Answer:
Explanation:
In standing wave pattern we find region of nodes where vibration is minimum or cold spots . The distance between any two consecutive node is half the wave length . There are 5 cold spot or node in between which is equal to 4 half wave length .
width of oven = 4 x half wave length
= 4 x (12 / 2 )
= 24 cm
Incandescent light bulbs contain a metallic filament inside. Metallic systems have allowed energy levels in a continuous range of energy, so electrons can make transitions of any energy within that range. In our lab, we will connect a light bulb to a variable AC voltage source (a Variac), which can deliver 0-140 V to the filament. The higher the voltage, the hotter we make the temperature, and the more energy we are giving the electrons in the metal, Suppose that at 20 V the filament is a dull red color, which means that most of the photons being emitted are red. When we increase the voltage, what color of light do you expect the filament to emit?
Answer:
Since at 20V most of the photons released are red then when the voltage keeps increasing the hotter the filament will be, therefore the color of light will be bright red.
Explanation:
The higher the energy the more the electrons in the molecules of the object will be excited, and when they de-excite to their ground states they release energy in the form of infrared light. The increase in voltage and higher temperatures make the object release brighter color and sometimes at the highest temperatures +1400 degrees Celsius, the color glows hot white.
A 12.0-N object is oscillating in simple harmonic motion at the end of an ideal vertical spring. Its vertical position y as a function of time t is given by y(t)=4.50cmcos[(19.5s−1)t−π/8].(a) What is the spring constant of the spring?
Answer:
465.6 N/m
Explanation:
We are given that
F=12 N
[tex]y(t)=4.50cos\left \{(19.5s^{-1}t-\frac{\pi}{8}\right \}[/tex]
We have to find the spring constant of the spring.
[tex]F=mg[/tex]
Where [tex]g=9.8 m/s^2[/tex]
Using the formula
[tex]12=m\times 9.8[/tex]
[tex]m=\frac{12}{9.8}[/tex]kg
Compare the given equation with
[tex]y(t)=Acos(\omega t-\phi)[/tex]
We get [tex]\omega=19.5[/tex]
[tex]k=m\omega^2[/tex]
Using the formula
Spring constant,[tex]k=\frac{12}{9.8}\times (19.5)^2=465.6 N/m[/tex]
The spring constant (k) for the object executing simple harmonic motion is approximately 463.29 N/m, calculated using the provided angular frequency and the mass of the object.
Explanation:To find the spring constant (k) of the spring for an object executing simple harmonic motion (SHM), we use the equation of motion provided:
y(t) = 4.50 cm cos[(19.5 s-1)t - π/8]
From the equation of motion, we can see that the angular frequency (ω) is 19.5 s-1. In SHM, the angular frequency ω is related to the spring constant k and the mass m by the following equation:
ω = sqrt(k/m)
To calculate k, we rearrange the equation to solve for k:
k = mω2
First, we find the mass (m) by using the weight of the object (12.0 N):
m = weight / g = 12.0 N / 9.81 m/s2 ≈ 1.22 kg
Then we calculate the spring constant (k):
k = (1.22 kg)(19.5 s-1)2 ≈ 463.29 N/m
Therefore, the spring constant of the spring is approximately 463.29 N/m.
An ideal gas is compressed isothermally to one-third of its initial volume. The resulting pressure will be
A) three times as large as the initial value.
B) less than three times as large as the initial value.
C) more than three times as large as the initial value.
D) equal to the initial value.
Answer:
A three times as large as the initial value
To model a tornado, superimpose the stream function for a vortex (Eq. 6.91) and a source (Eq. 6.83). Assume that the circulation is 8000 m2 /s and the pressure at the radius of 40 m is 2 kPa less than atmospheric
Answer:
The stream function for this potential flow is -4696.8
Explanation:
Given that,
Circulation = 8000 m²/s
Radius = 40 m
Pressure = 2 kPa
Suppose the determine the stream function for this potential flow
We need to calculate the stream function
Using formula of stream function
[tex]\Psi=-(\dfrac{\Gamma}{2\pi})ln(r)[/tex]
Where, Γ = circulation
r = radius
Put the value into the formula
[tex]\Psi=-\dfrac{8000}{2\pi}\times ln(40)[/tex]
[tex]\Psi=-4696.8[/tex]
Hence, The stream function for this potential flow is -4696.8
A disk of radius R (Fig. P25.73) has a nonuniform surface charge density s 5 Cr, where C is a constant and r is measured from the center of the disk to a point on the surface of the disk. Find (by direct integration) the electric potential at P.
Answer:
Explanation:
dV = k (dq)/d
dV = k (sigma da)/d
dV = k (5 C r (r dr d(phi)))/d
d = sqrt(r^2 + x^2)
dV = 5 k C (r^2/sqrt(r^2+x^2)) dr d(phi)
==> V = int{5 k C (r^2/sqrt(r^2+x^2)) dr d(phi)} ; from r=0 to r=R and phi=0 to phi=2pi
==> V = 5 k C int{(r^2/sqrt(r^2+x^2)) dr d(phi)} ; from r=0 to r=R and phi=0 to phi=2pi
==> V = 5 k C (2 pi) int{(r^2/sqrt(r^2+x^2)) dr} ; from r=0 to r=R
==> V = 5 k C (2 pi) (1/2) (R sqrt(R^2+x^2) - x^2 ln(sqrt(r^2+x^2) + r) + x^2 Ln(x))
"A 0.15 kg ball moving at 40 m/s is struck by a bat. The bat reverses the ball's direction and gives it a speed of 50 m/s. What average force does the bat apply to the ball if they are in contact for 6.0 ×10 -3 s?"
Final answer:
The average force exerted by the bat on the ball is 2250 N.
Explanation:
To calculate the average force exerted by the bat on the ball, we can use the impulse-momentum principle. The impulse experienced by the ball is equal to the change in its momentum. We can calculate the initial momentum by multiplying the mass of the ball by its initial velocity, and similarly, we can calculate the final momentum using the mass and final velocity. By subtracting the initial momentum from the final momentum, we get the change in momentum. Finally, dividing the change in momentum by the time of contact gives us the average force.
