution to the given points:
Given:
[tex]\to v_o =1980\ \frac{m}{s}\\\\[/tex]
[tex]\to a=g= -9.8 \frac{m}{s} \ \text{(Diection downwords )}[/tex]
Solution:
Using formula:
[tex]\to V= V_o +at\\\\[/tex]
[tex]\to v^2-v^2_0= 2ad \\\\[/tex]
For point a:
[tex]\to V= V_o +at\\\\[/tex]
[tex]=1980+ (-9.8) (20) \\\\[/tex]
[tex]= 1980-196 \\\\ =1784\ \frac{m}{s}\\\\[/tex]
The velocity at the conclusion of a [tex]20[/tex] second is [tex]\bold{1784\ \frac{m}{s}}\\\\[/tex].
For point b:
Using formula:
[tex]\to v^2-v^2_0= 2ad \\\\[/tex]
[tex]\to d=\frac{v^2_0 -V^2}{2a}\\\\[/tex]
[tex]= \frac{(1980)^2 - (1784)^2}{2\times 9.8} \\\\ =\frac{3920400-3182656}{19.6}\\\\= \frac{737744}{17.6}\\\\ =37640\ m \\\\ = 37.64\ km\\\\[/tex]
In [tex]20[/tex] seconds, the total distance traveled is [tex]401737 \ \ m (or \ 401.737\ \ km)[/tex].
For point c:
Please find the attached file.
A vertical bar magnet is dropped through the center of a horizontal loop of wire, with its north pole leading. At the instant when the midpoint of the magnet is in the plane of the loop, the induced current in the loop, viewed from above, is:
Answer:
When the midpoint of the magnet is in the plane of the loop, the induced current in the loop viewed from above is essentially zero.
Explanation:
At some point far away from the plane of the loop of the wire, there's no flux and hence, no induced emf, and no current in the loop. As the magnet descends and gets close to the entrance of the coil, some of the magnetic field from the magnet threads the top few imaginary disks of the coil. The time changing flux induces some EMF and subsequently some induced current.
As the magnet continues to fall, and enters the coil, more of its magnetic field is threading imaginary disks in the coil, so as it moves, the time rate of change of total flux increases, so the EMF goes up. Note that the field lines above and below the magnet's midpoint point in the same direction from the the North pole to the south.
As the bar moves through the plane of the coil, the North end first, flux is added by the motion of the magnet and flux is removed by the motion of South end.
At some point, the bar reaches the middle of the coil. At this point, the amount of flux added to the top half of the coil by a small motion of the magnet is equal to the amount of flux removed from the bottom half. Therefore, at this point the EMF is zero. And hence, there is no induced current observed in the wire at this point.
Hope this Helps!!!
When viewed from the above, there is no induced current observed in the wire at this point due to zero EMF.
The given problem is based on the concept of electric flux and induced emf. The electric flux is dependent on induced emf, which is produced to current. Since, point far away from the plane of the loop of the wire, there's no flux and hence, no induced emf, and no current in the loop.
As the magnet descends and gets close to the entrance of the coil, some of the magnetic field from the magnet threads the top few imaginary disks of the coil. The time changing flux induces some EMF and subsequently some induced current.When magnet continues to fall, and enters the coil, more of its magnetic field is threading imaginary disks in the coil, so as it moves, the time rate of change of total flux increases, so the EMF goes up.
Note: - The field lines above and below the magnet's midpoint point in the same direction from the the North pole to the south.
As the bar moves through the plane of the coil, the North end first, flux is added by the motion of the magnet and flux is removed by the motion of South end.
When the bar reaches the middle of the coil, then at this point the amount of flux added to the top half of the coil by a small motion of the magnet is equal to the amount of flux removed from the bottom half. Therefore, at this point the EMF is zero. And hence, there is no induced current observed in the wire at this point.
Thus, we conclude that when viewed from the above, there is no induced current observed in the wire at this point due to zero EMF.
Learn more about the electromagnetic induction here:
https://brainly.com/question/12536260
QUESTION 2
Objectives:
I . identify potential and kinetic energy in a situation and draw corresponding energy bar charts,
II . calculate gravitational and elastic potential energy,
III . draw & analyze potential energy functions, and (d) use conservation of energy to relate the total energy at one time to total energy at another time.
