Answer:
car1: a=3.1m/s^2 , car2: a=6.1m/s^2
(a) 1/2*3.1*t^2= 1/2*6.1*(t-0.9)^2
1.55t^2= 3.05(t^2-1.8t+0.81)= 3.05t^2-5.49t+2.4705
1.5t^2-5.49t+2.4705= 0
t= 3.13457 = 3.14[s] after.
(b) d= 1/2*3.1*3.13457^2= 15.23[m] approx.
(c) car1: v=at = 3.1*3.13457= 9.717m/s
car2: v=at = 6.1*(3.13457-0.9)= 13.631m/s
13.631-9.717= 3.914 = 3.91[m/s] faster than car1.
At the time after 3.14 seconds, car1 is overtaken by car2, this will happen after 15.23 meters and car2 is 3.914 m/s faster than car1.
What is Acceleration?In mechanics, acceleration is the measure of how quickly an object's velocity changes in relation to time. The quantity is a vector. An object's acceleration is determined by the direction of the net force exerted on it.
Acceleration is a vector quantity since it has a magnitude and a direction. A vector quantity is also velocity. Acceleration is defined as the change in velocity vector over a time interval divided by the time interval.
There are several types of acceleration :
Uniform AccelerationNon-Uniform AccelerationAverage Acceleration(a)
1/2 × 3.1× t²
⇒ 1/2 × 6.1 × (t-0.9)²
⇒1.55t²= 3.05 (t² - 1.8t + 0.81)
⇒ 3.05t² -5.49t + 2.4705
=1.5t² -5.49t +2.4705= 0
t= 3.14 seconds.
(b)
d= 1/2 × 3.1 × 3.14²
= 15.23 meters
(c)
car1: v=at
⇒3.1 × 3.14
= 9.717 m/s
car2:
v=at
⇒ 6.1 × (3.14 - 0.9)
= 13.631 m/s
13.631-9.717= 3.914 m/s
Hence, car2 is 3.914 m/s faster than car1.
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A 0.300 kg block is pressed against a spring with a spring constant of 8050 N/m until the spring is compressed by 6.00 cm. When released, the block slides along a horizontal surface that is frictionless except for a 7.00 m long rough section.
a) The block comes to a stop exactly at the end of the rough section. What is the coefficient of kinetic friction between the block and the rough section?
b) The rough section is now polished, so that the entire track is frictionless. The block slides along the surface and into the circular loop at the end. If the normal force between the loop and the block is 90.0 N when the block reaches point P (the exact top of the loop), what is the radius R of the circular loop?
Answer:
a) [tex]\mu_{k} = 0.704[/tex], b) [tex]R = 0.312\,m[/tex]
Explanation:
a) The minimum coeffcient of friction is computed by the following expression derived from the Principle of Energy Conservation:
[tex]\frac{1}{2}\cdot k \cdot x^{2} = \mu_{k}\cdot m\cdot g \cdot \Delta s[/tex]
[tex]\mu_{k} = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \Delta s}[/tex]
[tex]\mu_{k} = \frac{\left(8050\,\frac{N}{m} \right)\cdot (0.06\,m)^{2}}{2\cdot (0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (7\,m)}[/tex]
[tex]\mu_{k} = 0.704[/tex]
b) The speed of the block is determined by using the Principle of Energy Conservation:
[tex]\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}[/tex]
[tex]v = x\cdot \sqrt{\frac{k}{m} }[/tex]
[tex]v = (0.06\,m)\cdot \sqrt{\frac{8050\,\frac{N}{m} }{0.3\,kg} }[/tex]
[tex]v \approx 9.829\,\frac{m}{s}[/tex]
The radius of the circular loop is:
[tex]\Sigma F_{r} = -90\,N -(0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} ) = -(0.3\,kg)\cdot \frac{v^{2}}{R}[/tex]
[tex]\frac{\left(9.829\,\frac{m}{s}\right)^{2}}{R} = 309.807\,\frac{m}{s^{2}}[/tex]
[tex]R = 0.312\,m[/tex]
A student is studying simple harmonic motion of a spring. She conducts an experiment where she measures the amplitude and period of an undamped system to be 23 ± 2 mm and 0.40 ± 0.020 seconds, respectively. Using the equation for displacement as a function of time y(t) = Acos(ωt), what is the uncertainty of her displacement calculation in mm for t = 0.050 ± 0.0010 seconds?
Answer:
5.9*10^{-4}m
Explanation:
to find the uncertainty of the displacement it is necessary to compute the uncertainty for the angular frequency:
[tex]\omega=\frac{2\pi}{T}=\frac{2\pi}{0.40s}=15.707rad/s\\\\\frac{d \omega}{\omega}=\frac{dT}{T}\\\\d\omega=\omega \frac{dT}{T}=(15.707rad/s)\frac{0.020s}{0.40s}=0.785rad/s[/tex]
then, you can calculate the uncertainty in angular displacement:
[tex]\theta=\omega t\\\\\frac{d\theta}{\theta}=\sqrt{(\frac{d\omega}{\omega})^2+(\frac{dt}{t})^2}\\\\d\theta=\theta\sqrt{(\frac{d\omega}{\omega})^2+(\frac{dt}{t})^2}=0.0422[/tex]
finally, by using:
[tex]y=Acos(\omega t)\\\\dy=dAcos(\omega t)d(\omega t)=(dA)cos(\theta)d\theta=(0.002m)cos(0.785)(0.0422)\\\\dy=5.9*10^{-4}m[/tex]
The uncertainty of her displacement in mm is : 0.59 mm
Determine the uncertainty of her displacementFirst step : determine the uncertaintiy of the angular frequency
w = [tex]\frac{2\pi }{T}[/tex] = [tex]\frac{2\pi }{0.40} = 15.707 rad/s[/tex]
[tex]\frac{dw}{w} = \frac{dT}{T}[/tex]
therefore :
dw = 0.785 rad/s
Next step : determine the uncertainty of the angular displacement
θ = wt
dθ / θ = [tex]\sqrt{(\frac{dw}{w} )^2 + (\frac{dt}{t} )^2}[/tex]
therefore :
dθ = 0.0422
Final step : determine the uncertainty of displacement
y = Acos(wt)
dy = dAcos(wt)d(t) = (dA)cosθdθ
= ( 0.002m )cos (0.785)(0.0422)
= 5.9 * 10⁻⁴ m = 0.59 mm
Hence we can conclude that the uncertainty of her displacement in mm is : 0.59 mm
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The 1.18-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in the horizontal position θ = 0 where the spring is unstretched. If the bar is observed to momentarily stop in the position θ = 46°, determine the spring constant k. For your computed value of k, what is magnitude of the angular velocity of the bar when θ = 30°.