Using the given values, the initial momentum of the ball is (0.15 kg) × (-40 m/s) = -6 kg·m/s, and the final momentum is (0.15 kg) × (50 m/s) = 7.5 kg·m/s. The change in momentum is 7.5 kg·m/s - (-6 kg·m/s) = 13.5 kg·m/s. Dividing this by the time of contact, 6.0 × 10^-3 s, gives us an average force of 2250 N.
At the bottom of its path, the ball strikes a 2.30 kg steel block initially at rest on a frictionless surface. The collision is elastic. Find (a) the speed of the ball and (b) the speed of the block, both just after the collision.
Answer:
(a). The speed of the ball after collision is 2.01 m/s.
(b). The speed of the block after collision 1.11 m/s.
Explanation:
Suppose, A steel ball of mass 0.500 kg is fastened to a cord that is 50.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal.
Given that,
Mass of steel block = 2.30 kg
Mass of ball = 0.500 kg
Length of cord = 50.0 cm
We need to calculate the initial speed of the ball
Using conservation of energy
[tex]\dfrac{1}{2}mv^2=mgl[/tex]
[tex]v=\sqrt{2gl}[/tex]
Put the value into the formula
[tex]u=\sqrt{2\times9.8\times50.0\times10^{-2}}[/tex]
[tex]u=3.13\ m/s[/tex]
The initial speed of the ball [tex]u_{1}=3.13\ m/s[/tex]
The initial speed of the block [tex]u_{2}=0[/tex]
(a). We need to calculate the speed of the ball after collision
Using formula of collision
[tex]v_{1}=(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{1}+(\dfrac{2m_{2}}{m_{1}+m_{2}})u_{2}[/tex]
Put the value into the formula
[tex]v_{1}=(\dfrac{0.5-2.30}{0.5+2.30})\times3.13[/tex]
[tex]v_{1}=-2.01\ m/s[/tex]
Negative sign shows the opposite direction of initial direction.
(b). We need to calculate the speed of the block after collision
Using formula of collision
[tex]v_{2}=(\dfrac{2m_{1}}{m_{1}+m_{2}})u_{1}+(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{2}[/tex]
Put the value into the formula
[tex]v_{2}=(\dfrac{2\times0.5}{0.5+2.30})\times3.13+0[/tex]
[tex]v_{2}=1.11\ m/s[/tex]
Hence, (a). The speed of the ball after collision is 2.01 m/s.
(b). The speed of the block after collision 1.11 m/s.
3. A uniformly charged ring of radius 10.0 cm has a total T charge of 75.0 mC. Find the electric field on the axis of the ring at (a) 1.00 cm, (b) 5.00 cm, (c) 30.0 cm, and (d) 100 cm from the center of the ring.
To solve this problem we will apply the concepts related to the electric field in a ring. This concept is already standardized in the following mathematical expression, which relates the coulomb constant, the distance to the axis, the distance of the two points. Mathematically it is described as,
[tex]E = \frac{k_exq}{(x^2+r^2)^{(3/2)}} \hat{i}[/tex]
Here,
[tex]k_e[/tex] = Coulomb constant
q = Charge
x = Distance to the axis
r = Distance between the charges
Our values are given as,
[tex]r = 10.0cm = 10*10^{-2} m[/tex]
[tex]q = 75\mu C = 75*10^{-6} C[/tex]
[tex]x_a = 1.00cm[/tex]
[tex]x_b = 5.00cm[/tex]
[tex]x_c = 30cm[/tex]
[tex]x_d = 100cm[/tex]
So applying this to our 4 distances, we have
PART A)
[tex]E_a = \frac{(9*10^9)(1.00*10^{-2})(75*10^{-6})}{((1.00*10^{-2})^2+(10*10^{-2})^2)^{(3/2)}} \hat{i}[/tex]
[tex]E_a = 6.64*10^6N/C \hat{i}[/tex]
PART B)
[tex]E_b = \frac{(9*10^9)(5.00*10^{-2})(75*10^{-6})}{((5.00*10^{-2})^2+(10*10^{-2})^2)^{(3/2)}} \hat{i}[/tex]
[tex]E_b = 24.1*10^6N/C \hat{i}[/tex]
PART C)
[tex]E_c = \frac{(9*10^9)(30.00*10^{-2})(75*10^{-6})}{((30.00*10^{-2})^2+(10*10^{-2})^2)^{(3/2)}} \hat{i}[/tex]
[tex]E_c = 6.39*10^6N/C \hat{i}[/tex]
PART D)
[tex]E_d = \frac{(9*10^9)(100.00*10^{-2})(75*10^{-6})}{((100.00*10^{-2})^2+(10*10^{-2})^2)^{(3/2)}} \hat{i}[/tex]
[tex]E_d = 0.664*10^6N/C \hat{i}[/tex]
In a typical Van de Graaff linear accelerator, protons are accelerated through a potential difference of 20 MV. What is their kinetic energy if they started from rest? Give your answer in (a) eV, (b) keV,(c) MeV, (d) GeV, and (e) joules.
Answer:
a) 2 x10^7 eV
b) 2 x10^4 keV
c) 20 MeV
d) 0.02 Gev
e) 3.2 x 10^-12J
Explanation:
The potential difference = 20 x 10^6 V
The charge on the proton = 1.6 x10^-19
The work done to move the proton will be basically the proton will acquire if it accelerates.
Kinetic energy gained = ΔVq = 20 x10^6 x 1.6 x 10^-19
=3.2 x 10^-12J or 2 x10^7 eV
2 x10^7 eV = 2 x10^4 keV = 20 MeV = 0.02 Gev
Explanation:
Below is an attachment containing the solution.
