Answer:
Let's analyse the definition and applications of angular momentum, and its relation with torque.
First of all, it's important to consider that the angular momentum is a property of rotational dynamics. Also, it's the analogue of the linear momentum.
Mathematically, the angular momentum is defined as
[tex]L=r \times p[/tex]
Where [tex]L[/tex] represents the angular momentum vector, [tex]r[/tex] represents is the position vector and [tex]p[/tex] is the linear momentum vector.
Notice that the angular momentum is also a vector, which is the cross product of two vectorial magnitudes. In other words, the direction of the resulting vector (linear momentum) follows the right hand rule, which means that the resulting direction is according to the rotation direction, also means that the cross product is not commutative, which is a common assumption students make.
Now, the realtion between angular momentum and torque is that the change of the angular momentum with respect to time is equivalent to its torque:
[tex]\frac{d}{dt}[L]=\frac{d}{dt}[r \times p] =\tau[/tex]
Remember that torque is defined as [tex]\sum \tau = r \times \sum F[/tex], and the derivative of the cross product is
[tex]\frac{d}{dt}[L]=\frac{d}{dt}[r \times p] = \frac{dr}{dt} \times p + r \times \frac{dp}{dt}[/tex]
Then,
[tex]\frac{d}{dt}[L] =(v \times mv)+r \times \frac{dp}{dt}[/tex]
But, [tex]v \times mv = 0[/tex], because those vector are parallel.
So, [tex]\frac{d}{dt}[L] = r\times \sum F = \tau[/tex]
At this point, we demonstrate it the relation between torque and rotational momentum.
In words, the net torque on a particle is equal to the rate of change of the angular momentum with respect to time.
Now, the application of angular momentum can be seen in skating spins, notice that when the skater puts his arms closer to its body, he'll rotate faster. The reason of this phenomenon is because arms represents rotating mass and the axis is the body, so the postion of this arm mass changes to zero distance to the rotational axis, that will increase the angular momentum, making higher. If the angular momentum is higher, the torque will be also higher, that's way the skater increses its rotational velocity.
Angular momentum is the rotational equivalent of linear momentum, defined as L = Iω for extended bodies. It is conserved when net external torque is zero, similar to how linear momentum is conserved in the absence of external forces. Torque and angular momentum are related by net τ = ΔL/Δt.
In physics, the concept of angular momentum is the rotational equivalent of linear momentum. It is defined as L = r × p for a single particle, where r is the position vector and p is the linear momentum vector. For an extended body rotating about an axis of symmetry, angular momentum (L) is given by the product of the moment of inertia (I) and angular velocity (ω): L = Iω.
Just as linear momentum is conserved in the absence of external forces, angular momentum is conserved when the net external torque is zero. This principle is crucial in various applications, such as planetary motion and rotating machinery.
The relationship between torque (τ) and angular momentum is given by the equation: net τ = ΔL/Δt, meaning the net torque acting on a system is equal to the rate of change of angular momentum.
Angular momentum plays a significant role in rotational dynamics, similar to how linear momentum is essential for translational motion.In systems where no external torque acts, the initial angular momentum is equal to the final angular momentum, explaining phenomena like gyroscopic motion.A traveling wave on a string can be described by the equation : y = (5.26 ~\text{m}) \cdot \sin \big( (1.65 ~\frac{\text{rad}}{\text{m}})x - (4.64 ~\frac{\text{rad}}{\text{sec}})t +(1.33 ~\text{rad}) \big)y=(5.26 m)⋅sin((1.65 m rad )x−(4.64 sec rad )t+(1.33 rad)) How much time will it take for a peak on this traveling wave to propagate a distance of 5.00 meters along the length of the string?
Answer:
t = 1.77 s
Explanation:
The equation of a traveling wave is
y = A sin [2π (x /λ -t /T)]
where A is the oscillation amplitude, λ the wavelength and T the period
the speed of the wave is constant and is given by
v = λ f
Where the frequency and period are related
f = 1 / T
we substitute
v = λ / T
let's develop the initial equation
y = A sin [(2π / λ) x - (2π / T) t +Ф]
where Ф is a phase constant given by the initial conditions
the equation given in the problem is
y = 5.26 sin (1.65 x - 4.64 t + 1.33)
if we compare the terms of the two equations
2π /λ = 1.65
λ = 2π / 1.65
λ = 3.81 m
2π / T = 4.64
T = 2π / 4.64
T = 1.35 s
we seek the speed of the wave
v = 3.81 / 1.35
v = 2.82 m / s
Since this speed is constant, we use the uniformly moving ratios
v = d / t
t = d / v
t = 5 / 2.82
t = 1.77 s
what is the formula for braking force?