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Bungee Jump: you will step off with zero initial vertical velocity from a platform a height h above the ground. The bungee cord will act like a giant extensional spring that will, you hope, provide an upward force on becoming taut. After weighing you (you have a mass M), the operator has selected a bungee cordwith an un-stretched length of d and a spring constant of k.
Consider yourself to be a single point – i.e., use the particle model.
Choosing the ground as your origin (and the z-axis directed upwards), answer the following questions about your bungee jumping adventure in terms of M, h, d, k, z, and the gravitational field strength, g. Answer with variables.
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a)
Write an expression for the stretching ?L of the cord in terms of d and z and for the total potential energy U of the jumper-bungee-Earth system for each situation (consider the latter two situations together).
b)
Which type of potential energy ( UG or US ) is largest for large z (early in the fall)? For small z (late in the fall)?
c)
Sketch a graph of your gravitational potential energy, UG(z) vs your height, z, from z = 0 to z = h, on the left plot. Then sketch a graph of your elastic potential energy, US(z) on the center plot. Finally on the rightmost plot, sketch a graph of your total potential energy1 U(z) = UG(z) + US(z). Do these plots on your own without the help of a computer or calculator.
Answer:
See explaination and attachment please.
Explanation:
Potential energy is defined as the energy possessed by a body by virtue of its position relative to others, stresses within itself, electric charge, and other factors.
Kinetic energy on the other hand is defined as the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.
These understand will foster our knowledge to the best way of solving the question.
Please kindly check attachment for a step by step approach on the graph.
Final answer:
In a bungee jumping scenario, the stretching of the cord can be expressed in terms of length and height. Gravitational potential energy dominates early in the fall, while elastic potential energy is more significant later. Graphs can visually represent how potential energy changes with height during the jump.
Explanation:
a) Stretching of the cord: The stretching ?L of the cord can be expressed as ?L = d - z. The total potential energy U can be represented as U = UG + US, where UG is the gravitational potential energy and US is the elastic potential energy.
b) Largest potential energy: For large z (early in the fall), UG is largest. For small z (late in the fall), US is largest.
c) Graphs: Sketch a graph of UG(z) on the left plot, US(z) on the center plot, and U(z) = UG(z) + US(z) on the rightmost plot.
As time progresses a capacitor hooked up to a battery begins to act like a. a resistor draining power from the battery b. Another battery but working against the first battery c. Another battery aiding the first battery d. An inverse resistor giving power to the system
Answer:
D
Explanation:
An inverse resistor giving power to the system
A capacitor hooked up to a battery acts as another battery aiding the first battery as time progresses.
As time progresses, a capacitor hooked up to a battery begins to act as another battery aiding the first battery. The capacitor acts as a temporary storehouse of energy and can drive its collected charge through a second circuit.
6) The electron volt (eV) is a convenient unit of energy for expressing atomic-scale energies. It is the amount of energy that an electron gains when subjected to a potential of 1 volt; 1 eV = 1.602 × 10–19 J. Using the Bohr model, determine the energy, in electron volts, of the photon produced when an electron in a hydrogen atom moves from the orbit with n = 5 to the orbit with n = 2. Show your calculations.
Answer:
The energy of the photon is [tex]x = 2.86 eV[/tex]
Explanation:
From the question we are told that
The first orbit is [tex]n_1 = 5[/tex]
The second orbit is [tex]n_2 = 2[/tex]
According to Bohr model
The energy of difference of the electron as it moves from on orbital to another is mathematically represented as
[tex]\Delta E = k [\frac{1}{n^2 _1} + \frac{1}{n^2 _2} ][/tex]
Where k is a constant which has a value of [tex]k = -2.179 *10^{-18} J[/tex]
So
[tex]\Delta E = - 2.179 * 10^{-18} [\frac{1}{5^2 _1} + \frac{1}{2^2 _2} ][/tex]
[tex]= 4.576 *10^{-19}J[/tex]
Now we are told from the question that
[tex]1 eV = 1.602 * 10^{-19} J[/tex]
so x eV = [tex]= 4.576 *10^{-19}J[/tex]
Therefore
[tex]x = \frac{4.576*10^{-19}}{1.602 *10^{-19}}[/tex]
[tex]x = 2.86 eV[/tex]
The energy of the photon produced by an electron in a Hydrogen atom moving from the 5th orbit to the 2nd orbit is 2.85 electron Volts as calculated via the Rydberg formula.