Explain why the particle in a box and the harmonic oscillator are useful models for quantum mechanical systems: what chemically significant systems can they be used to represent
Answer:
Because it's used to find definite solutions to More complex quantum mechanical system. Harmonic oscillator helps in determining motion of small mass if strings.
Explanation:
The particle box is very precise when using it to calculate or find definite solutions in complex quantum mechanics in which particles might be trapped in a regions of electric potentials. Since we're dealing with electric potentials here, the harmonic oscillator is needed because it has to do squarely wit potential energy which is given as
V(x) = 0.5kx²
Where k is force constant and x is distance.
Final answer:
The particle in a box and the harmonic oscillator models are useful for understanding quantum mechanical systems. They can be used to represent various chemically significant systems, including blackbody radiators, atomic and molecular spectra, and optoelectronic devices.
Explanation:
The particle in a box and the harmonic oscillator are useful models for quantum mechanical systems because they provide insights into the behavior of particles in confined spaces and under the influence of potential energy. The particle in a box model describes a particle free to move in a small space surrounded by impenetrable barriers, and can be used to represent systems such as blackbody radiators, atomic and molecular spectra, and optoelectronic devices. The harmonic oscillator model, on the other hand, represents systems with periodic motion, such as molecular vibrations and wave packets in quantum optics.
Two objects are moving at equal speed along a level, frictionless surface. The second object has twice the mass of the first object. They both slide up the same frictionless incline plane. Which object rises to a greater height?
a. Object 1 rises to the greater height because it weighs less.
b. Object 2 rises to the greater height because it possesses a larger amount of kinetic energy.
c. Object 2 rises to the greater height because it contains more mass.
d. Object 1 rises to the greater height because it possesses a smaller amount of kinetic energy.
e. The two objects rise to the same height.
Answer:
Option E is correct.
The two objects rise to the same height.
Explanation:
Let the mass, velocity and height the bigger object will rise to be M, V and H respectively.
Let the mass, velocity and height the object will rise to be m, v and h respectively.
Note that V = v
Using the work energy theorem
The change in kinetic energy of a body between two points is equal to the work done on the body between those two points.
Change in kinetic energy of the bigger object = (final kinetic energy) - (initial kinetic energy) = 0 - (1/2)(M)(V²) = (-MV²/2)
(The final kinetic energy = 0 J because the object comes to rest at the final point)
Work done on the bigger object = work done by all the forces acting on the body
But the only force acting on the body is the force of gravity (since the inclined plane is frictionless)
Workdone on the body = work done by the force of gravity in moving the body up a height of H = - MgH
(-MV²/2) = - MgH
H = (V²/2g)
For the small body,
Change in kinetic energy of the bigger object = (final kinetic energy) - (initial kinetic energy) = 0 - (1/2)(m)(v²) = (-mv²/2)
Workdone on the body = work done by the force of gravity in moving the body up a height of H = - mgh
(-mv²/2) = - mgh
h = (v²/2g)
Since V = v as given in the question (Both bodies have the same speeds)
H = h = (V²/2g) = (v²/2g)
Hope this Helps!!!
A 0.140 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.900 m/s . It has a head-on collision with a 0.297 kg glider that is moving to the left with a speed of 2.25 m/s . Suppose the collision is elastic.
Answer:
(a). The final velocity of the first glider is −2.15 m/s.
(b). The final velocity of the second glider is -1.67 m/s.
Explanation:
Given that,
Mass of glider = 0.140 kg
Speed = 0.900 m/s
Mass of another glider = 0.297 kg
Speed =2 .25 m/s
Suppose, Find the magnitude of the final velocity of the first glider.
Find the magnitude of the final velocity of the second glider.
(a). We need to calculate the final velocity of the first glider
Using formula of collision
[tex]v_{1}=\dfrac{u_{1}(m_{1}-m_{2})+2m_{2}u_{2}}{m_{1}+m_{2}}[/tex]
Put the value into the formula
[tex]v_{1}=\dfrac{0.900(0.140+0.297)+2\times(-0.297)\times2.25}{0.140+0.297}[/tex]
[tex]v_{1}=-2.15\ m/s[/tex]
(b). We need to calculate the final velocity of the second glider
Using formula of collision
[tex]v_{2}=\dfrac{u_{2}(m_{2}-m_{1})+2m_{1}u_{1}}{m_{1}+m_{2}}[/tex]
Put the value into the formula
[tex]v_{1}=\dfrac{-2.25(0.297+0.140)+2\times(0.140)\times0.900}{0.140+0.297}[/tex]
[tex]v_{1}=-1.67\ m/s[/tex]
Hence, (a). The final velocity of the first glider is −2.15 m/s.
(b). The final velocity of the second glider is -1.67 m/s.
Answer with Explanation:
We are given that
Mass of glider=m=0.14 kg
Initial speed of first glider, u=0.9 m/s
m'=0.297 kg
Initial speed of second glider, u'=-2.25m/s
a.We have to find the magnitude of final velocity of 0.140 kg glider.
The collision is elastic then the final velocity of 0.140 kg glider
[tex]v=\frac{(m-m')u+2m'u'}{m+m'}[/tex]
Substitute the values
[tex]v=\frac{(0.140-0.297)\times 0.9+2\times 0.297\times (-2.25)}{0.140+0.297}=-3.38m/s[/tex]
Magnitude of final velocity of 0.140 kg glider=3.38 m/s
b.[tex]v'=\frac{u'(m'-m)+2mu}{m+m'}[/tex]
[tex]v'=\frac{-2.25(0.297-0.140)+2\times 0.140\times 0.9}{0.140+0.297}=-0.23 m/s[/tex]
Hence, the magnitude of final velocity of 0.297 kg glider=0.23 m/s
A proton moves perpendicular to a uniform magnetic field B at a speed of 1.00 x10^7 m/s and experiences an acceleration of 2.00 x10^13 m/s^2. in the positive x-direction when its velocity is in the positive z-direction. Determine the magnitude and direction of the field for which the magnitude of the field is a minimium
Answer:
So, Magnitude of the field (B) = 2.09 x 10^(-2) T.