Answer:
k = 11,564 N / m, w = 6.06 rad / s
Explanation:
In this exercise we have a horizontal bar and a vertical spring not stretched, the bar is released, which due to the force of gravity begins to descend, in the position of Tea = 46º it is in equilibrium;
let's apply the equilibrium condition at this point
Axis y
W_{y} - Fr = 0
Fr = k y
let's use trigonometry for the weight, we assume that the angle is measured with respect to the horizontal
sin 46 = [tex]W_{y}[/tex] / W
W_{y} = W sin 46
we substitute
mg sin 46 = k y
k = mg / y sin 46
If the length of the bar is L
sin 46 = y / L
y = L sin46
we substitute
k = mg / L sin 46 sin 46
k = mg / L
for an explicit calculation the length of the bar must be known, for example L = 1 m
k = 1.18 9.8 / 1
k = 11,564 N / m
With this value we look for the angular velocity for the point tea = 30º
let's use the conservation of mechanical energy
starting point, higher
Em₀ = U = mgy
end point. Point at 30º
[tex]Em_{f}[/tex] = K -Ke = ½ I w² - ½ k y²
em₀ = Em_{f}
mgy = ½ I w² - ½ k y²
w = √ (mgy + ½ ky²) 2 / I
the height by 30º
sin 30 = y / L
y = L sin 30
y = 0.5 m
the moment of inertia of a bar that rotates at one end is
I = ⅓ mL 2
I = ½ 1.18 12
I = 0.3933 kg m²
let's calculate
w = Ra (1.18 9.8 0.5 + ½ 11,564 0.5 2) 2 / 0.3933)
w = 6.06 rad / s
A ray of light is incident on a flat surface of a block of polystyrene, with an index of refraction of 1.49, that is submerged in water. The ray is split at the surface, where the angle of refraction of the transmitted ray is 19.1°. What is the angle of reflection (in degrees) of the reflected ray?
Answer:
21.5°
Explanation:
Given,
Refractive index of water, n₁ = 1.33
Refractive index of polystyrene, n₂ = 1.49
Angle of reflection = ?
Angle of refraction = 19.1°
Using Snell's law
n₁ sin θ₁ = n₂ sin θ₂
1.33 x sin θ₁ = 1.49 x sin 19.1°
sin θ₁ = 0.366
θ₁ = 21.5°
According to law of reflection angle of incidence is equal to angle of reflection.
Angle of reflection = 21.5°
Final answer:
The angle of reflection of the reflected ray in polystyrene that is submerged in water will be equal to the angle of refraction, so it is 19.1°.
Explanation:
The question is about the reflection and refraction of light as it strikes a flat surface of a polystyrene block that is in water. According to the Law of Reflection, the angle of reflection is equal to the angle of incidence.
Since we know the angle of refraction is 19.1°, and by Snell's Law (n₁ × sin(i) = n₂ × sin(r)), we can infer that the angle of incidence is also 19.1° because the boundary is between two different mediums (water and polystyrene), resulting in the light being split into reflected and refracted rays.
Therefore, the angle of reflection of the reflected ray is also 19.1°, identical to the angle of refraction given in the question, because the angle of incidence equals the angle of reflection.
A loud factory machine produces sound having a displacement amplitude in air of 1.00 μmμm, but the frequency of this sound can be adjusted. In order to prevent ear damage to the workers, the maximum pressure amplitude of the sound waves is limited to 10.0 PaPa. Under the conditions of this factory, the bulk modulus of air is 1.30×105 PaPa . The speed of sound in air is 344 m/s
What is the highest-frequency sound to which this machine can be adjusted without exceeding the prescribed limit?
Answer:
f =3.4*10^3 hz
Since f is in the range of (20 Hz - 20,000 Hz] which is the range of audible frequencies, the frequency is audible.
Explanation:
The relation that describes the pressure amplitude for a sound wave is
P_MAX = B*k*A (1)
Where the bulk modulus of the air is B = 1.30 x 10^5 Pa and the displacement amplitude of the waves produced by the machine is 1.00 μmμm.
Using (1) we can calculate k then we can use k to determine the wavelength A of the wave, and remember that λ = 2π/k.
So, substitute into (1) with 10 Pa for P_max, (1.30 x 10^5 Pa) for B and
1 x 10^-6 m for A
10 Pa = (1.30 x 10^5 Pa) x k x (1 x 10^-6 m)
k = 62.5 m^-1
We can use the following relation to calculate the wavelength
λ = 2π/k.
λ = 0.100 m
Finally, the relation between the wavelength and the frequency of a sound
wave is given by the following equation
f = v/ λ
=344/0.100 m
f =3.4*10^3 hz
Since f is in the range of (20 Hz - 20,000 Hz] which is the range of audible frequencies, the frequency is audible.
Which way do your feet need to push on the floor in order to
jump forward?
Answer: In order to walk on a floor (or any other surface), your foot must push backward on the floor (action force), so that the floor pushes you forward (reaction force).
Explanation:
In this experiment we will observe the magnetic fields produced by a current carrying wire. A long wire is suspended vertically, passing through a horizontal platform. The wire is connected to a power supply, allowing a current of 5.0 Amps to flow. A compass is placed on the platform stand near the vertical wire. When no current is present in the wire, the compass needle points in the same direction any where around the wire due to the Earth’s magnetic field. When a current flows through the wire, the compass needle deflects in a direction tangent to a circle, which is the direction of the magnetic field created by the current carrying wire.
Answer:
See explanation
Explanation:
Solution:-
Electric current produces a magnetic field. This magnetic field can be visualized as a pattern of circular field lines surrounding a wire. One way to explore the direction of a magnetic field is with a compass, as shown by a long straight current-carrying wire in. Hall probes can determine the magnitude of the field. Another version of the right hand rule emerges from this exploration and is valid for any current segment—point the thumb in the direction of the current, and the fingers curl in the direction of the magnetic field loops created by it.
Compasses placed near a long straight current-carrying wire indicate that field lines form circular loops centered on the wire. Right hand rule 2 states that, if the right hand thumb points in the direction of the current, the fingers curl in the direction of the field. This rule is consistent with the field mapped for the long straight wire and is valid for any current segment.
( See attachments )
- The equation for the magnetic field strength - B - (magnitude) produced by a long straight current-carrying wire is given by the Biot Savart Law:
[tex]B = \frac{uo*I}{2\pi *r}[/tex]
Where,
I : The current,
r : The shortest distance to the wire,
uo : The permeability of free space. = 4π * 10^-7 T. m/A
- Since the wire is very long, the magnitude of the field depends only on distance from the wire r, not on position along the wire. This is one of the simplest cases to calculate the magnetic field strength - B - from a current.