Explanation:The energy of the photon produced by the transition of an electron in a hydrogen atom from the 5th orbit to the 2nd orbit can be calculated using Rydberg's formula.
Rydberg's formula for energy is given as E = 13.6 * (1/n1^2 - 1/n2^2), where n1 and n2 are the initial and final energy levels respectively, and E is the energy difference in eV (electron volts). Here n1 = 2 and n2 = 5.
So, substituting these values in the formula E = 13.6 * (1/2^2 - 1/5^2) = -13.6 * (-0.21) = 2.85 eV. Notice that the energy is negative which signifies a transition down to a lower energy level (which is exothermic), however, we are interested in the magnitude of the energy which is 2.85 eV.
Learn more about Bohr model energy calculations here:https://brainly.com/question/37434791
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Objects 1 and 2 attract each other with a electrostatic force of 18.0 units. If the charge of Object 1 is one-third the original value AND the charge of object 2 is doubled AND the distance then the new electrostatic force will be _____ unit
Answer:
F'=(8/3)F
Explanation:
to find the change in the force you take into account that the electric force is given by:
[tex]F=k\frac{q_1q_2}{(18.0u)^2}=k\frac{q_1q_2}{324.0}u^2[/tex]
However, if q1'=1/3*q, q2'=2*q2 and the distance is halved, that is 18/=9.o unit:
[tex]F'=k\frac{q_1'q_2'}{(9u)^2}=k\frac{(1/3)q_1(2)q_2}{81.0u^2}=\frac{2}{3}k\frac{q_1q_2}{81.0u^2}[/tex]
if you multiply this result by 4 and divide by 4 you get:
[tex]F'=\frac{8}{3}k\frac{q_1q_2}{324.0u^2}=\frac{8}{3}F[/tex]
hence, the new force is 8/3 of the previous force F.
A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow has a 6.75 m radius, at how many revolutions per minute are the riders subjected to a centripetal acceleration equal to that of gravity
Answer: The riders are subjected to 11.5 revolutions per minute
Explanation: Please see the attachments below
Water at 45 C and 10 kPa enters an adiabatic pump and is discharged at a pressure of 8600 kPa.The pump efficiency is 75%. Calculate the work for the pump, the temperature change of the water and the entropy change. For water at 45C: V = 1010 cm3/kg, β= 485*10(-6) K-1and Cp=4.178kJ/(kgK)
Answer:
Explanation:
find the solution below
Why do plant cells need chloroplasts?
Answer:
Chloroplasts convert light energy into sugars that can be used by the cells. This conversion creates 'food' for the plant.
Explanation:
In a double-slit interference experiment, the slit separation is 2.41 μm, the light wavelength is 512 nm, and the separation between the slits and the screen is 4.45 m. (a) What is the angle between the center and the third side bright fringe? If we decrease the light frequency to 94.5% of its initial value, (b) does the third side bright fringe move along the screen toward or away from the pattern's center and (c) how far does it move?