It's in the negative direction since acceleration is in positive direction.
Explanation:
First of all, since acceleration is in positive x- direction, the magnetic field must be in negative y- direction.
We know that The magnitude of the Lorentz force F is; F = qvB sinθ
So, B = F/(qvsinθ)
F = ma.
Speed(v) = 1.00 x10^(7) m/s
acceleration (a) = 2.00 x10^(13) m/s^(2)
Mass of proton = 1.673 × 10^(-27) kilograms
q(elementary charge of proton) = 1.602×10^(−19)
Since right hand thumb rule, θ= 90°
So;B = [1.673 × 10^(-27) x 2.00 x10^(13)] / [ {1.602×10^(−19)} x {1.00 x10^(7)} x sin 90]
So,B = 2.09 x 10^(-2) T.
It's in the negative direction since acceleration is in positive direction.
The magnitude of the magnetic field that the proton experiences is 2.08 x 10^-4 Tesla. The direction of the magnetic field, determined by the right-hand rule, is in the negative y-direction.
Explanation:The force experienced by a charged particle moving within a magnetic field, like the proton in your question, can be calculated using the Lorentz force law: F = qvBsinθ, where F is the force, q is the charge of the particle, v is its velocity, B is the magnetic field, and θ is the angle between the velocity and the magnetic field. Because the proton is moving perpendicular to the field, θ = 90°, and sinθ = 1. Therefore, F = qvB. To find the magnetic field, B, we can rearrange this formula to give B = F/qv.
Now, we know from Newton's second law that F = ma, so we can replace F in our formula with the proton's mass (m, which is 1.67 x 10^-27 kg for a proton) times the given acceleration (a), yielding B = ma/qv.
Substitute the given values for acceleration, proton's charge, and velocity into our equation, we get: B = (1.67 x 10^-27 kg)(2 x 10^13 m/s^2) / (1.60 x 10^-19 C)(1 x 10^7 m/s) = 2.08 x 10^-4 Tesla (T).
The direction of the magnetic field is given by the right-hand rule as discussed in Essential Knowledge 2.D.1. If you arrange your right hand such that your thumb points in the direction of the proton's velocity (positive z-direction) and your fingers curl in the direction of the force (positive x-direction), your palm points in the direction of the magnetic field, which would be in the negative y-direction.
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Old cannons were built on wheeled carts, both to facilitate moving the cannon and to allow the cannon to recoil when fired. When a 150 kg cannon and cart recoils at 1.5 m/s, at what velocity would a 10 kg cannonball leave the cannon?
Answer:
22.5 m/s
Explanation:
Applying Newton's third law of motion
Momentum of the cannon and cart = momentum of the cannonball
MV = mv..................... Equation 1
Where M = mass of the cannon and the cart, V = Recoil velocity of the cannon and the cart, m = mass of the cannonball, v = velocity of the cannonball
make v the subject of the equation
v = MV/m............. Equation 2
Given: M = 150 kg, V = 1.5 m/s, m = 10 kg
Substitute into equation 2
v = 150(1.5)/10
v = 22.5 m/s
Hence the cannonball leave the cannon with a velocity of 22.5 m/s
Answer:
v2 = 22.5 m/s
Explanation:
Momentum is how hard to stop or turn a moving object . Generally, momentum measures mass in motion. Momentum is a vector quantity. Mathematically,
p = mass × velocity
The total momentum of an isolated system of bodies remains constant.
mometum before = 0
mass of the canon (m1) = 150 kg
mass of the ball (m2) = 10 kg
velocity of the ball (v2) = ?
velocity of the cannon(v1) = 1.5 m/s
momentum after = momentum before
m2v2 + m1v1 = 0
10v2 = 150 × 1.5
10v2 = 225
divide both sides by 10
v2 = 225/10
v2 = 22.5 m/s
Consider a sphere, an infinitely long cylinder, and a plane of infinite length and width (a, b and c below). Imagine that you can hover above each one in your own personal helicopter. In which case do you have the most freedom to move about without your view of the object changing? In other words, for each case consider if there are directions that you can move in without the objects distance or orientation, relative to you, changing.
Answer:
A plane of infinite length and width.
Explanation:
For the case of the sphere, we should consider spherical coordinates.
You can move around the sphere without changing your distance from the center of the sphere, however you can alter your azimuthal angle and polar angle, since they are symmetric in spherical coordinate system.
r --> Cannot change
Φ --> Free to change
θ --> Free to change
For the case of the infinite cylinder, we should consider cylindrical coordinates.
You can change your height and angular coordinate, but you cannot change your distance from the axis of the cylinder.
r --> Cannot change
θ --> Free to change
z --> Free to change
For the case of the infinite plane, we should consider cartesian coordinates.
Since the length and width of the plane is infinite, we cannot recognize whether we are getting closer or further away.
x --> Free to move
y --> Free to move
z --> Free to move
Therefore, in the case of infinite plane you have the most freedom to move about without your view of the object changing.
Final answer:
An infinite plane allows the most freedom of movement without changing the view, due to its infinite symmetry. In contrast, a sphere and a cylinder have more limited symmetrical properties, resulting in a restricted freedom of movement to keep the view unchanged.
Explanation:
When comparing the freedom to move above a sphere, an infinitely long cylinder, and an infinite plane, the plane provides the most freedom without changing your view of the object. On a sphere, any move results in a different orientation or distance from the surface. With the cylinder, movement along the axis does not change the view, but any other direction does. However, the infinite plane allows movement in any direction on a two-dimensional plane without changing the viewed geometry or orientation.
From a mathematical perspective, this question revolves around symmetry and geometric invariance under transformation. The infinite plane exhibits infinite symmetry; therefore, no matter how you move parallel to its surface, the view remains unchanged. Conversely, a sphere has rotational symmetry around its center, and a cylinder along its axis, limiting the directions in which one can move without altering the perceived distance or orientation.