- The magnetic field of a long straight wire has more implications than one might first suspect. Each segment of current produces a magnetic field like that of a long straight wire, and the total field of any shape current is the vector sum of the fields due to each segment. The formal statement of the direction and magnitude of the field due to each segment is called the Biot-Savart law. Integral calculus is needed to sum the field for an arbitrary shape current. The Biot-Savart law is written in its complete form as:
[tex]B = \frac{uo*I}{4\pi }*\int\frac{dl xr}{r^2}[/tex]
Where the integral sums over,
1) The wire length where vector dl = direction of current (in or out of plane)
2) r is the distance between the location of dl and the location at which the magnetic field is being calculated
3) r^ is a unit vector in the direction of r.
Answer:
B
Explanation:
If it is fixed at C and subjected to the horizontal 60-lblb force acting on the handle of the pipe wrench at its end, determine the principal stresses in the pipe at point A, which is located on the outer surface of the pipe. Express your answers, separated by a comma, to three significant figures.
The principal stress in the pipe at point A, which is located on the outer surface of the pipe is 132.773 psi
Finding the principal stress:It is given that l = 12 in
The force F = 60 lb
The radiuses are as follows:
outer radius, R = 3.90 in/2
R = 1.95 in
Inner radius, r = 3.65 in/2
r = 1.825 in
The angle between the force applied and the distanced from the axis is 30°
So we get the torque:
T = l×Fsin 30°
T = 12 × 60 × (0.5)
T = 360 lb-in
Now, the angle of rotation or angular displacement ωt is given as
ωt = π(R⁴ - r⁴)/(2R)
ωt = π((1.95 in)⁴ - (1.825 in)⁴)/(2×1.95 in)
ωt = 2.7114 in³
Then the principal stress in the pipe at point A is:
principal stress = T/ωt
principal stress = (360 lb-in)/(2.7114 in³)
principal stress = 132.773 lb/in²
principal stress = 132.773 psi
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Start with light that non polarized of intensity I (initial) It will travel through one polarizer that is set to 90 degrees to the x axis. It will then travel through 6 additional polarizers that decrease in angle size by 15 degrees. Polarizer 1 at 75 degrees Polarizer 2 at 60 degrees Polarizer 3 at 45 degrees Polarizer 4 at 30 degrees Polarizer 5 at 15 degrees Polarizer 6 at 0 degrees. Find the final intensity of light that travels through all polarizers as a percentage of the initial intensity
Answer:
35.35 is the final intensity
Explanation:
See attached file for calculation
Two spherical shells have their mass uniformly distrubuted over the spherical surface. One of the shells has a diameter of 2 meters and a mass of 1 kilogram. The other shell has a diameter of 1 meter. What must the mass mmm of the 1-meter shell be for both shells to have the same moment of inertia about their centers of mass
Answer: 4 kg
Explanation:
Given
Mass of the first shell, m1 = 1 kg
Diameter of the first shell, d1 = 2 m
Radius of the first shell, r1 = 1 m
Diameter of the second shell, d2 = 1 m
Radius of the second shell, r2 = 1/2 m
The moment of inertia of a spherical shell is given by the relation
I = mr²
This means that if two sphere's have the same moment of ineria:
I1 would be equal to I2. And thus
m1.r1² = m2.r2²
If we solve for the second mass m2
m2 = m1.r1²/r2²
m2 = m1 (r1 / r2)² and we substitute the values
m2 = 1 * (1 / 0.5)²
m2 = 2²
m2 = 4 kg
The needed mass of the second shell for their shells to have the same moment of inertia is 4 kg
What is the speed of a garbage truck that is 1.60×104 kg and is initially moving at 28.0 m/s just after it hits and adheres to a trash can that is 86.0 kg and is initially at rest
Answer:
Assuming the mass of the garbage truck is [tex]1.6*10^{4}[/tex] kg, the speed of the garbage truck is [tex]approx. = 27.8503\frac{m}{s}[/tex].
Explanation:
This is conservation of momentum where both objects stick at the end so they have the same final velocity so our equation is:
[tex]m_{1}v_{1initial} + m_{2}v_{2initial} = v_{both final}(m_{1} + m_{2})[/tex]
To solve for the final velocity, just divide by the sum of both masses:
[tex]\frac{ m_{1}v_{1initial} + m_{2}v_{2initial}}{(m_{1} + m_{2})} = v_{both final}[/tex]
So, plug in the known values (remember initial velocity for the trash can is 0):
[tex]\frac{ 1.6*10^{4}*28 + 86*0}{1.6+10^{4} + 86} = v_{both final}[/tex]
A 95.0 kg satellite moves on a circular orbit around the Earth at the altitude h=1.20*103 km. Find: a) the gravitational force exerted by the Earth on the satellite? b) the centripetal acceleration of the satellite? c) the speed of the satellite? d) the period of the satellite’s rotation around the Earth?
Answer:
a) [tex]F = 660.576\,N[/tex], b) [tex]a_{c} = 6.953\,\frac{m}{s^{2}}[/tex], c) [tex]v \approx 7255.423\,\frac{m}{s}[/tex], [tex]\omega = 9.583\times 10^{-4}\,\frac{rad}{s}[/tex], d) [tex]T \approx 1.821\,h[/tex]
Explanation:
a) The gravitational force exerted by the Earth on the satellite is:
[tex]F = G\cdot \frac{m\cdot M}{r^{2}}[/tex]
[tex]F = \left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot \frac{(95\,kg)\cdot (5.972\times 10^{24}\,kg)}{(7.571\times 10^{6}\,m)^{2}}[/tex]
[tex]F = 660.576\,N[/tex]
b) The centripetal acceleration of the satellite is:
[tex]a_{c} = \frac{660.576\,N}{95\,kg}[/tex]
[tex]a_{c} = 6.953\,\frac{m}{s^{2}}[/tex]
c) The speed of the satellite is:
[tex]v = \sqrt{a_{c}\cdot R}[/tex]
[tex]v = \sqrt{\left(6.953\,\frac{m}{s^{2}} \right)\cdot (7.571\times 10^{6}\,m)}[/tex]
[tex]v \approx 7255.423\,\frac{m}{s}[/tex]
Likewise, the angular speed is:
[tex]\omega = \frac{7255.423\,\frac{m}{s} }{7.571\times 10^{6}\,m}[/tex]
[tex]\omega = 9.583\times 10^{-4}\,\frac{rad}{s}[/tex]
d) The period of the satellite's rotation around the Earth is:
[tex]T = \frac{2\pi}{\left(9.583\times 10^{-4}\,\frac{rad}{s} \right)} \cdot \left(\frac{1\,hour}{3600\,s} \right)[/tex]
[tex]T \approx 1.821\,h[/tex]
The combination of a uniform flow and a source can be used to describe flow around a streamlined body called a half-body (see the Video). Assume that a certain body has the shape of a half-body with a thickness of 0.16 m. If this body is placed in an air stream moving at 20.6 m/s, what source strength is required to simulate flow around the body
Answer:
m = 41.39 m/s
Explanation:
For the source alone , we can write next equation.