Answer:
Explanation:
The braking force in the context of stopping a vehicle involves the frictional force exerted by the brakes, and it's related to the mass and deceleration of the vehicle. Calculations often use Newton's second law, F = ma, for force, or the work-energy principle, W = F * d * cos(θ), to determine stopping distances.
The formula for braking force isn't provided with a single standard equation, as it relates to several physical quantities, such as friction, mass, and acceleration. However, in the context of vehicle braking, we often analyze the frictional force exerted by the brakes on the car's wheels to determine stopping distance or to calculate the deceleration of the vehicle. The basic physics equation used in this context is Newton's second law, F = ma, where F is the force, m is the mass of the vehicle, and a is the acceleration (which will be negative when braking).
Based on the provided contexts, if a car has a mass (m) and applies a braking force (F), and you want to find the stopping distance (d), you can also use energy equations. The work done by the brakes (W) is equal to the change in the car's kinetic energy, which can be calculated using the equation W = F * d * cos(θ), where θ is the angle at which the force is applied - typically 0 degrees, as the force is in the opposite direction of motion.
A soft-drink manufacturer purchases aluminum cans from an outside vendor. A random sample of 70 cans is selected from a large shipment, and each is tested for strength by applying an increasing load to the side of the can until it punctures. Of the 70 cans, 53 meet the specification for puncture resistance.
Answer:
lol..WHAT IS THE QUESTION?!
Explanation:
A merry-go-round of radius 2.74 m and a moment of inertia of 340 kgm2 rotates without friction. It makes 1 revolution every 4.00 s. A child of mass 25.0 kg sitting at the center crawls out to the rim. Find (a) the new angular speed of the merry-go-round, and (b) the kinetic energy change during this process.
Answer:
a) [tex]\omega_{f} = 1.012\,\frac{rad}{s}[/tex], b) [tex]\Delta K = - 149.352\,J[/tex]
Explanation:
a) The initial angular speed of the merry-go-round is:
[tex]\omega_{o} = \left(\frac{1\,rev}{4\,s}\right)\cdot \left(\frac{2\pi\,rad}{1\,rev} \right)[/tex]
[tex]\omega_{o} \approx 1.571\,\frac{rad}{s}[/tex]
The final angular speed of the merry-go-round is computed with the help of the Principle of Angular Momentum:
[tex]\left(340\,\frac{kg}{m^{2}}\right)\cdot \left(1.571\,\frac{rad}{s} \right) = \left[340\,\frac{kg}{m^{2}}+(25\,kg)\cdot (2.74\,m)^{2} \right]\cdot \omega_{f}[/tex]
[tex]\omega_{f} = 1.012\,\frac{rad}{s}[/tex]
b) The change in kinetic energy is:
[tex]\Delta K = \frac{1}{2}\cdot \left\{\left[340\,\frac{kg}{m^{2}} + \left(25\,kg\right)\cdot \left(2.74\,m\right)^{2}\right]\cdot \left(1.012\,\frac{rad}{s} \right)^{2} - \left(340\,\frac{kg}{m^{2}} \right)\cdot \left(1.571\,\frac{rad}{s} \right)^{2} \right\}[/tex][tex]\Delta K = - 149.352\,J[/tex]
A guitar string vibrates at a frequency of 330Hz with wavelength 1.40m. The frequency and wavelength of this sound wave in air (20 C) when it reaches our ears is: Lower frequency, same wavelength Same frequency, same wavelength Higher frequency, same wavelength Same frequency, shorter wavelength
In the case when the frequency and wavelength of this sound wave in air (20 C) so it should be Same frequency, shorter wavelength.
Speed of wave:here the speed of the wave should be provided by the following equation.
v = fλ
λ = v/f
here,
f = Frequency
λ = Wavelength
Also, the wavelength is directly proportional to the velocity.
So, the frequency remains same and the wavelength shortens.
So based on this we can say that In the case when the frequency and wavelength of this sound wave in air (20 C) so it should be Same frequency, shorter wavelength.
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Final answer:
The sound wave produced by a vibrating guitar string will maintain the same frequency as it travels through the air to our ears because the frequency of a wave depends on its source and doesn't change in a uniform medium. Although the wavelength can change in different mediums, the question context does not specify a medium change; hence, the frequency and wavelength when heard would be the same as produced by the string.
Explanation:
The question is about the behavior of sound waves produced by a vibrating guitar string when they travel through the air. Specifically, it examines how the frequency and wavelength of the sound wave change once it reaches the listener's ear compared to when it is produced by the string.
It's important to note that when a guitar string vibrates at a certain frequency, it creates sound waves that propagate through the air with the same frequency. This is because the frequency of a wave is determined by its source and remains constant as it travels through a uniform medium. However, the wavelength may change if the speed of the wave changes, which depends on the medium's properties.