A train, traveling at a constant speed of 22.0 m/s, comes to an incline with a constant slope. While going up the incline, the train slows down with a constant acceleration of magnitude 1.40 m/s2. How far has the train traveled up the incline after 7.30 s
Answer:
123.30 m
Explanation:
Given
Speed, u = 22 m/s
acceleration, a = 1.40 m/s²
time, t = 7.30 s
From equation of motion,
v = u + at
where,
v is the final velocity
u is the initial velocity
a is the acceleration
t is time
V = at + U
using equation v - u = at to get line equation for the graph of the motion of the train on the incline plane
[tex]V_{x} = mt + V_{o}[/tex] where m is the slope
Comparing equation (1) and (2)
[tex]V = V_{x}[/tex]
a = m
[tex]U = V_{o}[/tex]
Since the train slows down with a constant acceleration of magnitude 1.40 m/s² when going up the incline plane. This implies the train is decelerating. Therefore, the train is experiencing negative acceleration.
a = - 1.40 m/s²
Sunstituting a = - 1.40 m/s² and u = 22 m/s
[tex]V_{x} = -1.40t + 22[/tex]
[tex]V_{x} = -1.40(7.30) + 22[/tex]
[tex]V_{x} = -10.22 + 22[/tex]
[tex]V_{x} = 11. 78 m/s[/tex]
The speed of the train at 7.30 s is 11.78 m/s.
The distance traveled after 7.30 sec on the incline is the area cover on the incline under the specific interval.
Area of triangle + Area of rectangle
[tex][\frac{1}{2} * (22 - 11.78) * (7.30)] + [(11.78 - 0) * (7.30)][/tex]
= 37.303 + 85.994
= 123. 297 m
≈ 123. 30 m
Answer:
123.297 m
Explanation:
A train, traveling at a constant speed of 22.0 m/s,
v = 22.0 m/s
the train slows down with a constant acceleration of magnitude 1.40 m/s².
[tex]a_s[/tex] = -1.4 m/s²
How far has the train traveled up the incline after 7.30 s
t =7.30 s
We can calculate the distance traveled up the incline after 7.30 s by using the formula:
[tex]x_f =x_i+v_xt+\frac{1}{2}a_st^2[/tex]
where;
[tex]x_f[/tex] = the distance traveled up
[tex]x_i[/tex] = 0
[tex]v_x[/tex] = speed of the train
[tex]a_s[/tex] = deceleration
t = time
Substituting our data; we have:
[tex]x_f = 0+(22m/s)(7.30s)+\frac{1}{2}(-1.4m/s^2)(7.30s)[/tex]
[tex]x_f =16.06 -37.303[/tex]
[tex]x_f[/tex] = 123.297 m
The electron drift speed in a metal wire is exceedingly slow. Yet, when you turnon the light switch the light begins to illuminate almost immediately. Explain whyt his is not a paradox?
Explanation:
The misunderstanding here is that the thing that turns on the light bulb is not the same electrons near the light switch. So, the electrons near the switch is not moving all the way across the circuit instantly. The electrons are distributed across the wire. When the light switch is turned on, the circuit is connected and there is a potential difference between the bulb and the source. This potential difference creates an electric field, and free electrons move under the influence of this electric field according to Coulomb's Law. When they start to move the electrons closest to the bulb causes the bulb to glow.
So, the important factor here is not the drift velocity of the electrons but the number of electrons and the strength of the electric field.
If 745-nm and 660-nm light passes through two slits 0.54 mm apart, how far apart are the second-order fringes for these two wavelengths on a screen 1.0 m away
Answer:
0.82 mm
Explanation:
The formula for calculation an [tex]n^{th}[/tex] bright fringe from the central maxima is given as:
[tex]y_n=\frac{n \lambda D}{d}[/tex]
so for the distance of the second-order fringe when wavelength [tex]\lambda_1[/tex] = 745-nm can be calculated as:
[tex]y_2 = \frac{n \lambda_1 D}{d}[/tex]
where;
n = 2
[tex]\lambda_1[/tex] = 745-nm
D = 1.0 m
d = 0.54 mm
substituting the parameters in the above equation; we have:
[tex]y_2 = \frac{2(745nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}[/tex]
[tex]y_2[/tex] = 0.00276 m
[tex]y_2[/tex] = 2.76 × 10 ⁻³ m
The distance of the second order fringe when the wavelength [tex]\lambda_2[/tex] = 660-nm is as follows:
[tex]y^'}_2 = \frac{2(660nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}[/tex]
[tex]y^'}_2[/tex] = 1.94 × 10 ⁻³ m
So, the distance apart the two fringe can now be calculated as:
[tex]\delta y = y_2-y^{'}_2[/tex]
[tex]\delta y[/tex] = 2.76 × 10 ⁻³ m - 1.94 × 10 ⁻³ m
[tex]\delta y[/tex] = 10 ⁻³ (2.76 - 1.94)
[tex]\delta y[/tex] = 10 ⁻³ (0.82)
[tex]\delta y[/tex] = 0.82 × 10 ⁻³ m
[tex]\delta y[/tex] = 0.82 × 10 ⁻³ m [tex](\frac{1.0mm}{10^{-3}m} )[/tex]
[tex]\delta y[/tex] = 0.82 mm
Thus, the distance apart the second-order fringes for these two wavelengths = 0.82 mm
Two long parallel wires 20 cm apart carry currents of 5.0 A and 8.0 A in the same direction. Is there any point between the two wires where the magnetic field is zero?
Answer:
x= 0.077 m from the wire carrying 5.0 A current.