V_r = m/2πr
Stagnation point will occur where r=b, thus;
V = m/2πb
Now we can find source strength, and therefore we have;
m = 2*π*b*V
b = 0.16/2*π
b = 0.0254
m = 2*π*0.32*20.6
m = 41.39 m/s
Final answer:
To simulate flow around a half-body with a thickness of 0.16 m in a 20.6 m/s air stream, the required source strength is 6.592 m^3/s, calculated using the formula Q = 2 * Vθ * d. = 6.592 m3/s
Explanation:
Calculation of Source Strength for Flow Simulation
To simulate flow around a streamlined half-body in a uniform air stream, a combination of a uniform flow and a source flow is used. In the given scenario, the thickness of the half-body is 0.16 m, and it is placed in an air stream moving at 20.6 m/s. The strength of the source, often represented by the symbol Q (and measured in m3/s), is the volume of fluid introduced into the flow per unit of time. For a two-dimensional flow over a half-body, the source strength required to simulate the flow pattern is proportional to the product of the freestream velocity and the thickness of the half-body.
To calculate the required source strength mathematically, you would use the formula:
Q = 2 * V∞ * d
Where V∞ is the freestream velocity, and d is the maximum thickness of the half-body. Substituting the given values:
Q = 2 * 20.6 m/s * 0.16 m
Q = 6.592 m^3/s
Thus, the source strength required to simulate flow around the half-body in uniform flow would be 6.592 m3/s. It is essential to ensure that the flow remains laminar and non-turbulent to maintain the accuracy of this simulation. Factors such as the Reynolds number would also be considered in a more complex analysis to determine the flow regime.
An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the earth with a speed of 0.99543c relative to the earth. A scientist at rest on the earth's surface measures that the particle is created at an altitude of 43.0 km .
(a) As measured by the scientist, how much time does it take the particle to travel the 45.0 km to the surface of the earth?
(b) Use the length-contraction formula to calculate the distance from where the particle is created to the surface of the earth as measured in the particle’s frame.
(c) In the particle’s frame, how much time does it take the particle to travel from where it is created to the surface of the earth? Calculate this time both by the time dilation formula and from the distance calculated in part (b). Do the two results agree?
Answer:
Check attachment
The question have two distance
I decided to use the one in the question "a" in attachment and I will use the other one here
Explanation:
Given that,
Speed of particle relative to the earth is
V = 0.99543c
Where c is speed of light
c = 3 × 10^8 m/s
Particle height as detected by scientist is 43km
The initial length is 43km
Lo = 43km
Lo = 43,000m
A. Time taken for the particle to reach the earth surface?
Speed = distance / Time
Time = distance / speed
t = L / V
t = 43,000 / 0.99543c
t = 43,000 / (0.99543 × 3 × 10^8)
t = 1.4399 × 10^-4 seconds
b. Initial Lenght is given as Lo = 43km
Using length contraction formula
L = Lo√(1 — u² / c²)
L = 43√[1 — (0.99543c)² / c²]
L = 43√[1 — 0.990881c² / c²]
L = 43√[1 — 0.990881]
L = 43 × √(9.1191 × 10^-3)
L = 43 × 0.095494
L = 4.1062 km
c. Using time dilation formula
∆to = ∆t√(1 — u² / c²)
∆t is gotten from question a
∆t = 1.4399 × 10^-4 seconds
∆to = ∆t√[1 — (0.99543c)² / c²]
∆to = ∆t√[1 — 0.990881c² / c²]
∆to = ∆t√[1 — 0.990881]
∆to = ∆t × √(9.1191 × 10^-3)
∆to = 1.4399 × 10^-4 × 0.095494
∆to = 1.375 × 10^-5 seconds
To check if the time dilation agree
t = L / V
t = 4.1062 × 1000 / 0.99543c
t = 4.1062 × 1000 / 0.99543 × 3 × 10^8
t = 1.375 × 10^-5 seconds
The time dilation agreed
A hydrogen atom is in state N = 4, where N = 1 is the lowest energy state. What is K+U in electron volts for this atomic hydrogen energy state? E4 = eV The hydrogen atom makes a transition to state N = 2. What is K+U in electron volts for this lower atomic hydrogen energy state? E2 = eV What is the energy in electron volts of the photon emitted in the transition from level N = 4 to N = 2? Ephoton = eV
Answer:
E₄ = - 0.85 eV
E₂ = - 3.4 eV
Ephoton = 2.55 eV
Explanation:
The sum of Kinetic Energy (K) and Potential Energy (U) of the Helium atom is equal to the total energy of Helium atom in the specified state N. From Bohr's atomic model, the energy of a hydrogen atom in state N is given as:
En = K + U = (-1/n²)(13.6 eV)
a)
Here,
n = 4
Therefore,
E₄ = (-1/4²)(13.6 eV)
E₄ = - 0.85 eV
b)
Here,
n = 2
Therefore,
E₂ = (-1/2²)(13.6 eV)
E₂ = - 3.4 eV
c)
The energy of photon emitted in the transition from level 4 to level 2 will be equal to the difference in the energy of both levels:
Ephoton = ΔE = E₄ - E₂
Ephoton = - 0.85 eV - (- 3.4 eV)
Ephoton = 2.55 eV
The electrons energy will be:
[tex]E_4 = -0.85 \ eV[/tex][tex]E_2 = -3.4 \ eV[/tex][tex]E_{photon} = 2.55 \ eV[/tex]As we know. the formula:
→ [tex]E_n = K+U[/tex]
[tex]= (-\frac{1}{n^2} ) (13.6 \ eV)[/tex]
(a)
Given:
n = 4then,
→ [tex]E_4 = (-\frac{1}{4^2} )(13.6 \ eV)[/tex]
[tex]= -0.85 \ eV[/tex]
(b)
Given:
n = 2then,
→ [tex]E_2 = (-\frac{1}{2^2} )(13.6 \ eV)[/tex]
[tex]= -3.4 \ eV[/tex]
(c)
The energy of photon emitted will be:
→ [tex]E_{photon} = \Delta E[/tex]
[tex]= E_4 -E_2[/tex]
[tex]=-0.85 \ eV-(-3.4 \ eV)[/tex]
[tex]= 2.55 \ eV[/tex]
Thus the above answers are appropriate.
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Problem: A disk of mass m and radius r is initially held at rest just above a larger disk of mass M and radius R that is rotating at angular speed wi. What is the final angular speed of the disks after the top one is dropped onto the bottom one and they stop slipping on each other? Note: the moment of inertia of a disk of mass M and radius R about an axis through its center and perpendicular to the disk is I = (1/2)MR2.
