To solve this problem, it is necessary to use the concepts related to the Force given in Newton's second law as well as the use of the kinematic equations of movement description. For this case I specifically use the acceleration as a function of speed and time.
Finally, we will describe the calculation of stress, as the Force produced on unit area.
By definition we know that the Force can be expressed as
F= ma
Where,
m= mass
a = Acceleration
The acceleration described as a function of speed is given by
[tex]a = \frac{\Delta v}{\Delta t}[/tex]
Where,
[tex]\Delta v =[/tex] Change in velocity
[tex]\Delta t =[/tex]Change in time
The expression to find the stress can be defined as
[tex]\sigma=\frac{F}{A}[/tex]
Where,
F = Force
A = Cross-sectional Area
Our values are given as
[tex]v= 80km/h\\t=5.8*10^3s\\m = 3kg \\A = 2.3*10^{-4}m^2[/tex]
Replacing at the values we have that the acceleration is
[tex]a = \frac{\Delta v}{\Delta t}[/tex]
[tex]a = \frac{80km/h(\frac{1h}{3600s})(\frac{1000m}{1km})}{5.8*10^3}[/tex]
[tex]a = 3831.41m/s^2[/tex]
Therefore the force expected is
[tex]F = ma\\F = 3*3831.41m/s^2 \\F = 11494.25N[/tex]
Finally the stress would be
[tex]\sigma = \frac{F}{A}[/tex]
[tex]\sigma = \frac{11494.25N}{2.3*10^{-4}}[/tex]
[tex]\sigma = 49.97*10^6 Pa = 49.97Mpa[/tex]
Therefore the compressional stress that the arm withstands during the crash is 49.97Mpa
Final answer:
The compressional stress is approximately 4.99787 × 10⁷ Pa
Explanation:
The question asks to find the compressional stress that the arm withstands during a crash, where the arm comes to a stop from an initial speed of 80 km/h in 5.8 ms, with an effective mass of 3.0 kg, and the total cross-sectional area of the load-bearing calcified portion of the forearm bones being approximately 2.3 cm².
First, convert the initial speed from km/h to m/s: 80 km/h = 22.22 m/s. To find the acceleration, use the formula a = ∆v / ∆t, where ∆v = -22.22 m/s (as it comes to rest) and ∆t = 5.8 ms or 0.0058 s. This gives an acceleration of approximately -3831.03 m/s².
The force experienced due to this acceleration can be calculated using Newton's second law, F = ma, with m = 3.0 kg and a = -3831.03 m/s², resulting in a force of approximately -11493.09 N. Finally, the compressional stress (σ) is found using the formula σ = F / A, where F = 11493.09 N and A = 2.3 cm² = 0.00023 m². This yields a compressional stress of approximately 4.99787 × 10⁷ Pa.
A 480 kg car moving at 14.4 m/s hits from behind another car moving at 13.3 m/s in the same direction. If the second car has a mass of 570 kg and a new speed of 17.9 m/s, what is the velocity of the first car after the collision?
Answer:
Velocity of the first car after the collision, [tex]v_1=8.93\ m/s[/tex]
Explanation:
It is given that,
Mass of the car, [tex]m_1 = 480\ kg[/tex]
Initial speed of the car, [tex]u_1 = 14.4\ m/s[/tex]
Mass of another car, [tex]m_2 = 570\ kg[/tex]
Initial speed of the second car, [tex]u_2 = 13.3\ m/s[/tex]
New speed of the second car, [tex]v_2 = 17.9\ m/s[/tex]
Let [tex]v_1[/tex] is the final speed of the first car after the collision. The total momentum of the system remains conserved, Using the conservation of momentum to find it as :
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]
[tex]m_1u_1+m_2u_2-m_2v_2=m_1v_1[/tex]
[tex]480\times 14.4+570\times 13.3-570\times 17.9=480v_1[/tex]
[tex]v_1=8.93\ m/s[/tex]
So, the velocity of the first car after the collision is 8.93 m/s. Hence, this is the required solution.
A 15-kN tensile load will be applied to a 50-m length of steel wire with E = 200 GPa. Determine the smallest diameter wire that can be used, knowing that the normal stress must not exceed 150 MPa and that the increase in length of the wire must not exceed 25 mm
Answer:
d=13.81 mm
Explanation:
Given that
P = 15 KN ,L = 50 m
E= 200 GPa
ΔL = 25 mm
σ = 150 MPa
Lets take d=Diameter
There are we have two criteria to find out the diameter of the wire
Case I :
According to Stress ,σ = 150 MPa
P = σ A
[tex]A=\dfrac{P}{\sigma}[/tex]
[tex]d=\sqrt{\dfrac{4P}{\pi \sigma}}[/tex]
By putting the values
[tex]d=\sqrt{\dfrac{4\times 15000}{\pi \times 150}}[/tex]
d= 11.28 mm
Case II:
According to elongation ,ΔL = 25 mm
[tex]\Delta L=\dfrac{PL}{AE}[/tex]
[tex]A=\dfrac{PL}{E\Delta L}[/tex]
[tex]A=\dfrac{4PL}{\pi E\Delta L}[/tex]
[tex]d=\sqrt{\dfrac{4\times 15000\times 50000}{\pi \times 200\times 1000\times 25}}[/tex]
d=13.81 mm
Therefore the answer will be 13.81 mm .Because it satisfy both the conditions.
The diameter is 13.81 mm
Details required to determine the diameter:A 15-kN tensile load will be applied to a 50-m length of steel wire with E = 200 GPa.
Calculation of the diameter:As per the Stress ,σ = 150 MPa
P = σ A
[tex]d = \sqrt{\frac{4\times 15000}{\pi \times 150} }[/tex]
= 11.28mm
Now
According to elongation ,ΔL = 25 mm
[tex]= \sqrt{\frac{4\times 15000 \times 50000}{\pi \times 200\times 1000 \times 25} }[/tex]
= 13.81 mm
For determining the wire of the smallest diameter, the above formulas should be used.
Therefore, we can conclude that the diameter is 13.81 mm.
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The Sun delivers an average power of 1150 W/m2 to the top of the Earth’s atmosphere. The permeability of free space is 4π × 10−7 T · N/A and the speed of light is 2.99792 × 108 m/s. Find the magnitude of Em for the electromagnetic waves at the top of the atmosphere. Answer in units of N/C.