Explanation:
If the wire can be approximated as an infinite one, and we can neglect the diameter of the wire, we can find the magnetic field B at a distance d from the wire, with the following expression:[tex]B =\frac{\mu_{0} * I}{2*\pi*d}[/tex]
Due to the currents are in the same direction, this means that the magnetic field lines (taking the shape of circumferences) will have opposite directions between the wires.So, if we assume that at some distance from both wires, the magnetic field will be 0, we can write the following equation:[tex]\frac{\mu_{0} * I_{1}}{2*\pi*x} - \frac{\mu_{0} * I_{2} }{2*\pi*(d-x)} = 0[/tex]
where I₁ = 5.0A, I₂= 8.0A and d = 0.2 mSimplifying common terms, we can solve for x, as follows:[tex]\frac{I_{1} }{x} = \frac{I_{2} }{(d-x)} \\ \frac{5.0A}{x(m)} = \frac{8.0A}{(0.2m-x(m))}[/tex]
⇒ [tex]x =\frac{1m}{13} = 0.077 m = 7.7 cm[/tex]
A girl whirls a stone in a horizontal circle 1.50 m above the ground by means of a string 165 cm long. The string breaks, and the stone flies of horizontally and strike the ground 6.90 m away. What is the centripetal acceleration of the stone while in circular motion
Answer:
95.5ms-2
Explanation:
First we obtain the time taken for the motion from the equation of motion. Secondly we obtain the horizontal velocity of the stone. This can now be used to calculate the centripetal acceleration. Note that the length of the chord is the radius of the circle around which the stone moves.
All details are contained in the detailed step-by-step solution attached to this answer.
A charge of 8.473 nC is uniformly distributed along the x-axis from −1 m to 1 m . What is the electric potential (relative to zero at infinity) of the point at 5 m on the x-axis? The value of the Coulomb constant is 8.98755 × 10^9 N · m2 /C^2 . Answer in units of V.
The electric potential at 5 meters on the x-axis due to a uniformly distributed charge of 8.473 nC along the x-axis from -1m to 1m is approximately 51.84 volts.
Explanation:To determine the electric potential at a point on the x-axis created by a uniformly distributed charge, we can use the formula for electric potential due to continuous charge distribution. For a continuous charge distribution along the x-axis, the electric potential (V) at a point (P) is given by the integral of [tex]\(k \cdot \frac{dq}{r}\), where \(k\)[/tex] is Coulomb's constant, (dq) is the small charge element, and \(r) is the distance between the charge element and the point (P).
In this scenario, we integrate the expression across the entire distribution of charge from -1m to 1m along the x-axis to find the total potential at point 5m on the x-axis. Since the charge is uniformly distributed, we can represent (dq) as [tex]\(\lambda dx\),[/tex]where [tex]\(\lambda\)[/tex] is the charge per unit length and \(dx\) is the small element of length along the axis.
By integrating this expression and plugging in the given values into the formula, we obtain the result of approximately 51.84 volts for the electric potential at the point 5m away from the charge distribution on the x-axis.
This value signifies the work done per unit charge to bring a positive test charge from infinity to the specified point, considering the influence of the continuous charge distribution along the x-axis.
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In the Rutherford model of hydrogen the electron (mass=9.11x10-31 kg) is in a planetary type orbit around the proton (mass=1.67x10-27 kg). Assume that the radius of the orbit is 2.5 Angstroms (10-10 m). (a) What is the magnitude of the gravitational force of attraction between electron and proton? AN (b) What is the magnitude of the electric force of attraction between electron and proton? N
Answer:
(a) [tex]F_g=1.62*10^{-48}N[/tex]
(b) [tex]F_e=3.68*10^{-9}N[/tex]
Explanation:
(a) We use Newton's law of universal gravitation, in order to calculate the gravitational force between electron and proton:
[tex]F_g=-G\frac{m_1m_2}{r^2}[/tex]
Where G is the Cavendish gravitational constant, [tex]m_1[/tex] and [tex]m_2[/tex] are the masses of the electron and the proton respectively and r is the distance between them:
[tex]F_g=-6.67*10^{-11}\frac{N\cdot m^2}{kg^2}\frac{(9.11*10^{-31}kg)(1.67*10^{-27}kg)}{(2.5*10^{-10}m)^2}\\F_g=-1.62*10^{-48}N[/tex]
The minus sing indicates that the force is repulsive. Thus, its magnitude is:
[tex]F_g=1.62*10^{-48}N[/tex]
(b) We use Coulomb's law, in order to calculate the electric force between electron and proton, here k is the Coulomb constant and e is the elementary charge:
[tex]F_e=-k\frac{e^2}{r^2}\\F_e=-8.99*10^{9}\frac{N\cdot m^2}{C^2}\frac{(1.6*10^{-19}C)^2}{(2.5*10^{-10}m)^2}\\F_e=-3.68*10^{-9}N[/tex]
Its magnitude is:
[tex]F_e=3.68*10^{-9}N[/tex]
A magnetic field has a magnitude of 1.2 \times 10^{-3} T, and an electric field has a magnitude of 4.6 \times 10^{3}N/C. Both fields point in the same direction. A positive 1.8 \mu C charge moves at a speed of 3.1 \times 10^6 m/s in a direction that is perpendicular to both fields. Determine the magnitude of the net force that acts on the charge.
Answer: F = 113.4.[tex]10^{-3}[/tex]N
Explanation: Net Force is the total forces acting in an object. In this case, there are two forces acting on the charge: one due to magnetic field (Fm) and another due to electric field (Fe). So, net force is
F = Fe + Fm
Force due to electric field
To determine this force:
Fe = q.E, where q is the charge and E is electric field.
Calculating:
Fe = q.E
Fe = 1.8.[tex]10^{-6}[/tex].4.6.[tex]10^{3}[/tex]
Fe = 8.28.[tex]10^{-3}[/tex]N
Force due to magnetic field: It can only happens when the charge is in movement, so
Fm = q.(v×B), where v represents velocity and B is magnetic field
The cross product indicates that force is perpendicular to the velocity and the field.