Answer:
The final angular velocity is [tex]w_f = \frac{MR^2}{MR^2+ mr^2} w[/tex]
Explanation:
From the question we are told that
The mass of the first disk is m
The radius of the first disk is r
The mass of second disk is M
The radius of second disk is R
The speed of rotation is w
The moment of inertia of second disk is [tex]I = \frac{1}{2} MR^2[/tex]
Since the first disk is at rest initially
The initial angular momentum would be due to the second disk and this is mathematically represented as
[tex]L_i = Iw = \frac{1}{2} MR^2 w[/tex]
Now when the first disk is then dropped the angular momentum of the whole system now becomes
[tex]L_f = (I_1 + I_2 ) w_f= ( \frac{1}{2} MR^2 + \frac{1}{2} m^2 r^2) w_f[/tex]
This above is because the formula for moment of inertia is the same for every disk
According to the law conservation of angular momentum
[tex]L_f = L_i[/tex]
[tex]( \frac{1}{2} MR^2 + \frac{1}{2} m^2 r^2) w_f = \frac{1}{2} MR^2 w[/tex]
=> [tex]w_f = \frac{\frac{1}{2} MR^2 w }{( \frac{1}{2} MR^2 + \frac{1}{2} m^2 r^2)}[/tex]
[tex]w_f = \frac{MR^2}{MR^2+ mr^2} w[/tex]
What is an Amplitude
Amplitude is a measurement of the magnitude of displacement (or maximum disturbance) of a medium from its resting state, as diagramed in the peak deviation example below (it can also be a measurement of an electrical signal's increased or decreased strength above or below a nominal state).
Suppose you had a collection of a large number of hypothetical quantum objects, each of whose individual energy levels were -4.6 eV, -3.0 eV, -2.1 eV, and -1.6 eV. If nearly all of these identical objects were in the ground state, what would be the energies of dark spectral lines in an absorption spectrum if visible white light (1.8 to 3.1 eV) passes through the material? Enter the energies in order of increasing energy, followed by entering 0 in any later boxes for which there is no dark line within the visible spectrum. (That is, if your answers were 1, 2, and 3 eV, you would enter 1 in the first box, 2 in the second box, and 3 in the third box. If your answers were 1 and 2 eV, you would enter 1 in the first box, 2 in the second box, and 0 in the third box. If your answer is just 1 eV, you would enter 1 in the first box, 0 in the second box, and 0 in the third box.) Smallest energy of a dark line: eV Next larger energy of a dark line (or 0): eV Next larger energy of a dark line (or 0): eV Additional Materials
The energy differences between the levels of the quantum objects and the photons in the visible spectrum determine the dark spectral lines in an absorption spectrum.
Explanation:In order to determine the energies of the dark spectral lines in an absorption spectrum, we need to consider the energy differences between the energy levels of the quantum objects and the energy levels of the photons in the visible spectrum. The energy levels of the quantum objects are -4.6 eV, -3.0 eV, -2.1 eV, and -1.6 eV. The range of photon energies for visible light is 1.63 to 3.26 eV. Only photons with energies matching the energy differences between the levels of the quantum objects will be absorbed, resulting in dark lines in the spectrum.
Based on the given information, the dark spectral lines within the visible spectrum would be:
Smallest energy of a dark line: -4.6 eVNext larger energy of a dark line (or 0): -3.0 eVNext larger energy of a dark line (or 0): -2.1 eVLearn more about Energy differences and absorption spectrum here:https://brainly.com/question/31230618
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A 1.50-m string of weight 0.0125 N is tied to the ceil- ing at its upper end, and the lower end supports a weight W. Ignore the very small variation in tension along the length of the string that is produced by the weight of the string. When you pluck the string slightly, the waves traveling up the string obey the equation. Assume that the tension of the string is constant and equal to W.
(a) How much time does it take a pulse to travel the full length of the string?
(b) What is the weight W?
(c) How many wavelengths are on the string at any instant of time?
(d) What is the equation for waves traveling ?down? the string?
The wave equation is missing and it is y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)
Answer:
A) 0.0534 seconds
B) 0.67N
C) 41
D) (8.50 mm)cos(172 rad/m x + 4830 rad/s t)
Explanation:
we are given weight of string = 0.0125N
Thus, since weight = mg
Then, mass of string = 0.0125/9.8
Mass of string = 1.275 x 10⁻³ kg
Length of string; L= 1.5 m .
mass per unit length; μ = (1.275 x 10⁻³)/1.5
μ = 0.85 x 10⁻³ kg/m
We are given the wave equation: y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)
Now if we compare it to the general equation of motion of standing wave on a string which is:
y(x,t) = Acos(Kx − ω t)
We can deduce that
angular velocity;ω = 4830 rad/s
Wave number;k = 172 rad/m
A) Velocity is given by the formula;
V = ω/k
Thus, V = 4830/172 m/s
V = 28.08 m /s
Thus time taken to go up the string = 1.5/28.08 = 0.0534 seconds
B) We know that in strings,
V² = F/μ
Where μ is mass per unit length and V is velocity.
Thus, F = V²*μ =28.08² x 0.85 x 10⁻³
F = 0.67N
C) Formula for wave length is given as; wave length;λ = 2π /k
λ = 2 x π/ 172
λ = 0.0365 m
Thus, number of wave lengths over whole length of string
= 1.5/0.0365 = 41
D) The equation for waves traveling down the string
= (8.50 mm)cos(172 rad/m x + 4830 rad/s t)
The required time is: 0.395 s. Weight and wavelengths are 0.013 and 1.5. The equation for waves traveling down the string is: y(x, t) = (8.50 mm)cos(172 rad/m)x+4830 rad/s)t).
a) To find the time it takes for a pulse to travel the full length of the string, we need to determine the velocity of the wave. The velocity of a wave on a string is given by the equation [tex]v = \sqrt{(T/\mu)[/tex], where T is the tension and μ is the linear mass density of the string.
Given that the weight is 0.0125 N and the length is 1.50 m.
Here, mass = w/g = 0.0125/9.8 = 1.275 x 10⁻³ kg
we can calculate μ as follows: μ = m/L = (1.275 x 10⁻³)/(1.50 m) = 0.00085 kg/m.
Substituting the values into the equation:
we get v = [tex]\sqrt{W/0.00085}[/tex]. As it is given that the tension of the string is constant and equal to W.
Thus, v = [tex]\sqrt{14.7}[/tex] = 3.8 m/s
The time it takes for a pulse to travel the full length of the string can be calculated using the formula time = distance/velocity = 1.50/3.8. So, time = 0.395 s.
b) In case of string, V² = F/μ
μ x V² = F
This implies:
F = 0.00085 x [tex](3.8)^2[/tex] = 0.0123
C) We know, f (frequency) = 1/t = 1/0.395 = 2.53
Also, v = λf. This implies:
λ = v/f = 3.8/2.53 = 1.5
D) Finally, to find the equation for waves traveling down the string, we need to adjust the sign inside the cosine function to account for the wave traveling in the opposite direction. It is given that:
y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)
So, the equation becomes y(x, t) = (8.50 mm)cos(172 rad/m)x+4830 rad/s)t).