Answer:
E=930.84 N/C
Explanation:
Given that
I = 1150 W/m²
μ = 4Π x 10⁻⁷
C = 2.999 x 10⁸ m/s
E= C B
C=speed of light
B=Magnetic filed ,E=Electric filed
Power P = I A
A=Area=4πr² ,I=Intensity
[tex]I=\dfrac{CB^2}{2\mu_0}[/tex]
[tex]I=\dfrac{CE^2}{2\mu_0 C^2}[/tex]
[tex]E=\sqrt{{2I\mu_0 C}}[/tex]
[tex]E=\sqrt{{2\times 1150\times 4\pi \times 10^{-7}(2.99792\times 10^8)}}[/tex]
E=930.84 N/C
Therefore answer is 930.84 N/C
To find the magnitude Em of the electromagnetic waves at the top of the earth's atmosphere, we use the intensity of electromagnetic wave and solving the equation Em = sqrt(2Icμo), we can find the magnitude of Em in units of N/C.
Explanation:To find the magnitude Em of the electromagnetic waves at the top of the Earth's atmosphere, we use the fact that the power received per unit area is the intensity I of the electromagnetic wave. According to the given information, this intensity is 1150 W/m2. The relationship between the intensity and electromagnetic fields is given by the equation I = 0.5 * E²/c * μo. Solving for Em, we get Em = sqrt(2Icμo), where μo = 4π × 10-7 T N/A² is the permeability of free space and c = 2.99792 × 10⁸ m/s is the speed of light.
Subbing in the given values, we can compute Em as:
Em = sqrt[2 * 1150 W/m² * 2.99792 × 10⁸ m/s * 4π × 10-7 T N/A²]
This computation will give the strength of the electric field at the top of the earth’s atmosphere in units of N/C.
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The Americium nucleus, 241 95 Am, decays to a Neptunium nucleus, 237 93 Np, by emitting an alpha particle of mass 4.00260 u and kinetic energy 5.5 MeV. Estimate the mass of the Neptunium nucleus, ignoring its recoil, given that the Americium mass is 241.05682 u
Answer:
The mass of Neptunium is 237.054 u.
Explanation:
Given that,
Mass of Americium = 241.05682 u
Mass of alpha particle = 4.00260 u
The equation is,
[tex]_{95}^{241}Am\rightarrow _{93}^{237}NP+\alpha\ particle+5.5 MeV[/tex]
Let the mass of Neptunium is m.
Since the mass remain same.
We need to calculate the mass of Neptunium
Using formula of mass
Mass of Neptunium = Mass of Americium -Mass of alpha particle
Put the value into the formula
[tex]m=241.05682 -4.00260[/tex]
[tex]m=237.054\ u[/tex]
Hence, The mass of Neptunium is 237.054 u.
The mass of the Neptunium nucleus is estimated to be 237.05422 u.
The Americium nucleus has a given mass of 241.05682 u and decays by emitting an alpha particle with a mass of 4.00260 u and kinetic energy of 5.5 MeV. To estimate the mass of the Neptunium nucleus, ignoring its recoil, we initially consider the conservation of mass-energy principle.
Mass of Neptunium = Mass of Americium -Mass of alpha particle
The mass of the Neptunium nucleus can be calculated simply by subtracting the mass of the alpha particle from the original Americium nucleus, which results in -:
m = 241.05682 u - 4.00260 u = 237.05422 u
An engineering student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A and a voltage of 110 V, a lamp that contains a 100-W bulb, an overhead light with a 60-W bulb, and various other small devices adding up to 3.00 W.
a) Assuming the power plant that supplies 110 V electricity to the dorm is 10 km away and the two aluminum transmission cables use 0-gauge wire with a diameter of 8.252 mm, estimate the total power supplied by the power company that is lost in the transmission. (Hint: Find the power needed by the student for all devices. Use this to find the current necessary to be running through the transmission cable)
(b) What would be the result is the power company delivered the electric power at 110 kV?
Answer:
a) [tex]P_{L}=199.075W[/tex]
b) [tex]P_{L}=1.991x10^{-4}W[/tex]
Explanation:
1) Notation
Power on the refrigerator: [tex]P=IV=3Ax110V=330W[/tex]
Voltage [tex]V=110V[/tex]
[tex]D=8.252mm[/tex], so then the radius would be [tex]r=\frac{8.252}{2}=4.126mm[/tex]
[tex]L=2x10km=20km=20000m[/tex], representing the length of the two wires.
[tex]\rho=2.65x10^{-8}\Omega m[/tex], that represent the resistivity for the aluminum founded on a book
[tex]P_L[/tex] power lost in the transmission.
2) Part a
We can find the total power adding all the individual values for power:
[tex]P_{tot}=(330+100+60+3)W=493W[/tex]
From the formula of electric power:
[tex]P=IV[/tex]
We can solve for the current like this:
[tex]I=\frac{P}{V}[/tex]
Since we know [tex]P_{tot}[/tex] and the voltage 110 V, we have:
[tex]I=\frac{493W}{110V}=4.482A[/tex]
The next step would be find the cross sectional are for the aluminum cables with the following formula:
[tex]A=\pi r^2 =\pi(0.004126m)^2=5.348x10^{-5}m^2[/tex]
Then with this area we can find the resistance for the material given by:
[tex]R=\rho \frac{L}{A}=2.65x10^{-8}\Omega m\frac{20000m}{5.348x10^{-5}m^2}=9.910\Omega[/tex]
With this resistance then we can find the power dissipated with the following formula:
[tex]P_{L}=I^2 R=(4.482A)^2 9.910\Omega=199.075W[/tex]
And if we want to find the percentage of power loss we can use this formula
[tex]\% P_{L}=\frac{P_L}{P}x100[/tex]
3) Part b
Similar to part a we just need to change the value for V on this case to 110KV.
We can solve for the current like this:
[tex]I=\frac{P}{V}[/tex]
Since we know [tex]P_{tot}[/tex] and the voltage 110 KV=110000V, we have:
[tex]I=\frac{493W}{110000V}=4.482x10^{-3}A[/tex]
The cross sectional area is the same
The resistance for the material not changes.
With this resistance then we can find the power dissipated with the following formula:
[tex]P_{L}=I^2 R=(4.482x10^{-3}A)^2 9.910\Omega=1.991x10^{-4}W[/tex]
Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Initially ball A is moving upward along the y axis at 2.0 m/s and ball B is moving to the right along the x axis with speed 3.7 m/s. After the collision ball B is moving along the positive y axis. a. What is the speed of ball A and B after the collision? b. What is the direction of motion of ball A after the collision? c. What is the total momentum and kinetic energy of the two balls after the collision?