Calculating:
Fm = q.v.B.senθ
As θ=90°,
Fm = q.v.B
Fm = 1.8.[tex]10^{-6}[/tex].3.1.[tex]10^{6}[/tex].1.2.[tex]10^{-3}[/tex]
Fm = 6.696.[tex]10^{-3}[/tex]N
F, Fm and Fe make a triangle. So, using Pythagorean theorem:
F = [tex]\sqrt{Fe^{2} + Fm^{2} }[/tex]
F = [tex]\sqrt{(8.28.10^{-3} )^{2} +(6.696.10^{-3} )^{2} }[/tex]
F = 113.4.[tex]10^{-3}[/tex]N
The net force acting on the charge is F = 113.4.[tex]10^{-3}[/tex]N
Two light bulbs are operated at a potential difference of 110 V. Light bulb A produces 60 W of power and light bulb B produces 100 W of power. Which statement below is correct?
a. The 60 W bulb has a greater resistance and smaller current than the 100 W bulb.
b. The 60 W bulb has a smaller resistance and smaller current than the 100 W bulb.
c. The 60 W bulb has a greater resistance and greater current than the 100 W bulb.
d. The 60 W bulb has a smaller resistance and greater current than the 100 W bulb.
Answer:a
Explanation:
Given
Potential difference [tex]V=110\ V[/tex]
Power of bulb A [tex]P_A=60\ W[/tex]
Power of bulb B [tex]P_B=100\ W[/tex]
If voltage is same for both the bulbs then Power is given by
[tex]P=\dfrac{V^2}{R}[/tex]
[tex]P_A=\dfrac{(110)^2}{R_A}[/tex]
[tex]60=\dfrac{110^2}{R_A}[/tex]
[tex]R_A=201.66\ \Omega[/tex]
similarly
[tex]R_B=\dfrac{110^2}{100}[/tex]
[tex]R_B=121\ \Omega[/tex]
[tex]R_A>R_B[/tex]
so current in bulb A is smaller than B
Thus the 60 W bulb has a greater resistance.
Thus option (a) is correct
The correct statement is that a) the 60 W bulb has a greater resistance and smaller current than the 100 W bulb, since both bulbs are operated at the same voltage but the 60 W bulb has lower power.
To determine the correct statement about the resistance and current for two light bulbs operated at a potential difference of 110 V, we need to consider the power ratings of the bulbs and know that power (P) is the product of current (I) and voltage (V), P = IV. Furthermore, by using Ohm's law (V = IR), we can infer that Power can also be expressed as P = I2R or P = V²/R, which relates power to resistance (R).
Both bulbs operate at the same voltage (110 V), which means their power differences are due to differences in current and resistance.
For bulb A (60 W): P = V²/R, so R = V²/P = (110 V)2/60 W.
For bulb B (100 W): R = V²/P = (110 V)²/100 W.
Comparing these two, bulb A has higher resistance because it has lower power for the same voltage. Also, since P = IV, and bulb A has a lower power at the same voltage, it must also have a smaller current. Therefore, the correct answer is: a. The 60 W bulb has a greater resistance and smaller current than the 100 W bulb.
A racquet ball with mass m = 0.238 kg is moving toward the wall at v = 12.4 m/s and at an angle of θ = 31° with respect to the horizontal. The ball makes a perfectly elastic collision with the solid, frictionless wall and rebounds at the same angle with respect to the horizontal. The ball is in contact with the wall for t = 0.078 s. 1)What is the magnitude of the initial momentum of the racquet ball?
Answer:
[tex]||\vec p || = 2.951\,\frac{kg\cdot m}{s}[/tex]
Explanation:
The initial momentum of the racquet ball is:
[tex]||\vec p || = (0.238\,kg)\cdot (12.4\,\frac{m}{s} )[/tex]
[tex]||\vec p || = 2.951\,\frac{kg\cdot m}{s}[/tex]
A 0.60-kg particle has a speed of 2.0 m/s at point A and a kinetic energy of 7.5 J at point B. What is (a) its kinetic energy at A? (b) Its speed at point B ? (c) The total work done on the particle as it moves from A to B ?
Answer:
Explanation:
note:
solution is attached due to error in mathematical equation. please find the attachment
Answer:
(a) 1.2 J
(b) 5 m/s
(c) 6.3 J
Explanation:
(a) Kinetic energy at A
Ek = 1/2mv².................. Equation 1
Where Ek = Kinetic energy at A, m = mass of the particle, v = velocity at A.
Given: m = 0.6 kg, v = 2.0 m/s
Substitute into equation 1
Ek = 1/2(0.6)(2²)
Ek = 0.6(2)
Ek = 1.2 J.
(b)
Speed at point B
Ek' = 1/2mv'²............... Equation 2
Make v' the subject of the equation,
v' = √(2Ek'/m).................. Equation 3
Where, Ek' = kinetic energy at B, v' = velocity at B.
Given: Ek' = 7.5 J, m = 0.6 kg,
Substitute into equation 3
v' = √[(2×7.5)/0.6]
v' =√(15/0.6)
v' = √25
v' = 5 m/s.
(c)
Wt = Δ kinetic energy from A to B
Where Wt = total work done as the particle moves from A to B.
Wt = 1/2m(v'²-v²)
Wt = 1/2(0.6)(5²-2²)
Wt = 0.3(25-4)
Wt = 0.3(21)
Wt = 6.3 J
A typical laboratory centrifuge rotates at 3700 rpm . Test tubes have to be placed into a centrifuge very carefully because of the very large accelerations. Part A What is the acceleration at the end of a test tube that is 10 cm from the axis of rotation
Answer:
Explanation:
acceleration of test tube
= ω² R
= (2πn)² R
= 4π²n²R
n = no of rotation per second
= 3700 / 60
= 61.67
R = .10 m
acceleration
= 4π²n²R
= 4 x 3.14² x 61.67² x .10
= 14999 N Approx
Final answer:
The acceleration at the end of a test tube 10 cm from the axis of rotation in a centrifuge spinning at 3700 rpm is calculated to be 15051.2 m/s², demonstrating the significant centrifugal forces generated by such devices.