Complete Question:
Question
A 1.50-m string of weight 0.0125 N is tied to the ceil-ing at its upper end, and the lower end supports a weight W. Ignore the very small variation in tension along the length of the string that is produced by the weight of the string. When you pluck the string slightly, the waves traveling up the string obey the equation. Assume that the tension of the string is constant and equal to W.
y(x, t) = (8.50 mm)cos(172 rad/m)x+4830 rad/s)t).
(a) How much time does it take a pulse to travel the full length of the string?
(b) What is the weight W?
(c) How many wavelengths are on the string at any instant of time?
(d) What is the equation for waves traveling ?down? the string?
In a physics lab, light with a wavelength of 540 nm travels in air from a laser to a photocell in a time of 16.5 ns. When a slab of glass with a thickness of 0.820 m is placed in the light beam, with the beam incident along the normal to the parallel faces of the slab, it takes the light a time of 21.5 ns to travel from the laser to the photocell.
What is the wavelength of the light in the glass? Use 3.00×10^8 m/s for the speed of light in a vacuum.
Answer:
189.47nm
Explanation:
We can solve this problem by taking into account the time that light takes in crossing the distance between the laser and the photocell, and the time in crossing the slab.
By using the values of c and 16.5ns we can calculate the value of d
[tex]d=(3*10^{8}m/s)(16.5*10^{-9}s)=4.95m[/tex]
We have to compute the time that light takes in crossing d-0.820m:
[tex]4.95m-0.820m=4.13m\\\\t=\frac{4.13m}{3*10^8m/s}=6.22*10^{-8}s=13.7ns[/tex]
Now, we can calculate the speed of the light in the slab by using the time difference between 21.5 ns and 13.7ns:
[tex]\Delta t=21.5ns-13.7ns=7.8ns\\\\v_l=\frac{0.82m}{7.8ns}=1.05*10^8m/s[/tex]
Then, the index of refraction will be:
[tex]n=\frac{c}{v_l}=2.85[/tex]
Finally, we have that:
[tex]\frac{\lambda_2}{\lambda_1}=\frac{n_1}{n_2}\\\\\lambda_2=\lambda_1 \frac{n_1}{n_2}=(540nm)\frac{1.00}{2.85}=189.47nm[/tex]
hope this helps!!
A battery with emf 10.30 V and internal resistance 0.50 Ω is inserted in the circuit at d, with its negative terminal connected to the negative terminal of the 8.00-V battery. What is the difference of potential Vbc between the terminals of the 4.00-V battery now?
Answer:
3.78V
Explanation:
The resistance in the circuit is
The potential difference across the 4.00 V battery, Vbc, in a circuit where it is in parallel with a series combination of 10.30 V and 8.00 V batteries, is equal to the total emf of the two batteries in series minus the potential drop due to their current through the 0.50 Ω internal resistance.
Explanation:To determine the potential difference Vbc across the terminals of the 4.00 V battery, we first need to understand the circuit's behavior. The circuit now consists of two batteries connected in series - one with 10.30 V (internal resistance 0.50 Ω) and another of 8.00 V - and the 4.00 V battery is in parallel with this series combination.
Because the batteries in series share the same current, we can calculate the total current (I) using Ohm's law and considering the total emf and resistance:
I= (10.30 V + 8.00 V)/( R + 0.50 Ω)
Note that R includes any circuit resistance (which hasn't been mentioned) and the 4.00 V battery's own internal resistance. Therefore, Vbc, the potential difference across the 4.00 V battery will be equal to the total potential provided by the series batteries minus the potential drop due to their current through the 0.50 Ω internal resistance.
Vbc = (10.30 V + 8.00 V) - I * 0.50 Ω
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Consider two straight wires lying on the x-axis, separated by a gap of 4 nm. The potential energy in the gap is about 3 eV higher than the energy of a ondu tion ele tron in either wire. What is the probability that a ondu tion ele tron in one wire arriving at the gap will pass through the gap into the other wire?
Answer: 1.3 ×10^-31
Explanation:
the required probability is P = e^(-2αL)
Firstly, evaluate (-2αL)
α= 1/hc √2mc^2 (U - E)
h= modified planck's constant
where,
(-2αL)
= -(2L)/(h/2π ) ×√2mc^2 (U - E)
= -(2L) / (hc^2/π )×√2mc^2 (U - E)
(hc^2/2pi) = 197*eV.nm (standard constant)
2*L = 8 nm
mc^2 = 0.511×10^6 eV
Where m = mass electron
C= speed of light
(-2αL) = [-8nm/(197 eV.nm)] × (1.022× 10^6 eV*×3 eV)^0.5
(-2αL) = -71.1
Probability = e^(-2αL) = e^-71.1 = 1.3 ×10^-31
Therefore, the probability that a ondu tion ele tron in one wire arriving at the gap will pass through the gap into the other wire 1.3 ×10^-31.
In your research lab, a very thin, flat piece of glass with refractive index 1.30 and uniform thickness covers the opening of a chamber that holds a gas sample. The refractive indexes of the gases on either side of the glass are very close to unity. To determine the thickness of the glass, you shine coherent light of wavelength λ0 in vacuum at normal incidence onto the surface of the glass. Whenλ0= 496 nm, constructive interference occurs for light that is reflected at the two surfaces of the glass. You find that the next shorter wavelength in vacuum for which there is constructive interference is 386 nm.
1) Use these measurements to calculate the thickness of the glass.
2) What is the longest wavelength in vacuum for which there is constructive interference for the reflected light?
Answer:
A) The thickness of the glass is 868 nm
B) The wavelength is 3472 nm
Explanation:
We are given;
Refractive index = 1.30
Wavelength = 496 nm
Next wavelength = 386 nm
Now, we need to calculate the thickness of the glass
Using formula for constructive interference which is given as;
2nt = (m + ½)λ
Where;
m is the order of the interference
λ is wavelength
t is thickness
Now, for the first wavelength, we have;
2nt = (m + ½)496 - - - - eq1
for the second wavelength, we have;
2nt = (m + 1 + ½)386
2nt = (m + 3/2)386 - - - - eq2
Thus, combining eq1 and eq2, we have;
(m + ½)496 = (m + 3/2)386
496m + 248 = 386m + 579
496m - 386m = 579 - 248
110m = 331
m = 331/110
m = 3
Put 3 for m in eq 1;
2nt = (3 + ½)496
2nt = 1736
t = 1736/(2 x 1)
t = 868 nm
B) now we need to calculate the longest wavelength.