Answer:
Using equation 2dsinФ=n*λ
given d=2.41*10^-6m
λ=512*10^-12m
θ=52.64 degrees
Answer:
a
The angle between the center and the third side bright fringe is
[tex]\theta = 39.60^o[/tex]
b
The third side bright fringe move away from the pattern's center
c
The distance by which it moves away is [tex]\Delta z=0.3906 m[/tex]
Explanation:
From the question the
The wavelength is [tex]\lambda = 512nm[/tex]
In the first question we a asked to obtain the angle between the center and the third side bright fringe
since we are considering the third side of the bright fringe the wavelength of light on the three sides would be evaluated as
[tex]\lambda_{3} = 3 * 512nm[/tex]
The slit separation is given as [tex]d = 2.41 \mu m[/tex]
The angle between the center and the third side bright fringe is
[tex]\theta = sin^{-1} (\frac{\lambda_3}{d} )[/tex]
[tex]\theta = sin^{-1} (\frac{3 *512*10^{-9}}{2.24*10^{-6}} )[/tex]
[tex]= sin^{-1} (0.6374)[/tex]
[tex]\theta = 39.60^o[/tex]
When the frequency of the light is reduced the wavelength is increased
i.e [tex]f = \frac{c}{\lambda}[/tex]
and this increase would cause the third side bright to move away from the pattern's center
Now from the question frequency is reduce to 94.5% this mean that the wavelength would also increase by the same as mathematically represented below
[tex]\lambda_{new} = \frac{512 *10^{-9}}{0.945}[/tex]
[tex]= 0.542 \mu m[/tex]
The angle between the center and the third side bright fringe is for new wavelength
[tex]\theta = sin^{-1} (\frac{3 *512*10^{-9}}{2.41*10^{-6}} )[/tex]
[tex]= 42.46^o[/tex]
The distance traveled away from the pattern's center is mathematically represented as
[tex]z = A tan \theta[/tex]
Where A is the separation between the slits and the screen
[tex]\Delta z = 4.45(tan 42.46 - tan39.60 )[/tex]
[tex]\Delta z=0.3906 m[/tex]
Please help me. Calculate the average travel time for each distance, and then use the results to calculate.
A 6-column table with 3 rows in the second, third, and fifth columns and 1 row in the other columns. The first column labeled Number of Washers has entry 1 washer mass = 4.9 grams. The second column labeled Trial has entries trial number 1, trial number 2, trial number 3. The third and fourth columns labeled Time to travel 0.25 meters t subscript 1 (seconds) have entries in the third column 2.24, 2.21, 2.23 and average in the fourth column. The fifth and sixth columns labeled time to travel 0.5 meters t subscript 2 (seconds) have entries in the fifth column 3.16, 3.08, 3.15 and in the sixth column average.
The average time that it takes for the car to travel the first 0.25m is
s.
The average time to travel just between 0.25 m and 0.50 m is
s.
Given the time taken to travel the second 0.25 m section, the velocity would be
m/s.
Answer:
What is the average velocity of the car over the first 0.25m?
⇒ 0.11 m/s
What is the average velocity of the car over the second 0.25m?
⇒ 0.28 m/s
Explanation:
Just did this problem! :)
Answer:
⇒ 2.23
⇒ 0.90
⇒ 0.28
Explanation:
Just did the problem (⌐■_■)
How are linear measurements usually expressed in the building industry?
In the building industry, linear measurements are expressed in units like miles, feet, and inches, or in the metric system, meters and millimeters. The choice of unit depends on the precision needed for the task, with smaller units used for finer details and larger ones for general dimensions.
Linear Measurements in the Buildings Industry
In the building industry, linear measurements are typically expressed in units that provide the most practical and precise information for the specific dimension being measured. Linear dimensions refer to measurements that can be expressed using linear units such as miles, feet, or inches; and in the metric system, meters or millimeters (mm). These measurements are essential for architects and builders to communicate the sizes of different components of a structure, like the length of a wall or the size of a window. Typically, for smaller measurements such as the dimensions of timber or the size of rooms, feet or meters are used, while millimeters might be preferred for finer details.
For example, when measuring a specific length like the width of a staircase, the builders would commonly use millimeters in the metric system to ensure precision. Similarly, to describe to a European the dimensions of "two-by-four" lumber used in the US, conversions from inches to centimeters and from feet to meters are required. The key is to use the most appropriate unit of measurement that ensures clarity and precision for constructing a building accurately.
When dealing with linear measurements, especially in scientific contexts or in detailed building plans, precise units like millimeters, centimeters, and meters are often used. The context will dictate whether a larger unit such as kilometers or smaller units like micrometers are more appropriate.
A roller coaster at the Six Flags Great America amusement park in Gurnee, Illinois, incorporates some clever design technology and some basic physics. Each vertical loop, instead of being circular, is shaped like a teardrop. The cars ride on the inside of the loop at the top, and the speeds are fast enough to ensure that the cars remain on track. The biggest loop is 40.0m high. Suppose the speed at the top is 14.4m/s and the corresponding centripetal acceleration is 2g.