The momentum conservation allows to find the results for the questions about the collision of the balls are:
A) The velocities after the collision are: vₐ = 3.7 i m/s and [tex]v_b[/tex] = 2 j m/s
B) The direction of ball A is: x-axis
C) The total moment is: [tex]p_{total}[/tex] = 4.2m and the direction is θ = 28.4º
The total kinetic energy is: [tex]K_{total}[/tex] = 8.84m J
Given parameters
Ball A initially moves y-axis at [tex]v_{oay}[/tex] = 2.0 m / s Ball B moves end x axis [tex]v_[obx}[/tex]vobx = 3.7 m / s Ball B moves on the y-axis after collisionTo find
A) The speeds after the collision.
B) direction of the ball A.
C) The moment and energy after the crash.
Momentum is defined by the product of mass and velocity. When a system is isolated, there is no external force, the momentum is conserved. This is an important conservation law of physics.
p = m v
Where the bold letters indicate vectors, m is the mass and v the velocity.
A) The system is formed by the two balls, therefore it is isolated and the momentum is conserved. Since the momentum is a vector quantity we write the components on each axis, see attached for a schematic.
Initial instant. Before the crash.
They indicate that ball B moves in the axis and after the collision.
x-axis
p₀ₓ = m [tex]v_{obx}[/tex]vobx
y-axis
[tex]p_{oy} = m v_{oay}[/tex]
Final moment. After the crash.
x-axis
[tex]p_{fx} = m v_{fax}[/tex]
y- axis
[tex]p_{fy} = m v_{fay} + m v_{fby}[/tex]
The momentum is preserved
x- axis
p₀ₓ = [tex]p_{fx}[/tex]
m [tex]v_{obx}[/tex] = m [tex]v_{fax}[/tex]
[tex]v_{fax} = v _{obx}\\v_{fax} = 3.7 m/s[/tex]
y- axis
[tex]p_{oy} = p_[fy} \\m v_{oay} = m v_{fay} + m v_[fby}\\v_{oay} = v_{fay} + v_{fby}[/tex]
2 = [tex]v_{fay} + v_fby}[/tex]
Ball A moves in the x-axis therefore it has no final velocity in the y-axis.
[tex]v_{fay}=0[/tex]
[tex]v_{fby}[/tex] = 2 m / s
B) Ball A moves on the x-axis.
C) the total momentum is the sum of the momentum of each particle
[tex]p_{total} = p_a+p_b[/tex]
To find the module let's use the Pythagorean theorem.
[tex]p_{total} = m \sqrt{2^2 + 3.7^2}[/tex]
p = 4.2m
Let's use trigonometry for the direction.
tan θ = [tex]\frac{p_{fb}}{p_{fa}}[/tex]
θ = tan⁻¹ [tex]\frac{p_{fb}x}{p_{fa}}[/tex]
θ = tam⁻¹ [tex]\frac{2}{3.7}[/tex]
θ = 28.4º
The total kinetic energy is
[tex]K_{total}= K_1+K_2[/tex]
[tex]K_{total}[/tex] = ½ m [tex]v_a^2[/tex] + ½ m [tex]v_b^2[/tex]
[tex]K_{total}[/tex] = [tex]\frac{m}{2}[/tex] (3.7² + 2²)
[tex]K_{total}[/tex] = 8.845 m
In conclusion using momentum conservation we can find the results for the questions about the collision of the balls are:
A) The velocities after the collision [tex]v_a[/tex] = 3.7 i m/s and [tex]v_b[/tex] = 2 j m/s
B) The direction of ball A is: x-axis
C) The total momentum is: [tex]p_{total}[/tex] = 4.2m and the direction is θ = 28.4º
The total kinetic energy is: K_ {total} = 8.84m J
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What is the maximum speed that a car can maneuver over a circular turn of radius r = 75.0 m without sliding if the coefficient of static friction is µs = 0.780? (b) What is the maximum speed that a car can maneuver over a circular turn of radius r = 25.0 m without sliding if the coefficient of static friction is µs = 0.120?
Answer:
Explanation:
Given
Radius of Track [tex]r_1=75 m[/tex]
coefficient of Static Friction [tex]\mu _s=0.78[/tex]
Here centripetal Force is Balanced by Friction Force
thus
[tex]\frac{mv^2}{r}=\mu _sg[/tex]
[tex]\frac{v^2}{r}=\mu _sg[/tex]
[tex]v=\sqrt{\mu _srg}[/tex]
[tex]v=\sqrt{0.78\times 75\times 9.8}[/tex]
[tex]v=23.94 m/s[/tex]
(b)For [tex]r_2=25 m[/tex]
[tex]\mu _s=0.12[/tex]
[tex]v=\sqrt{\mu _sr_2g}[/tex]
[tex]v=\sqrt{0.12\times 25\times 9.8}[/tex]
[tex]v=5.42 m/s[/tex]
Final answer:
The maximum speed that a car can maneuver over a circular turn without sliding depends on the coefficient of static friction and the radius of the turn. This can be calculated using the formula v = sqrt(μs * g * r). For the given scenarios, the maximum speeds are 30.5 m/s and 7.36 m/s respectively.
Explanation:
The maximum speed that a car can maneuver over a circular turn without sliding depends on the coefficient of static friction and the radius of the turn. The maximum speed can be calculated using the formula:
v = sqrt(μs * g * r)
where v is the maximum speed, μs is the coefficient of static friction, g is the acceleration due to gravity (9.8 m/s²), and r is the radius of the turn.
For the first scenario with r = 75.0 m and μs = 0.780, the maximum speed is:
v = sqrt(0.780 * 9.8 * 75.0) = 30.5 m/s
For the second scenario with r = 25.0 m and μs = 0.120, the maximum speed is:
v = sqrt(0.120 * 9.8 * 25.0) = 7.36 m/s
The Balmer series is formed by electron transitions in hydrogen that
a.begin on the n = 1 shell.
b. end on the n = 2 shell.
c. end on the n = 1 shell.
d. are between the n = 1 and n = 3 shells.
e. begin on the n = 2 shell.
Answer:
b. end on the n = 2 shell.