Explanation:
To calculate the acceleration at the end of a test tube that is 10 cm from the axis of rotation in a centrifuge spinning at 3700 rpm, we first need to convert the rotational speed to radians per second. The formula to convert revolutions per minute (rpm) to radians per second (ω in rad/s) is ω = (2π×rpm)/60. Thus, ω = (2π× 3700)/60 = 387.98 rad/s. Next, we use the formula for centripetal acceleration, a = ω²×r, where r is the radius of the circle (distance from the center of rotation to the point of interest) in meters. Given that r = 10 cm = 0.1 m, the acceleration a = (387.98)^2× 0.1 = 15051.2 m/s².
This centripetal acceleration is much larger than Earth's gravitational acceleration, indicating the extreme forces at play in a centrifuge's operation, which is crucial for its role in laboratory settings for sedimentation of materials.
The top of a swimming pool is at ground level. If the pool is 3.00 m deep, how far below ground level does the bottom of the pool appear to be located for the following conditions? (The index of refraction of water is 1.333.)
(a) The pool is completely filled with water.
______m below ground level
(b) The pool is filled halfway with water.
______m below ground level
Answer:
a) 2.25 m
b) 2.625 m
Explanation:
Refraction is the name given to the phenomenon of the speed of light changing the the boundary when it moves from one physical medium to the other.
Refractive index is the ratio of the speed of light in empty vacuum (air is an appropriate substitution) to the speed of light in the medium under consideration.
In terms of real and apparent depth, the refractive index is given by
η = (real depth)/(apparent depth)
a) Real depth = 3.00 m
Apparent depth = ?
Refractive index, η = 1.333
1.333 = 3/(apparent depth)
Apparent depth = 3/1.3333 = 2.25 m.
Hence the bottom of the pool appears to be 2.25 m below the ground level.
b) Real depth = 1.5 m
Apparent depth = ?
Refractive index, η = 1.333
1.3333 = 1.5/(apparent depth)
Apparent depth = 1.5/1.3333 = 1.125 m
But the pool is half filled with water, there is a 1.5 m depth on top of the pool before refraction starts.
So, apparent depth of the pool = 1.5 + 1.125 = 2.625 m below the ground level
The apparent depth of a swimming pool is measured by considering the water's refractive index. With the pool completely filled, the bottom appears to be 2.25m deep. When halfway filled, the pool appears to be 2.625m deep.
Explanation:When light travels from a medium with a high refractive index to one with a lower refractive index, the light is refracted, or bent, making objects appear closer than they actually are. We can calculate this apparent depth by using the formula d' = d / n, where d' is the apparent depth, d is the actual depth, and n is the refractive index.
(a) If the pool is completely filled with water, for a person looking from above ground, the bottom of the pool appears to be closer than it actually is. Substituting the given values into the formula, we get the apparent depth: d' = 3.00m / 1.333 = 2.25 m below ground level.
(b) If the pool is halfway filled with water, the apparent depth of the water is calculated in the same way. However, the depth beneath the water is below the refractive index border and is not subject to refraction. Therefore, the apparent total depth of the partially filled pool is the sum of the actual depth of the air part (1.50m) and the apparent depth of the water part (1.50m / 1.333 = 1.125m). This gives us 2.625m.
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Bulb ܣ is rated for 20W at 12V. Bulb ܤ is rated for 20W at 120V. The two bulbs are mistakenly mixed up so that bulb ܣ is connected to a 120V line and bulb ܤ is connected to a 12V line. a. Determine the power dissipated by Bulb ܣ while connected to the 120Vline. b. Determine the power dissipated by Bulb ܤ while connected to the 12Vline. c. Which bulb, if any, is more likely to burn out and why?
Answer:
a. 2000W
b.0.2W
c. Bulb 1
Explanation:
Data given:
bulb 1 rating =20w,12v
bulb 2 rating =20w,120v.
we calculate the resistance for each bulb
[tex]bulb 1: R=\frac{V^2}{P}\\ bulb 1: R=\frac{12^2}{20}\\ R=7.2 ohms[/tex]
for bulb 2
[tex]bulb 2: R=\frac{V^2}{P}\\ bulb 2: R=\frac{120^2}{20}\\ R=720 ohms[/tex]
when miss connection occur we need to calculate the dissipated power from wash bulb
a. for bulb 1 with R=7.2ohms and V=120v,
the power is calculated as
[tex]P=\frac{v^2}{R}\\ P=\frac{120^2}{7.2}\\ P=2000W[/tex]
b. for bulb 2 with R=720 ohms and V=12v,
the power is calculated as
[tex]P=\frac{v^2}{R}\\ P=\frac{12^2}{720}\\ P=0.2W\\[/tex]
c. Bulb 1 is more likely to burn out because it is operating above the rated power.
Two coils have the same number of circular turns and carry the same current. Each rotates in a magnetic field acting perpendicularly to its axis of rotation. Coil 1 has a radius of 5.6 cm and rotates in a 0.24-T field. Coil 2 rotates in a 0.44-T field. Each coil experiences the same maximum torque. What is the radius (in cm) of coil 2?
Answer:
4.14 cm.
Explanation:
Given,
For Coil 1
radius of coil, r₁ = 5.6 cm
Magnetic field, B₁ = 0.24 T
For Coil 2
radius of coil, r₂ = ?
Magnetic field, B₂ = 0.44 T
Using formula of maximum torque
[tex]\tau_{max}= NIAB[/tex]
Since both the coil experience same maximum torques
now,
[tex] NIA_1B_1 = NIA_2B_2[/tex]
[tex]A_1B_1 = A_2B_2[/tex]
[tex] r_1^2 B_1 = r_2^2 B_2[/tex]
[tex] 5.6^2\times 0.24= r_2^2\times 0.44[/tex]
[tex]r_2 = 4.14\ cm[/tex]
Radius of the coil 2 is equal to 4.14 cm.