In the context of sound waves in air at 20°C, the speed of sound is approximately 343 m/s. Considering this, the frequency of the sound wave when it reaches our ears remains the same as the frequency of the vibrating string, because the frequency of a wave only changes if the speed of the wave's source relative to the observer changes, which is not the case here.
The wavelength of the sound wave in the air might differ from the wavelength on the string due to the different speeds of wave propagation in different mediums (the string versus air), but for the purpose of this specific question, such details are beyond the scope provided.
Therefore, the correct answer is 'Same frequency, same wavelength', under the assumption that both the measurements were taken in the air or that the provided wavelength already accounts for the transition from the string to the air.
The switch in the circuit has been in the left position for a long time. At t=0 it moves to the right position and stays there.a. Write the expression for the capacitor voltage v(t), fort≥0. b. Write the expression for the current through the 2.4kΩ resistor, i(t), fort≥0+.
Answer:
Pls refer to attached file
Explanation:
When the temperature is at 30∘C, the A-36 steel pipe fits snugly between the two fuel tanks. When fuel flows through the pipe, the temperature at ends A and B rise to 130∘C and 80∘C, respectively. If the temperarture drop along the pipe is linear, determine the average normal stress developed in the pipe. Assume the walls of each tank acts as a spring, each having a stiffness of k=900 MN/m.
In a system where an A-36 steel pipe is snugly fit between two fuel tanks, the rise in fuel temperature will cause thermal stresses in the pipe due to the restraint from the fuel tanks. This is related to the thermal properties of the material and the rate of change due to temperature rising, from which the average normal stress can be calculated.
Explanation:Determining the average normal stress developed in an A-36 steel pipe when fuel flows through it at varying temperatures requires knowledge of thermal expansion and associated stress in materials. This is related to thermal properties and the rate of change due to temperature rise.
When temperature rises along the pipe such as mentioned, from room temperature to 130∘C and 80∘C, the steel pipe expands. However, the pipe is restrained by the fuel tanks acting as springs, leading to development of stress within the pipe. The average normal stress can be calculated by dividing the force exerted by the expansion (or contraction) by the cross-sectional area of the pipe:
F/A = σ
Where, F is the force and A is the cross-sectional area of the pipe. The force can be obtained from Hooke's law for springs (F = k Δx), where k is the stiffness of the tank walls acting as springs, and Δx is the change in length due to thermal expansion.
The average normal stress is a measure of the extent to which the pipe is going through physical changes due to thermal variations.
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Consider the steel spring in the illustration.
(a) Find the pitch, solid height, and number of active turns.
(b) Find the spring rate. Assume the material is A227 HD steel.
(c) Find the force Fs required to close the spring solid.
(d) Find the shear stress in the spring due to the force Fs.
Answer:
Explanation:
find the answer below
Pease circle the statements incompatible with the Kelvin-Planck Statement. (A) No heat engine can have a thermal efficiency of 100%. (B) It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work without rejecting waste heat to a cool reservoir. (C) The impossibility of having a 100% efficient heat engine is always due to friction or other dissipative effects such as the system is perfectly designed or the material needed for the system design is not available. (D) Any device that violates the Kelvin-Planck statement also violates the Clausius statement, and vice versa.
Answer:
(C) The impossibility of having a 100% efficient heat engine is always due to friction or other dissipative effects such as the system is perfectly designed or the material needed for the system design is not available.
Explanation:
The above option was never stated in the law
Find the corresponding initial value problem of the following spring-mass systems, that if solved, would give the position u of the mass at any time t. Use 9.8 m/s 2 for the acceleration due to gravity. DO NOT SOLVE THE IVP. A mass of 60 kg stretches a spring 19.6 m. The mass is acted on by an external force of 6 sin(4t) N and moves in a medium that imparts a viscous force of 8 N when the speed of the mass is 16 cm/s. The mass is pushed up 5 cm and set in motion an initial downward velocity of 2 cm/s. Write the IVP so that u would be in meters if solved.
Answer:
Explanation:
The picture attached shows the full explanation
The resistance of physiological tissues is quite variable. The resistance of the internal tissues of humans, primarily composed of salty solutions, is quite low. Here the resistance between two internal points in the body is on the order of 100 ohms. Dry skin, however, can have a very high resistance, with values ranging from thousands to hundreds of thousands of ohms. However, if skin is wet, it is far more conductive, and so even contact with small voltages can create large, dangerous currents though a human body. (For example, although there is no specific minimum current that is lethal, currents generally exceeding a couple tenths of Amps may be deadly.)
Assuming that electrocution can be prevented if currents are kept below 0.1 A, and assuming the resistance of dry skin is 100,000 ohms, what is the maximum voltage with which a person could come into contact while avoiding electrocution? (Of course, all bets are off and things become far more dangerous if this person's skin is wet, which can reduce the resistance by more than a factor of 100.)