Explanation:
When hydrogen atoms move from higher energy level to lower energy level then it shows spectral lines and these lines are known as Balmer series. The only four lines are visible and other liens are not in the visible range.
The Balmer series formed by hydrogen electron and it ends when n = 2.
Therefore the answer is b.
b. end on the n = 2 shell.
Suppose that the speed of an electron traveling 2.0 km/s is known to an accuracy of 1 part in 105 (i.e., within 0.0010%). What is the least possible uncertainty within which we can determine the position of this electron? (melectron = 9.11 × 10-31 kg, h = 6.626 × 10-34 J ∙ s)
Answer: [tex]2.89(10)^{-3} m[/tex]
Explanation:
The Heisenberg uncertainty principle postulates that the fact each particle has a wave associated with it, imposes restrictions on the ability to determine its position and speed at the same time.
In other words:
It is impossible to measure simultaneously (according to quantum physics), and with absolute precision, the value of the position and the momentum (linear momentum) of a particle. Thus, in general, the greater the precision in the measurement of one of these magnitudes, the greater the uncertainty in the measure of the other complementary variable.
Mathematically this principle is written as:
[tex]\Delta x \geq \frac{h}{4 \pi m \Delta V}[/tex] (1)
Where:
[tex]\Delta x[/tex] is the uncertainty in the position of the electron
[tex]h=6.626(10)^{-34}J.s[/tex] is the Planck constant
[tex]m=9.11(10)^{-31}kg[/tex] is the mass of the electron
[tex]\Delta V[/tex] is the uncertainty in the velocity of the electron.
If we know the accuracy of the velocity is [tex]0.001\%[/tex] of the velocity of the electron [tex]V=2 km/s=2000 m/s[/tex], then [tex]\Delta V[/tex] is:
[tex]\Delta V=2000 m/s(0.001\%)[/tex]
[tex]\Delta V=2000 m/s(\frac{0.001}{100})[/tex]
[tex]\Delta V=2(10)^{-2} m/s[/tex] (2)
Now, the least possible uncertainty in position [tex]\Delta x_{min}[/tex] is:
[tex]\Delta x_{min}=\frac{h}{4 \pi m \Delta V}[/tex] (3)
[tex]\Delta x_{min}=\frac{6.626(10)^{-34}J.s}{4 \pi (9.11(10)^{-31}kg) (2(10)^{-2} m/s)}[/tex] (4)
Finally:
[tex]\Delta x_{min}=2.89(10)^{-3} m[/tex]
The least possible uncertainty within which we can determine the position of this electron is 0.29 micrometers. This value is derived using the Heisenberg's Uncertainty Principle, which states that the more precisely one measures the position of a particle, the less precisely one can measure its speed, and vice versa.
Explanation:The subject question is dealing with the principle of uncertainty in quantum mechanics. According to Heisenberg's Uncertainty Principle, the position and momentum of a particle cannot both be accurately measured at the same time. The more precisely one measures the position of a particle, the less precisely one can measure its speed, and vice versa.
The accuracy is given as 1 part in 105, that is 0.001% or 1e-5. The speed of the electron is known to be 2.0 km/s so the uncertainty in velocity (Δv) would be 2.0 km/s * 1e-5 = 2e-8 m/s. The mass of the electron (melectron) is given to be 9.11 × 10-31 kg.
Momentum is the product of mass and velocity, so Δp = melectron x Δv = (9.11 × 10-31 kg) * (2e-8 m/s) = 1.822e-38 kg m/s. According to Heisenberg's Uncertainty Principle (ΔxΔp ≥ h/4π) the least uncertainty in position (Δx) = h / (4πΔp), where h is Planck's constant 6.626 × 10-34 J.s. Substituting the values, Δx = 6.626 × 10-34 J.s / (4π * 1.822e-38 kg m/s) = 2.9e-7 m or 0.29 μm.
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A grocery cart with a mass of 20 kg is pushed at constant speed along an aisle by a force F = 12 N. The applied force acts at a 20° angle to the horizontal.
Find the work done by each of the forces on the cart if the aisle is 18 m long.
Answer:
Explanation:
Given
mass of cart m=20 kg
Force applied F=12 N at angle of [tex]\theta =20^{\circ}[/tex] with horizontal
Force can be resolved in two components i.e. [tex]F\cos \theta [/tex]and [tex]F\sin \theta [/tex]
[tex]F\sin \theta [/tex]is will act Perpendicular to the motion of cart and [tex]F\cos \theta [/tex]will help to move the cart .
Since cart is moving with constant velocity therefore net force is zero on the cart as friction is opposing the [tex]F\cos \theta [/tex]
Work done by [tex]F\cos \theta [/tex]is
[tex]W_1=F\cos \theta \cdot L[/tex]
[tex]W_1=12\cos (20)\cdot 18=202.978 J[/tex]
Work done by Friction will be same but in Opposite Direction
[tex]W_f=-202.978 J[/tex]
Work done by [tex]F\sin \theta [/tex] is
[tex]W_2=0[/tex] because direction of force and motion is Perpendicular to each other.
A wheel rotates with a constant angular acceleration of 3.45 rad/s^2. Assume the angular speed of the wheel is 1.85 rad/s at ti = 0.
(a) Through what angle does the wheel rotate between t = 0 and t = 2.00 s
Answer:
θ =607.33°
Explanation:
Given that
Angular acceleration α = 3.45 rad/s²
Initial angular speed ,ω = 1.85 rad/s
The angle rotates by wheel in time t
[tex]\theta=\omega t +\dfrac{1}{2}\alpha t^2[/tex]
Now by putting the values
[tex]\theta=\omega t +\dfrac{1}{2}\alpha t^2[/tex]
[tex]\theta=1.85\times 2 +\dfrac{1}{2}\times 3.45\times 2^2[/tex]
θ = 10.6 rad
[tex]\theta=\dfrac{180}{\pi}\times 10.6\ degree[/tex]
θ =607.33°
Therefore angle turn by wheel in 2 s is θ =607.33°
One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has a length of 0.80 m and is uniform. It is hanging vertically straight downward. The end of the rod nearest the floor is given a linear speed v0, so that the rod begins to rotate upward about the pivot. What must be the value of v0, such that the rod comes to a momentary halt in a straight-up orientation, exactly opposite to its initial orientation?
Answer:
6.86 m/s
Explanation:
This problem can be solved by doing the total energy balance, i.e:
initial (KE + PE) = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}
Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.
Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.
Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s
The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²
This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)
Now, 1/6m(v0)² = mgL ⇒ v0 = [tex]\sqrt{6gL}[/tex]
Hence, v0 = 6.86 m/s
To find the value of v0 that will cause the rod to come to a momentary halt in a straight-up orientation, we need to consider conservation of energy and angular momentum. The equation (3/2)gh = 4(v0)^2 can then be used to solve for v0. The value of v0 is equal to the square root of (3/8)gh.
Explanation:To find the value of v0 that will cause the rod to come to a momentary halt in a straight-up orientation, we need to consider conservation of energy and angular momentum. When the rod is released, its potential energy is converted to kinetic energy and rotational kinetic energy. At the moment the rod comes to a halt, all of its initial kinetic energy will be converted back into potential energy. Since the rod is initially hanging vertically downward, we can equate the initial potential energy to the potential energy at the moment of momentary halt:
mgh = (1/2)Iω2
Where m is the mass of the rod, g is the acceleration due to gravity, h is the height of the rod, I is the moment of inertia of the rod about the pivot point, and ω is the angular velocity of the rod at the moment of momentary halt. The moment of inertia of a rod rotating about one end is given by I = (1/3)mL2, where L is the length of the rod. Therefore, we can rewrite the equation as:
mgh = (1/2)(1/3)mL2ω2
Simplifying the equation, we have:
(3/2)gh = ω2L2
To find ω, we need to relate it to the linear speed v0. Since the rod is rotating about a pivot point, the linear speed of a point on the rod is related to the angular velocity by the equation v = ωr, where r is the distance of the point from the pivot. In this case, the distance r is equal to half the length of the rod, so r = L/2. Substituting this into the equation, we have:
v0 = ω(L/2)
From this equation, we can solve for ω:
ω = (2v0)/L
Substituting this into the previous equation, we get:
(3/2)gh = ((2v0)/L)2L2
Simplifying further, we have:
(3/2)gh = 4(v0)2
Finally, solving for v0, we get:
v0 = √((3/8)gh)
A helium-filled balloon has a volume of 1.6 m3 . As it rises in the Earth’s atmosphere, its volume changes. What is its new volume if its original temperature and pressure are 16◦C and 1.3 atm and its final temperature and pressure are −41◦C and 1.2 atm? Answer in units of m
Answer: 1.39 m3
Explanation:
If we consider the helium to be an ideal gas, at any condition, we can apply the Ideal Gas Equation as follows:
P V = n R T
Taking the initial state as (1), we can write:
P1V1 = n R T1 (1)
In the initial state, we have P= 1.3 atm, V=1.6 m3, and T= 273 K + 16 K =289 K.
Let’s call (2) to the final state, so we can write as follows:
P2V2 = n R T2 (2)
In the final state, our givens are P= 1.2 atm, and T= 273 K -41 K = 232 K
So, dividing both sides in (1) and (2), we can solve for V2, as follows:
V2 = (1.3 atm. 1.6 m3.232 K) / 1.2 atm. 289 K = 1.39 m3
The average distance between collisions for atoms in a real gas is known as the mean free path (see pp. 298-9 in McKay). Which of the answers below best represents the mean free path for gas molecules at STP with atomic radii on the order of 1.0 x 10-10 m? The average distance between collisions for atoms in a real gas is known as the mean free path (see pp. 298-9 in McKay). Which of the answers below best represents the mean free path for gas molecules at STP with atomic radii on the order of 1.0 x 10-10 m?
A. 30 nm
B. 3000 nm
C. 300 nm
D. 3 nm
Answer:
300 nm
Explanation:
R = Gas constant = 8.314 J/molK
r = Atomic radii = [tex]1\times 10^{-10}\ m[/tex]
d = Atomic diameter = [tex]2r=2\times 10^{-10}\ m[/tex]
At STP
T = Temperature = 273.15 K
P = Pressure = 100 kPa
[tex]N_A[/tex] = Avogadro's number = [tex]6.022\times 10^{23}[/tex]
The mean free path is given by
[tex]\lambda=\frac{RT}{\sqrt2d^2N_AP}\\\Rightarrow \lambda=\frac{8.314\times 273.15}{\sqrt2 \pi \times (2\times 10^{-10})^2\times 6.022\times 10^{23}\times 100000}\\\Rightarrow \lambda=2.12165\times 10^{-7}\ m=212.165\times 10^{-9}\ m=212.165\ nm[/tex]
The answer that best represents the mean free path for gas molecules is 300 nm
A golfer starts with the club over her head and swings it to reach maximum speed as it contacts the ball. Halfway through her swing, when the golf club is parallel to the ground, does the acceleration vector of the club head point straight down, parallel to the ground, approximately toward the golfer's shoulders, approximately toward the golfer's feet, or toward a point above the golfer's head?
Answer:
a) parallel to the ground True
c) parallel to the ground towards man True
Explanation:
To examine the possibilities, we propose the solution of the problem.
Let's use Newton's second law
F = m a
The force is exerted by the arm and the centripetal acceleration of the golf club, which in this case varies with height.
In our case, the stick is horizontal in the middle of the swing, for this point the centripetal acceleration is directed to the center of the circle or is parallel to the arm that is also parallel to the ground;
Ask the acceleration vector
a) parallel to the ground True
b) down. False
c) parallel to the ground towards True men
d) False feet
e) the head. False
Suppose that the strings on a violin are stretched with the same tension and each has the same length between its two fixed ends. The musical notes and corresponding fundamental frequencies of two of these strings are G (196 Hz) and E (659.3 Hz). The linear density of the E string is 3.40 × 10-4 kg/m. What is the linear density of the G string?
Answer:
0.00384 kg/m
Explanation:
The fundamental frequency of string waves is given by
[tex]f=\frac{1}{2L}\sqrt{\frac{F}{\mu}}[/tex]
For some tension (F) and length (L)
[tex]f\propto\frac{1}{\mu}[/tex]
Fundamental frequency of G string
[tex]f_G=196\ Hz[/tex]
Fundamental frequency of E string
[tex]f_E=659.3\ Hz[/tex]
Linear mass density of E string is
[tex]\mu_E=3.4\times 10^{-4}\ kg/m[/tex]
So,
[tex]\frac{F_G}{F_E}=\sqrt{\frac{\mu_E}{\mu_G}}\\\Rightarrow \frac{F_G^2}{F_E^2}=\frac{\mu_E}{\mu_G}\\\Rightarrow \mu_G=3.4\times 10^{-4}\times \frac{659.3^2}{196^2}\\\Rightarrow \mu_G=0.00384\ kg/m[/tex]
The linear density of the G string is 0.00384 kg/m
) Force F = − + ( 8.00 N i 6.00 N j ) ( ) acts on a particle with position vector r = + (3.00 m i 4.00 m j ) ( ) . What are (a) the torque on the particle about the origin, in unit-vector notation, and (b) the angle between the directions of r and F ?