From earlier, we saw that ;
2nt = (m + ½)λ
Making wavelength λ the subject, we have;
λ = 2nt/(m + ½)
The longest wavelength will be at m = 0
Thus,
λ = (2 x 1 x 868)/(0 + ½)
λ = 3472 nm
Two identical wheels, wheel 1 and wheel 2, initially at rest begin to rotate with constant angular accelerations α. After rotating through the same angular displacement, Δθ0, the angular velocity of wheel 1 is ω1 and the angular velocity of wheel 2 is ω2=3ω1. How does the angular acceleration of wheel 2, α2, compare to the angular acceleration of wheel 1, α1?a. a2 = a1b. a2 = a1/3c. a2 = 3a1d. a2 = 9a1
Answer:
d. a2 = 9a1
Explanation:
We can apply the following equation of motion to calculate the angular acceleration:
[tex]\omega^2 - \omega_0^2 = 2\alpha\theta[/tex]
Since both wheel starts from rest, their [tex]\omega_0 = 0 rad/s[/tex]
[tex]\omega^2 = 2\alpha\theta[/tex]
We can take the equation for the 1st wheel, divided by the equation by the 2nd wheel:
[tex]\frac{\omega_1^2}{\omega_2^2} = \frac{2\alpha_1\theta_1}{2\alpha_2\theta_2}[/tex]
As they were rotating through the same angular displacement [tex]\theta_1 = \theta_2[/tex], these 2 cancel out
[tex]\left(\frac{\omega_1}{\omega_2}\right)^2 = \frac{\alpha_1}{\alpha_2}[/tex]
[tex]\left(\frac{1}{3}\right)^2 = \frac{\alpha_1}{\alpha_2}[/tex]
[tex]\frac{1}{9} = \frac{\alpha_1}{\alpha_2}[/tex]
[tex]\alpha_2 = 9\alpha_1[/tex]
So d is the correct answer
The angular acceleration of the first wheel is four times higher than that of the second. Option D is correct.
What is angular acceleration?It can be defined as the rate of change in the angular velocity of an object or body. It can be calculated by the equation of motion:
[tex]\omega ^2 - \omega _0^2 = 2\alpha \theta[/tex]
Since initial angular rotation is zero for both the wheels,
[tex]\omega ^2 = 2\alpha \theta[/tex]
Compare the angular acceleration of both wheels,
[tex]\dfrac {\omega_1^2}{\omega_2^2} = \dfrac {2\alpha_1 \theta}{2\alpha_2\theta }[/tex]
Put the values,
[tex]\begin{aligned} (\dfrac 13)^2&= \dfrac {\alpha_1 }{\alpha_2 }\\\\ \dfrac 19 &= \dfrac {\alpha_1 }{\alpha_2 }\\\\\alpha_2 &= 9\alpha_1 \end {aligned}[/tex]
Therefore, The angular acceleration of the first wheel is four times higher than that of the second. Option D is correct.
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Summarize ocean acidification in one sentence.
Answer:
The ocean absorbs a significant portion of carbon dioxide (CO2) emissions from human activities, equivalent to about one-third of the total emissions for the past 200 years from fossil fuel combustion, cement production and land-use change (Sabine et al., 2004). Uptake of CO2 by the ocean benefits society by moderating the rate of climate change but also causes unprecedented changes to ocean chemistry, decreasing the pH of the water and leading to a suite of chemical changes collectively known as ocean acidification. Like climate change, ocean acidification is a growing global problem that will intensify with continued CO2 emissions and has the potential to change marine ecosystems and affect benefits to society.
The average pH of ocean surface waters has decreased by about 0.1 unit—from about 8.2 to 8.1—since the beginning of the industrial revolution, with model projections showing an additional 0.2-0.3 drop by the end of the century, even under optimistic scenarios (Caldeira and Wickett, 2005).1 Perhaps more important is that the rate of this change exceeds any known change in ocean chemistry for at least 800,000 years (Ridgewell and Zeebe, 2005). The major changes in ocean chemistry caused by increasing atmospheric CO2 are well understood and can be precisely calculated, despite some uncertainty resulting from biological feedback processes. However, the direct biological effects of ocean acidification are less certain
image
1 “Acidification” does not mean that the ocean has a pH below neutrality. The average pH of the ocean is still basic (8.1), but because the pH is decreasing, it is described as undergoing acidification.
Page 2
Suggested Citation:"Summary." National Research Council. 2010. Ocean Acidification: A National Strategy to Meet the Challenges of a Changing Ocean. Washington, DC: The National Academies Press. doi: 10.17226/12904. ×
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and will vary among organisms, with some coping well and others not at all. The long-term consequences of ocean acidification for marine biota are unknown, but changes in many ecosystems and the services they provide to society appear likely based on current understanding (Raven et al., 2005).
In response to these concerns, Congress requested that the National Research Council conduct a study on ocean acidification in the Magnuson-Stevens Fishery Conservation and Management Reauthorization Act of 2006. The Committee on the Development of an Integrated Science Strategy for Ocean Acidification Monitoring, Research, and Impacts Assessment is charged with reviewing the current state of knowledge and identifying key gaps in information to help federal agencies develop a program to improve understanding and address the consequences of ocean acidification (see Box S.1 for full statement of task). Shortly after the study was underway, Congress passed another law—the Federal Ocean Acidification Research and Monitoring (FOARAM) Act of 2009—which calls for, among other things, the establishment of a federal ocean acidification program; this report is directed to the ongoing strategic planning process for such a program.
Although ocean acidification research is in its infancy, there is already growing evidence of changes in ocean chemistry and ensuing biological impacts. Time-series measurements and other field data have documented the decrease in ocean pH and other related changes in seawater chemistry (Dore et al., 2009). The absorption of anthropogenic CO2 by the oceans increases the concentration of hydrogen ions in seawater (quanti-
Explanation:
Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore frictional losses. (a) What is the height of the hill? (b) Released from rest at the same height, a can of frozen juice rolls to the bottom of the same hill. What is the translational speed of the frozen juice can when it reaches the bottom?