(a) What is the radius of the arc of the teardrop at the top?
(b)If the total mass of a car plus the riders is M, what force does the rail exert on the car at the top?
(c) Suppose the roller coaster had a circular loop of radius 21.4 m. If the cars have the same speed, 14.4 m.s at the top, what is the centipetal acceleration at the top?
Answer:
a)radius of the arc of the teardrop at the top is 10.58m
b)T = mg
c) the centipetal acceleration at the top is 14.5m/s
Explanation:
Part A
Given that roller coaster is of tear drop shape
So the speed at the top is given as
v = 14.4 m/s
acceleration at the top is given as
a = 2 g
[tex]a = 2(9.8) = 19.6 m/s^2[/tex]
now we know the formula of centripetal acceleration as
[tex]a = \frac{v^2}{R}\\[/tex]
[tex]19.6 = \frac{14.4^2}{R}\\R = 10.58 m[/tex]
Part B
now
the total mass of the car and the ride is M
Let the force exerted by the track be n
By Newton law
[tex]n +Mg =\frac{Mv^2}{r} \\\\n=\frac{Mv^2}{r} -Mg\\\\=M(\frac{v^2}{r}-g )\\\\=M(2g-g)\\\\T=Mg[/tex]
Part C
If the radius of the loop is 21.4 m
speed is given by same v = 14.4 m/s
now the acceleration is given as
[tex]a = \frac{v^2}{R}[/tex]
[tex]a = \frac{14.4^2}{21.4} \\\\= 9.69 m/s^2[/tex]
Now for normal force at the top is given by force equation
[tex]F_n + mg = ma\\F_n = m(a-g)[/tex]
The force exerted by the rail is less than zero because acceleration is less than 9.69m/s²
So the normal force would have to point away from the centre, For safe ride this normal force must be positive i.e [tex]a \prec g[/tex]
[tex]\frac{v^2}{r} \prec \sqrt{g} \\\\v = \sqrt{rg} \\\\v = \sqrt{21.4 \times 9.8} \\\\v = 14.5m/s[/tex]
A machine part has the shape of a solid uniform sphere of mass 240 g and diameter 2.50 cm . It is spinning about a frictionless axle through its center, but at one point on its equator it is scraping against metal, resulting in a friction force of 0.0200 N at that point. Part A Find its angular acceleration. Let the direction the sphere is spinning be the positive sense of rotation.
Answer:
[tex]-16.6 rad/s^2[/tex]
Explanation:
The torque exerted on a rigid body is related to the angular acceleration by the equation
[tex]\tau = I \alpha[/tex] (1)
where
[tex]\tau[/tex] is the torque
I is the moment of inertia of the body
[tex]\alpha[/tex] is the angular acceleration
Here we have a solid sphere: the moment of inertia of a sphere rotating about is centre is
[tex]I=\frac{2}{5}MR^2[/tex]
where
M = 240 g = 0.240 kg is the mass of the sphere
[tex]R=\frac{2.50}{2}=1.25 cm = 0.0125 m[/tex] is the radius of the sphere
Substituting,
[tex]I=\frac{2}{5}(0.240)(0.0125)^2=1.5\cdot 10^{-5} kg m^2[/tex]
The torque exerted on the sphere is
[tex]\tau = Fr[/tex]
where
F = -0.0200 N is the force of friction
r = 0.0125 m is the radius of the sphere
So
[tex]\tau=(-0.0200)(0.0125)=-2.5\cdot 10^{-4} Nm[/tex]
Substituting into (1), we find the angular acceleration:
[tex]\alpha = \frac{\tau}{I}=\frac{-2.5\cdot 10^{-4}}{1.5\cdot 10^{-5}}=-16.6 rad/s^2[/tex]
Two concentric circular coils of wire lie in a plane. The larger coil has N1 = 7200 turns and radius of a = 75.50 cm. The smaller coil has N2 = 920 turns and radius of b = 1.00 cm. Ultimately we will find the Mutual Inductance, M, and then the induced emf, ϵmf. So let's take it step by step.
First, what is the magnitude of the B-field at the smaller coil due to the larger coil?