To develop this problem it is necessary to apply the concepts related to the Cross Product of two vectors as well as to obtain the angle through the magnitude of the angles.
The vector product between the Force and the radius allows us to obtain the torque, in this way,
[tex]\tau = \vec{F} \times \vec{r}[/tex]
[tex]\tau = (8i+6j)\times(-3i+4j)[/tex]
[tex]\tau = (8*4)(i\times j)+(6*-3)(j\times i)[/tex]
[tex]\tau = 32k +18k[/tex]
[tex]\tau = 50 k[/tex]
Therefore the torque on the particle about the origen is 50k
PART B) To find the angle between two vectors we apply the definition of the dot product based on the vector quantities, that is,
[tex]cos\theta = \frac{r\cdot F}{|\vec{r}|*|\vec{F}|}[/tex]
[tex]cos\theta = \frac{(8*-3)+(4*3)}{\sqrt{(-3)^2+4^2}*\sqrt{8^2+6^2}}[/tex]
[tex]cos\theta = -0.24[/tex]
[tex]\theta = cos^{-1} (-0.24)[/tex]
[tex]\theta = 103.88\°[/tex]
Therefore the angle between the ratio and the force is 103.88°
The eccentricity of an asteroid's orbit is 0.0442, and the semimajor axis is 1.12 x 1011 m. The Sun's center is at one focus of the asteroid's orbit. (a) How far from this focus is the other focus in meters? (b) What is the ratio of this distance to the solar radius, 6.96 x 108 m?
Answer:
a. [tex]d=99x10^{8}m[/tex]
b. [tex]r=14.22 R_s[/tex]
Explanation:
The eccentricity of an asteroid's is 0.0442 so
a.
to find the focus distance between both focus is
[tex]d=2*e*a[/tex]
[tex]e=1.12x10^{11}m\\a=0.0442[/tex]
So replacing numeric
[tex]d=2*1.12x10^{11}m*0.0442=9900800000[/tex]
[tex]d=99x10^{8}m[/tex]
b.
Now to find the ratio of that distance between the solar radius and the distance
[tex]r=\frac{d_1}{d_s}[/tex]
[tex]r=R_s*\frac{99x10^8m}{6.96x10^8m}[/tex]
[tex]r=14.22 R_s[/tex]
What trend is visible in the sandstone beds as they are traced from east to west?
a There is no information on how the sandstone beds change from east to west.
b. The sandstone beds grade into conglomerate from east to west.
c. The sandstone beds become thinner to the west.
d. The sandstone beds become thicker to the west.
Answer: d.
Explanation:
The sandstone beds become thicker to the west.
Based on the information provided, there is no explicit trend mentioned in the sandstone beds from east to west. To find a trend, one would need to look for changes in grain size, sediment type, or sedimentary structures, none of which are detailed in the given data.
Explanation:The trend visible in the sandstone beds as they are traced from east to west is not explicitly listed in the provided information. However, looking at the descriptions, it seems that no direct trend such as grading into conglomerate, becoming thinner, or becoming thicker is mentioned.
To determine such a trend, geologists might look for changes in the grain size, sediment type, or sedimentary structures within the sandstone layers. For example, if the beds were observed to contain increasingly larger grain sizes, or a transition to a different rock type such as conglomerate, that might indicate a trend in energy conditions and depositional environments.
Since no such trend is described in the information given, the most appropriate answer based on the provided options would be There is no information on how the sandstone beds change from east to west, which corresponds to option a.
Calculate the energy (in MeV) released when α decay converts uranium 232U (atomic mass = 232.037146 u) into thorium 228Th (atomic mass = 228.028731 u). The atomic mass of an α particle is 4.002603 u.
To solve this problem it is necessary to apply the concepts related to the energy released through the mass defect.
Mass defect can be understood as the difference between the mass of an isotope and its mass number, representing binding energy.
According to the information given we have that the reaction presented is as follows:
[tex]^{232}U_{92} \Rightarrow ^{228}Th_{90}+^4He_2[/tex]
The values of the atomic masses would then be:
Th = 232.037146 u
Ra = 228.028731 u
He = 4.0026
The mass difference of the reaction would then be represented as
[tex]\Delta m = 232.037146 u - (228.028731 u + 4.002603 u )[/tex]
[tex]\Delta m = 0.005812 u[/tex]
From the international measurement system we know that 1 atomic mass unit is equivalent to 931.5 MeV,
[tex]\Delta m = 0.005812 u (\frac{931.5MeV}{1u})[/tex]
[tex]\Delta m = 5.414MeV[/tex]
Therefore the energy is 5.414MeV
During a double-slit experiment, you have the option of using either a green laser or a blue laser for your light source. Given the same configuration of your equipment, which one of these lasers would produce an interference pattern that has the largest spacing between bright spots?
Answer:
Green laser
Explanation:
In the double slit experiment, the distance between two consecutive bright spots or two consecutive dark spots is defined as the fringe width
We are assuming that the two slits are on a vertical line
The formula for fringe width is (D× wavelength of light) ÷ d
where D is the distance from the slit to the screen where the interference pattern is observed
d is the distance between the two slits
As it is given that both lasers have same configurations which means they both have same value of D and d
∴ Fringe width directly depends on the wavelength of the light
As green laser has more wavelength when compared to blue laser
∴ Green laser would produce an interference pattern that has the largest spacing between bright spots when compared to blue laser
The magnetic field around the head has been measured to be approximately 3.00×10−8 gauss . Although the currents that cause this field are quite complicated, we can get a rough estimate of their size by modeling them as a single circular current loop 16.0 cm (the width of a typical head) in diameter.What is the current needed to produce such a field at the center of the loop? Answer in A.