Answer:
a) h=3.16 m, b) v_{cm }^ = 6.43 m / s
Explanation:
a) For this exercise we can use the conservation of mechanical energy
Starting point. Highest on the hill
Em₀ = U = mg h
final point. Lowest point
[tex]Em_{f}[/tex] = K
Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere
K = ½ m [tex]v_{cm }^{2}[/tex] + ½ [tex]I_{cm}[/tex] w²
angular and linear speed are related
v = w r
w = v / r
K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²
Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)
as there are no friction losses, mechanical energy is conserved
Em₀ = Em_{f}
mg h = ½ v_{cm }^{2} (m + I_{cm} / r²) (1)
h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)
for the moment of inertia of a basketball we can approximate it to a spherical shell
I_{cm} = ⅔ m r²
we substitute
h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)
h = ½ v_{cm }^{2}/g 5/3
h = 5/6 v_{cm }^{2} / g
let's calculate
h = 5/6 6.1 2 / 9.8
h = 3.16 m
b) this part of the exercise we solve the speed of equation 1
v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)
in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia
I_{cm} = ½ m r²
we substitute
v_{cm } = √ [2gh / (1 + ½)]
v_{cm } = √(4/3 gh)
let's calculate
v_{cm } = √ (4/3 9.8 3.16)
v_{cm }^ = 6.43 m / s
A 62.2-kg person, running horizontally with a velocity of 3.80 m/s, jumps onto a 19.7-kg sled that is initially at rest. (a) Ignoring the effects of friction during the collision, find the velocity of the sled and person as they move away. (b) The sled and person coast 30.0 m on level snow before coming to rest. What is the coefficient of kinetic friction between the sled and the snow
Answer:
a) [tex]v = 2.886\,\frac{m}{s}[/tex], b) [tex]\mu_{k} = 0.014[/tex]
Explanation:
a) The final speed is determined by the Principle of Momentum Conservation:
[tex](62.2\,kg)\cdot (3.80\,\frac{m}{s} ) = (81.9\,kg)\cdot v[/tex]
[tex]v = 2.886\,\frac{m}{s}[/tex]
b) The deceleration experimented by the system person-sled is:
[tex]a = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(2.886\,\frac{m}{s} \right)^{2}}{2\cdot (30\,m)}[/tex]
[tex]a = -0.139\,\frac{m}{s^{2}}[/tex]
By using the Newton's Laws, the only force acting on the motion of the system is the friction between snow and sled. The kinetic coefficient of friction is:
[tex]-\mu_{k}\cdot m\cdot g = m\cdot a[/tex]
[tex]\mu_{k} = -\frac{a}{g}[/tex]
[tex]\mu_{k} = -\frac{\left(-0.139\,\frac{m}{s^{2}} \right)}{9.807\,\frac{m}{s^{2}} }[/tex]
[tex]\mu_{k} = 0.014[/tex]
A large fraction of the ultraviolet (UV) radiation coming from the sun is absorbed by the atmosphere. The main UV absorber in our atmosphere is ozone, O3O3. In particular, ozone absorbs radiation with frequencies around 9.38×1014 HzHz . What is the wavelength λλlambda of the radiation absorbed by ozone? Express your answer in nanometers.
λ = ___.
Answer:
λ = 3.2 x 10⁻⁷ m = 320 nm
Explanation:
The relationship between the velocity of electromagnetic waves (UV rays) and the their frequency is:
v = fλ
where,
v = c = speed of the electromagnetic waves (UV rays) = speed of light
c = 3 x 10⁸ m/s
f = frequency of the electromagnetic waves (UV rays) = 9.38 x 10¹⁴ Hz
λ = wavelength of the electromagnetic waves (UV rays) = ?
Therefore, substituting the values in the relation, we get:
3 x 10⁸ m/s = (9.38 x 10¹⁴ Hz)(λ)
λ = (3 x 10⁸ m/s)/(9.38 x 10¹⁴ Hz)
λ = 3.2 x 10⁻⁷ m = 320 nm
So, the radiation of 320 nm wavelength is absorbed by Ozone.
Answer:
0.03052 nm
Explanation:
From c = f¥
Where c = speed of electromagnetic wave in vacuum = 3x10^8 m/s
f = frequency of the wave
¥ = wavelenght of the wave.
¥ = c/f = (3x10^8)/(9.38×10^14)
= 3.052x10^-7 m
= 0.03052 nm
A 910-kg object is released from rest at an altitude of 1200 km above the north pole of the earth. If we ignore atmospheric friction, with what speed does the object strike the surface of the earth? (G = 6.67 × 10-11 N ∙ m2/kg2, Mearth = 5.97 × 1024 kg, the polar radius of the earth is 6357 km) 2.7 km/s 3.2 km/s 4.8 km/s 4.5 km/s 2.2 km/s
The object strikes the surface of the Earth with a speed of approximately 11.2 km/s.
Ignoring atmospheric friction, the object's total mechanical energy (the sum of its gravitational potential energy and kinetic energy) is conserved throughout its fall. We can use this principle to calculate the object's speed at the Earth's surface.
Here's how:
Calculate the gravitational potential energy (PE) at the initial altitude:
PE_i = -G * M_earth * m / (R_earth + h)
where:
G is the gravitational constant (6.67 × 10^-11 N m²/kg²)
M_earth is the mass of the Earth (5.97 × 10^24 kg)
m is the mass of the object (910 kg)
R_earth is the Earth's polar radius (6357 km)
h is the initial altitude (1200 km)
Set the final PE (PE_f) to 0:
Since all the PE will be converted to kinetic energy when the object reaches the surface, PE_f = 0.
Apply the conservation of mechanical energy:
E_i = PE_i + KE_i = PE_f + KE_f = KE_f (as PE_i = PE_f = 0)
where:
KE_i is the initial kinetic energy (0 J, as the object is at rest)
KE_f is the final kinetic energy
Solve for the final velocity (v_f):
KE_f = 1/2 * m * v_f^2
v_f = sqrt(2 * E_i / m)
Plug in the values and calculate v_f:
v_f = sqrt(2 * (-G * M_earth * m / (R_earth + h)) / m)
v_f ≈ 11,180 m/s
Convert m/s to km/s:
v_f ≈ 11.2 km/s
Astronomers observe two separate solar systems each consisting of a planet orbiting a sun. The two orbits are circular and have the same radius R. It is determined that the planets have angular momenta of the same magnitude L about their suns, and that the orbital periods are in the ratio of three to one; i.e., T1 = 3T2. The ratio m1/m2 of the masses of the two planets is
(A) 1
(C) (3)^1/2
(C) 2
(D) 3
(E) 9
Answer:
(D) 3
Explanation:
The angular momentum is given by:
[tex]\vec{L}=\vec{r}\ X \ \vec{p}[/tex]
Thus, the magnitude of the angular momenta of both solar systems are given by:
[tex]L_1=Rm_1v_1=Rm_1(\omega R)=R^2m_1(\frac{2\pi}{T_1})=2\pi R^2\frac{m_1}{T_1}\\\\L_2=Rm_2v_2=2\pi R^2\frac{m_2}{T_2}[/tex]
where we have taken that both systems has the same radius.
By taking into account that T1=3T2, we have
[tex]L_1=2\pi R^2\frac{m_1}{3T_2}=\frac{1}{3}2\pi R^2\frac{1}{T_2}m_1=\frac{1}{3}\frac{L_2}{m_2}m_1[/tex]
but L1=L2=L:
[tex]L=\frac{1}{3}L\frac{m_1}{m_2}\\\\\frac{m_1}{m_2}=3[/tex]
Hence, the answer is (D) 3
HOPE THIS HELPS!!