Note: The values of the current and the radii are not given, substitute whatever the value of the current and radius are to the given solution to obtain the magnitude.
Answer:
magnitude of the B-field at the smaller coil due to the larger coil is [tex]B = \frac{7200\mu_{0}I }{2a}[/tex]
Explanation:
N₁ = 7200 turns
N₂ = 920 turns
a = 75.50 cm = 0.755 m
b = 1.00 cm = 0.01 m
From the given data, b<<a, and it is at the center of the larger coil,
so we are safe to assume that the magnetic field at the smaller coil is constant.
The formula for magnetic field due to circular loop at the center of a coil is given by:
[tex]B = \frac{N\mu_{0}I }{2R}[/tex]
Therefore, magnetic field in the smaller coil due to the larger coil of radius a will be given by:
[tex]B = \frac{7200\mu_{0}I }{2a}[/tex]
Where [tex]\mu_{0} = 4\pi * 10^{-7} Wb/A-m[/tex]
I = current in the larger coil
The magnitude of the B-field at the smaller coil due to the larger coil will be "4π × 10⁻⁷ Wb/A-m".
Magnetic fieldAccording to the question,
Number of turns, N₁ = 7200 turns
N₂ = 920 turns
Radius, a = 75.50 cm or,
= 0.755 m
b = 1.00 cm or,
= 0.01 m
We know the formula,
Magnetic field, B = [tex]\frac{N \mu_0 I}{2R}[/tex]
here, μ₀ = 4π × 10⁻⁷ Wb/A-m
By substituting the values,
= [tex]\frac{7200 \mu_0 I}{2a}[/tex]
Thus the above answer is correct.
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In Example 10.1 (p. 234), if the pebble was instead launched at a 45 degree angle above the horizontal, how would the total mechanical energy of the system change compared to when the pebble was launched directly upwards?
Answer:
The total mechanical energy of the system would stay the same
Explanation:
The law of conservation energy states:
In a closed system, (a system that isolated from its surroundings) the total energy of the system is conserved.
For instance, the air resistance is negligible. So non conservative force is acting on the system and so the energy is conserved for the system.
The total initial energy must be equal to the total final energy , if the energy is conserved.
Therefore, when the pebbles was launched at 45 degrees to the horizontal the total mechanical energy stay the same.
A solenoid with 500 turns has a radius of 0.040 m and is 40 cm long. If this solenoid carries a current of 12 A, what is the magnitude of the magnetic field near the center of the solenoid? (μ 0 = 4π × 10-7 T · m/A)
Answer:
[tex]18.8\times 10^{-3} T[/tex]
Explanation:
We are given that
Number of turns,N=500
Radius,r=0.04 m
Length of solenoid,L=40 cm=[tex]\frac{40}{100}=0.4 m[/tex]
1 m=100 cm
Current,I=12 A
We have to find the magnitude of magnetic field near the center of the solenoid.
Number of turns per unit length,n=[tex]\frac{500}{0.4}=1250[/tex]
Magnetic field near the center of the solenoid,B=[tex]\mu_0 nI[/tex]
Where [tex]\mu_=0=4\pi\times 10^{-7}Tm/A[/tex]
[tex]B=4\pi\times 10^{-7}\times 1250\times 12=18.8\times 10^{-3} T[/tex]
[tex]B=18.8\times 10^{-3} T[/tex]
An explosion in a rigid pipe shoots three balls out of its ends. A 6 gam ball comes out the right end. A 4 gram ball comes out the left end with twice the speed of the 6 gram ball. From which end does the third ball emerge?
Answer:
The third ball emerges from the right side.
Explanation:
This is a conservation of Momentum problem
In an explosion or collision, the momentum is always conserved.
Momentum before explosion = Momentum after explosion
Since the rigid pipe was initially at rest,
Momentum before explosion = 0 kgm/s
- Taking the right end as the positive direction for the velocity of the balls
- And calling the speed of the 6 g ball after explosion v
- This means the velocity of the 4 g ball has to be -2v
- Mass of the third ball = m
- Let the velocuty of the third ball be V
Momentum after collision = (6)(v) + (4)(-2v) + (m)(V)
Momentum before explosion = Momentum after explosion
0 = (6)(v) + (4)(-2v) + (m)(V)
6v - 8v + mV = 0
mV - 2v = 0
mV = 2v
V = (2/m) v
Note that since we have established that the sign on m and v at both positive, the sign of the velocity of the third ball is also positive.