Answer:
[tex]3.81972\times 10^{-7}\ A[/tex]
Explanation:
B = Magnetic field = [tex]3\times 10^{-8}\ G[/tex]
d = Diameter of loop = 16 cm
r = Radius = [tex]\frac{d}{2}=\frac{16}{2}=8\ cm[/tex]
i = Current
[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi \times 10^{-7}\ H/m[/tex]
The magnetic field of a loop is given by
[tex]B=\frac{\mu_0i}{2r}\\\Rightarrow i=\frac{B2r}{\mu_0}\\\Rightarrow i=\frac{3\times 10^{-8}\times 10^{-4}\times 2\times 0.08}{4\pi\times 10^{-7}}\\\Rightarrow i=3.81972\times 10^{-7}\ A[/tex]
The current needed to produce such a field at the center of the loop is [tex]3.81972\times 10^{-7}\ A[/tex]
In a carnival ride, passengers stand with their backs against the wall of a cylinder. The cylinder is set into rotation and the floor is lowered away from the passengers, but they remain stuck against the wall of the cylinder. For a cylinder with a 2.0-m radius, what is the minimum speed that the passengers can have for this to happen if the coefficient of static friction between the passengers and the wall is 0.25?
Answers
2.3 m/s
3.0 m/s
4.9 m/s
8.9 m/s
It depends on the mass of the passengers.
To solve the problem, it is necessary to apply the related concepts to Newton's second law as well as the Normal and Centripetal Force experienced by passengers.
By Newton's second law we understand that
[tex]F = mg[/tex]
Where,
m= mass
g = Gravitational Acceleration
Also we have that Frictional Force is given by
[tex]F_r = \mu N[/tex]
In this particular case the Normal Force N is equivalent to the centripetal Force then,
[tex]N = \frac{mv^2}{r}[/tex]
Applying this to the information given, and understanding that the Weight Force is statically equivalent to the Friction Force we have to
[tex]F = F_r[/tex]
[tex]mg = \mu N[/tex]
[tex]mg = \mu \frac{mv^2}{r}[/tex]
Re-arrange to find v,
[tex]v= \sqrt{\frac{gr}{\mu}}[/tex]
[tex]v = \sqrt{\frac{(9.8)(2)}{0.25}}[/tex]
[tex]v = 8.9m/s[/tex]
From the last expression we can realize that it does not depend on the mass of the passengers.
The minimum speed required for passengers to remain stuck to the wall of a cylindrical carnival ride with a radius of 2.0 meters and a coefficient of static friction of 0.25 is d) 8.9 m/s.
To determine the minimum speed required to prevent passengers from sliding down the wall of a cylindrical carnival ride with a 2.0-m radius, we can use the principles of centripetal force and friction.
The centripetal force needed to keep a rider in circular motion is provided by the normal force (N) which acts horizontally. This force is balanced by the frictional force (f) acting vertically upwards to counteract the gravitational force (mg) pulling the rider down.
The frictional force is given by:
f = μN
Where μ is the coefficient of static friction (0.25). The normal force is equivalent to the centripetal force needed for circular motion:
N = mv² / r
Thus, the frictional force equation becomes:
μ(mv² / r) = mg
Solving for v:
μv² / r = g
v² = rg / μ
[tex]v = \sqrt{\frac{rg}{\mu}}[/tex]
Substituting the given values (r = 2.0 m, g = 9.8 m/s², μ = 0.25):
[tex]v = \sqrt{\frac{2.0 \cdot 9.8}{0.25}}[/tex]
v = √(78.4)
v = 8.9 m/s
Therefore, the correct option is d) as the minimum speed required is 8.9 m/s.
Suppose that the hatch on the side of a Mars lander is built and tested on Earth so that the internal pressure just balances the external pressure. The hatch is a disk 50.0 cm in diameter. When the lander goes to Mars, where the external pressure is 650 N/m^2, what will be the net force (in newtowns and pounds) on the hatch, assuming that the internal pressure is the same in both cases? Will it be an inward or outward force? Answer= 19.8 kN, 4440 lb. Outward force. Please explain how to get these answers. Equation in textbook : Fnet=(P2-P1)A
Answer:
[tex]F=19.8kN=4442lb[/tex]
Explanation:
On Earth the atmospheric presure is [tex]P_E=101325 N/m^2[/tex]. This will be the pressure inside the lander. Outside, the pressure on Mars will be [tex]P_M=650 N/m^2[/tex]. This means that the net force will be outward (since inside the pressure is higher) and, since the area of the hatch is [tex]A=\pi r^2[/tex], of value:
[tex]F=(P_E-P_M)\pi r^2=(101325N/m^2-650N/m^2)\pi (\frac{0.5m}{2})^2=19768N=19.8kN[/tex]
Since 1lb in weight is equal to 4.45N, we can write:
[tex]F=19768N=19768N\frac{1lb}{4.45N}=4442lb[/tex]
1) An object is hung using a metal spring. If now a current is passed through the spring, what will happen to this system?
A) The spring will begin to swing like a pendulum.
B) The spring will extend, lowering the weight.
C) The weight will not move.
D) The spring will contract, raising the weight.
E) None of these are true.
Answer:
D) The spring will contract, raising the weight.
Explanation:
According to the statement there is current that will enter the current through the metal ions that it has in its stratum. The passage of the current will generate within the spring a magnetic field that travels in a loop. That is, while the upper part of the spring which is also that of the spring acts as a north pole, the lower part of the spring and the magnetic field will act as the south pole. The position of the poles will generate an opposition effect that will generate an attraction to each other which will generate a contraction in the spring and an increase in weight on it.
The volume of water in the Pacific Ocean is about 7.0 × 10 8 km 3 . The density of seawater is about 1030 kg/m3. (a) Determine the gravitational potential energy of the Moon–Pacific Ocean system when the Pacific is facing away from the Moon. (b) Repeat the calculation when Earth has rotated so that the Pacific Ocean faces toward the Moon. (c) Estimate the maximum speed of the water in the Pacific Ocean due to the tidal influence of the Moon. For the sake of the calculations, treat the Pacific Ocean as a pointlike object (obviously a very rough approximation)
To solve the problem it is necessary to consider the concepts related to Potential Energy and Kinetic Energy.