Hence, the velocity of the third ball according to our convention is to the right.
Hope this Helps!!!
Answer:
the third ball emerge from right end .
Explanation:
initially the three balls are at rest so u = 0
Given m1 = 6g
v1 = v(i) (Right end indicates positive X axis so it is represented as (i))
m2 = 4g
v2 = 2(v)(-i) (left end indicates negative X axis so it is represented as (-i))
m3 = m
v3 = v'
Applying conservation of momentum
m1(u1) + m2(u2) + m3(u3) = m1(v1) + m2(v2) + m3(v3)
u1 = u2 = u3 = u = 0 ( at rest)
0 = (6×10⁻³)(v)(i) + (4×10⁻³)(2v)(-i) + m(v')
m(v') = (2×10⁻³)(v)(i)
v' = (2×10⁻³)(v)(i)/(m)
So positive (i) indicates right end
So the third ball emerge from right end .
Block A weighs 1.20 N and block Bweighs 3.60 N. The coefficient of kinetic frictionbetween all surfaces is 0.300. Find the magnitudeof the horizontal force F necessary to drag blockB to the left at constant speed a) if A rests on Band moves with it and b) if A is held at rest.
Answer:
Explanation:
a ) In this case A rests on B and both rest on horizontal surface
Both moves together .
Total reaction due to weight of both of them
R = 1.2 + 3.6 = 4.8 N
friction force by horizontal surface = μ R , μ is coefficient of friction .
= .3 x 4.8 = 1.44 N
Force equal to frictional force will be required to put both of them in uniform motion .
b ) If A is held stationary , friction force will arise at both , the upper and lower surface of B .
At upper surface friction force = μ x weight of A
= .3 x 1.2 = .36 N
At lower surface friction force = μ x weight of A +B
= .3 x 4.8
= 1.44
Total frictional force on B
= 1.8 N N .
So 1.8 N force will be required to put B in uniform keeping A stationary.
To find the magnitude of the horizontal force F necessary to drag block B to the left at constant speed, we need to consider two scenarios: a) when block A rests on block B and moves with it, and b) when block A is held at rest. In both scenarios, the friction force between block B and the surface is given by the equation F_friction = μ * N, where μ is the coefficient of friction and N is the normal force. The normal force is the sum of the weights of the blocks in scenario a) and the weight of block B in scenario b).
Explanation:To find the magnitude of the horizontal force F necessary to drag block B to the left at constant speed, we need to consider two scenarios: a) when block A rests on block B and moves with it, and b) when block A is held at rest.
a) When block A rests on block B and moves with it, the friction force between the blocks and the surface is the force needed to overcome the friction. The friction force is given by the equation:
F_friction = μ * N
Here, μ is the coefficient of kinetic friction and N is the normal force between the blocks and the surface. In this case, the normal force N is the sum of the weights of both blocks, so N = m_A * g + m_B * g. Substituting this value into the equation, we get:
F_friction = μ * (m_A * g + m_B * g)
b) When block A is held at rest, the friction force between block B and the surface is the force needed to overcome the friction. The friction force is the product of the coefficient of static friction and the normal force:
F_friction = μ * N
Here, μ is the coefficient of static friction and N is the normal force between block B and the surface. The normal force N is equal to the weight of block B, so N = m_B * g. Substituting this value into the equation, we get:
F_friction = μ * (m_B * g)
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A wheel rotating with a constant angular acceleration turns through 23 revolutions during a 3 s time interval. Its angular velocity at the end of this interval is 16 rad/s. What is the angular acceleration of the wheel
Answer
Given,
Revolution of wheel = 23 rev.
= 23 x 2π = 46 π
Time = 3 s
final angular velocity = 16 rad/s
angular acceleration of wheel = ?
Now, Calculating the initial angular speed of the wheel
Angular displacement = [tex]\dfrac{1}{2}[/tex](initial velocity + final velocity) x time.
[tex]46 π = \dfrac{\omega_o+16}{2}\times 3[/tex]
[tex]\omega_0 = 80.29\ rad/s[/tex]
now, angular acceleration
[tex]\alpha = \dfrac{\omega-\omega_0}{t}[/tex]
[tex]\alpha = \dfrac{16-80.29}{3}[/tex]
[tex]\alpha = -21.43\ rad/s^2[/tex]
Hence, the angular acceleration of wheel is negative means wheel is decelerating.
The angular acceleration of the wheel is negative (-).
Angular acceleration:The angular acceleration would be the temporal ratio during which the angular speed changes and therefore is commonly denoted by alpha (α) as well as written throughout radians/sec.
According to the question,
Revolutions, 23 rev or,
23 × 2π = 46π
Time, 3 seconds
Final angular velocity, 16 rad/s
We know the formula,
→ Angular displacement = [tex]\frac{1}{2}[/tex] (Initial velocity + Final velocity) × Time
By substituting the values,
46 = [tex]\frac{\omega_o +16}{2}[/tex] × 3
[tex]\omega_o[/tex] = 80.29 rad/s
hence,
The angular acceleration will be:
→ α = [tex]\frac{\omega - \omega_o}{T}[/tex]
= [tex]\frac{16-80.29}{3}[/tex]
= -21.43 rad/s²
Thus the above response is correct.
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Which of the following is the least important property of a mineral?
A. color
B. luster
C. streak
D. hardness
Answer:
Color
Explanation:
The most obvious property of a mineral, its color, and is unfortunately also the least diagnostic.
Answer:
color
Explanation:
A mass of 148 g stretches a spring 13 cm. The mass is set in motion from its equlibrium position with a downward velocity of 10 cm/s and no damping is applied.
(a) Determine the position u of the mass at any time t. Use 9.8 m/s as the acceleration due to gravity. Pay close attention to the units.
(b) When does the mass first return to its equilibrium position?
Answer:
[tex]u(t)=1.15 \sin (8.68t)cm[/tex]
0.3619sec
Explanation:
Given that
Mass,m=148 g
Length,L=13 cm
Velocity,u'(0)=10 cm/s
We have to find the position u of the mass at any time t
We know that
[tex]\omega_0=\sqrt{\frac{g}{L}}\\\\=\sqrt{\frac{980}{13}}\\\\=8.68 rad/s[/tex]
Where [tex]g=980 cm/s^2[/tex]
[tex]u(t)=Acos8.68 t+Bsin 8.68t[/tex]
u(0)=0
Substitute the value
[tex]A=0\\u'(t)=-8.68Asin8.68t+8.68 Bcos8.86 t[/tex]
Substitute u'(0)=10
[tex]8.68B=10[/tex]
[tex]B=\frac{10}{8.68}=1.15[/tex]
Substitute the values
[tex]u(t)=1.15 \sin (8.68t)cm[/tex]
Period =T = 2π/8.68
After half period
π/8.68 it returns to equilibruim
π/8.68 = 0.3619sec
A wire loop with a current flowing through it is also within a uniform magnetic field. While the loop may have a net force equal to zero, in most cases, it will have an induced __________ on it.
Answer:
Current
Explanation:
A wire loop with a current flowing through it is also within a uniform magnetic field. While the loop may have a net force equal to zero, in most cases, it will have an induced current on it.
The pupil of a cat's eye narrows to a vertical slit of width 0.540 mm in daylight. Assume the average wavelength of the light is 471 nm. What is the angular resolution for horizontally separated mice
Answer:
Resolution of the light will be equal to [tex]872.22\times 10^{-6}radian[/tex]
Explanation:
We have given a pupil of cat width of the slit d = 0.540 mm = [tex]0.540\times 10^{-3}m[/tex]
Average wavelength of the light [tex]\lambda =471nm=471\times 10^{-9}m[/tex]
We have to find the angular resolution for horizontally separated mice
Angular resolution is given by [tex]\Theta =\frac{\lambda }{d}=\frac{471\times 10^{-9}}{0.540\times 10^{-3}}=872.22\times 10^{-6}radian[/tex]
So resolution of the light will be equal to [tex]872.22\times 10^{-6}radian[/tex]