Potential Energy because of a planet would be given by the equation,
[tex]PE=\frac{GMm}{r}[/tex]
Where,
G = Gravitational Universal Constant
M = Mass of Ocean
M = Mass of Moon
r = Radius
From the data given we can calculate the mass of the ocean water through the relationship of density and volume, then,
[tex]m = \rho V[/tex]
[tex]m = (1030Kg/m^3)(7*10^8m^3)[/tex]
[tex]m = 7.210*10^{11}Kg[/tex]
It is necessary to define the two radii, when the ocean is far from the moon and when it is facing.
When it is far away, it will be the total diameter from the center of the earth to the center of the moon.
[tex]r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m[/tex]
When it's near, it will be the distance from the center of the earth to the center of the moon minus the radius,
[tex]r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m[/tex]
PART A) Potential energy when the ocean is at its furthest point to the moon,
[tex]PE_1 = \frac{GMm}{r_1}[/tex]
[tex]PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}[/tex]
[tex]PE_1 = 9.05*10^{15}J[/tex]
PART B) Potential energy when the ocean is at its closest point to the moon
[tex]PE_2 = \frac{GMm}{r_2}[/tex]
[tex]PE_2 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.776*10^8}[/tex]
[tex]PE_2 = 9.361*10^{15}J[/tex]
PART C) The maximum speed. This can be calculated through the conservation of energy, where,
[tex]\Delta KE = \Delta PE[/tex]
[tex]\frac{1}{2}mv^2 = PE_2-PE_1[/tex]
[tex]v=\sqrt{2(PE_2-PE_1)/m}[/tex]
[tex]v = \sqrt{\frac{2*(9.361*10^{15}-9.05*10^{15})}{7.210*10^{11}}}[/tex]
[tex]v = 29.4m/s[/tex]
A uranium nucleus is traveling at 0.94 c in the positive direction relative to the laboratory when it suddenly splits into two pieces. Piece A is propelled in the forward direction with a speed of 0.43 c relative to the original nucleus. Piece B is sent backward at 0.35 c relative to the original nucleus. Part A Find the velocity of piece A as measured by an observer in the laboratory. Do the same for piece B.
Final answer:
To find the velocities of pieces A and B as measured by an observer in the laboratory, use the relativistic velocity addition formula.
Explanation:
To find the velocities of pieces A and B as measured by an observer in the laboratory, we need to use the relativistic velocity addition formula. Let's call the initial velocity of the uranium nucleus as v. Piece A is moving forward with a speed of 0.43c relative to the original nucleus and piece B is moving backward at 0.35c relative to the original nucleus.
The velocity of piece A as measured by an observer in the laboratory is given by vA = (v + vA') / (1 + v*vA'/c^2), where vA' is the velocity of piece A relative to the original nucleus. Plugging in the values, we get vA = (v + 0.43c) / (1 + v*0.43c/c^2).
The velocity of piece B as measured by an observer in the laboratory is given by vB = (v - vB') / (1 - v*vB'/c^2), where vB' is the velocity of piece B relative to the original nucleus. Plugging in the values, we get vB = (v - 0.35c) / (1 - v*0.35c/c^2).
A 0.18-kg turntable of radius 0.32 m spins about a vertical axis through its center. A constant rotational acceleration causes the turntable to accelerate from 0 to 24 revolutions per second in 8.0 s.Calculate the rotational acceleration.
Answer:
Angular acceleration will be [tex]18.84rad/sec^2[/tex]
Explanation:
We have given that mass m = 0.18 kg
Radius r = 0.32 m
Initial angular velocity [tex]\omega _i=0rev/sec[/tex]
And final angular velocity [tex]\omega _f=24rev/sec[/tex]
Time is given as t = 8 sec
From equation of motion
We know that [tex]\omega _f=\omega _i+\alpha t[/tex]
[tex]24=0+\alpha \times 8[/tex]
[tex]\alpha =3rev/sec^2=3\times 2\times \pi rad/sec^2=18.84rad/sec^2[/tex]
So angular acceleration will be [tex]18.84rad/sec^2[/tex]
A spring with an 8-kg mass is kept stretched 0.4 m beyond its natural length by a force of 32 N. The spring starts at its equilibrium position and is given an initial velocity of 4 m/s. Find the position of the mass at any time t. (Use x for the displacement in meters from the equilibrium position.)
Answer:
x = 1.26 sin 3.16 t
Explanation:
Assume that the general equation of the displacement given as
x = A sinω t
A=Amplitude ,t=time ,ω=natural frequency
We know that speed V
[tex]V=\dfrac{dx}{dt}[/tex]
V= A ω cosωt
Maximum velocity
V(max)= Aω
Given that F= 32 N
F = K Δ
K=Spring constant
Δ = 0.4 m
32 =0.4 K
K = 80 N/m
We know that ω²m = K
8 ω² = 80
ω = 3.16 s⁻¹
Given that V(max)= Aω = 4 m/s
3.16 A = 4
A= 1.26 m
Therefore the general equation of displacement
x = 1.26 sin 3.16 t
Answer:
Explanation:
mass, m = 8 kg
extension, y = 0.4 m
force, F = 32 N
maximum velocity, v = 4 m/s
maximum velocity , v = ω A
where, ω be the angular velocity and A be the amplitude
4 = ω x 0.4
ω = 10 rad/s
position
x = A Sin ωt
x = 0.4 Sin 10 t
A tank of gasoline (n = 1.40) is open to the air (n = 1.00). A thin film of liquid floats on the gasoline and has a refractive index that is between 1.00 and 1.40. Light that has a wavelength of 626 nm (in vacuum) shines perpendicularly down through the air onto this film, and in this light the film looks bright due to constructive interference. The thickness of the film is 290 nm and is the minimum nonzero thickness for which constructive interference can occur. What is the refractive index of the film?
nfilm = 1Your answer is incorrect.I did t=(m)(wavelengthfilm)/(2) solving for the wavelegth of filmThenI did: wavelength film = wavelength vaccum / n and my n comes out as 1.18 which is the wrong answer can anyone help??
Answer:
1.08
Explanation:
This is the case of interference in thin films in which interference bands are formed due to constructive interference of two reflected light waves , one from upper layer and the other from lower layer . If t be the thickness and μ be the refractive index then
path difference created will be 2μ t.
For light coming from rarer to denser medium , a phase change of π occurs additionally after reflection from denser medium, here, two times, once from upper layer and then from the lower layer , so for constructive interference
path diff = nλ , for minimum t , n =1
path diff = λ
2μ t. = λ
μ = λ / 2t
= 626 / 2 x 290
= 1.08