Given Information:
Current = I = 0.1 A
Resistance = R = 100 kΩ
Required Information:
Voltage = V = ?
Answer:
Voltage = V = 1000 V
Step-by-step explanation:
We know that electrocution depends upon the amount of current flowing through the body and the voltage across the body.
V = IR
Where I is the current flowing through the body and R is the resistance of body.
If electrocution can be avoided when the current is below 0.1 A then
V = 0.1*10×10³
V = 1000 Volts
Therefore, 1000 V is the maximum voltage with which a person could come into contact while avoiding electrocution, any voltage more than 1000 V may result in fatal electrocution.
Also note that human body has very low resistance when the body is wet therefore, above calculated value would not be applicable in such case.
A light beam is directed parallel to the axis of a hollow cylindrical tube. When the tube contains only air, the light takes 8.72 ns to travel the length of the tube, but when the tube is filled with a transparent jelly, the light takes 1.82 ns longer to travel its length. What is the refractive index of this jelly?
Answer:
1.208
Explanation:
L = Length of tube
c = Speed of light in air
v = Speed of light in jelly
Time taken by light in tube
[tex]\dfrac{L}{c}=8.72[/tex]
Time taken when jelly is present
[tex]\dfrac{L}{v}=8.72+1.82\\\Rightarrow \dfrac{L}{v}=10.54\ ns[/tex]
Dividing the above equations we get
[tex]\dfrac{v}{c}=\dfrac{8.72}{10.54}\\\Rightarrow \dfrac{c}{v}=\dfrac{10.54}{8.72}\\\Rightarrow n=1.208[/tex]
The refractive index of the jelly is 1.208
The power liens that run through your neighborhood carry alternating currents that reverse direction 120 times per second. As current changes so does magnetic field. If you put a loop of wire up near the power line to extract power by tapping the magnetic field, sketch how you would orient the coil of wire next to a power line to develop the max emf in the coilo If magnetic flux through a loop changes, induced emf is produced
o If the area of the loop is parallel to the field, the flux through the loop is minimum
• So no emf is produced
o To get the most of the flux through the loop you place it closer to the power lines and in orientations so the plane of the loop also contains power lines
• Flux would be max and result in greater induced emf as field goes from max to zero to max and then in other direction
Answer:
the normal to the area of the loop parallel to the wire to induce the maximum electromotive force
Explanation:
Faraday's law is
ε = - dΦ / dt
where Ф magnetic flow
the flow is
Ф = B. dA = B dA cos θ
therefore, to obtain the maximum energy, the cosine function must be maximum, for this the direction of the fluctuating magnetic field and the normal direction to the area must be parallel.
The magnetic field in a cable through which current flows is circular, so the loop must be perpendicular to the wire, this is the normal to the area of the loop parallel to the wire to induce the maximum electromotive force
an outline is shown in the attachment
the correct answer is b
Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, the first having a mass of 260000 kg and a velocity of 0.32 m/s in the horizontal direction, and the second having a mass of 52500 kg and a velocity of -0.14 m/s in the horizontal direction. What is their final velocity?
Answer:
0.243 m/s
Explanation:
From law of conservation of motion,
mu+m'u' = V(m+m')................. Equation 1
Where m = mass of the first car, m' = mass of the second car, initial velocity of the first car, u' = initial velocity of the second car, V = Final velocity of both cars.
make V the subject of the equation
V = (mu+m'u')/(m+m')................. Equation 2
Given: m = 260000 kg, u = 0.32 m/s, m' = 52500 kg, u' = -0.14 m/s
Substitute into equation 2
V = (260000×0.32+52500×(-0.14))/(260000+52500)
V = (83200-7350)/312500
V = 75850/312500
V = 0.243 m/s
A magnetic field is uniform over a flat, horizontal circular region with a radius of 1.50 mm, and the field varies with time. Initially the field is zero and then changes to 1.50 T, pointing upward when viewed from above, perpendicular to the circular plane, in a time of 125 ms
what is the average induced emf around the border of the circular region?
Answer:
The average induced emf around the border of the circular region is [tex]8.48\times 10^{-5}\ V[/tex].
Explanation:
Given that,
Radius of circular region, r = 1.5 mm
Initial magnetic field, B = 0
Final magnetic field, B' = 1.5 T
The magnetic field is pointing upward when viewed from above, perpendicular to the circular plane in a time of 125 ms. We need to find the average induced emf around the border of the circular region. It is given by the rate of change of magnetic flux as :
[tex]\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=A\dfrac{-d(B'-B)}{dt}\\\\\epsilon=\pi (1.5\times 10^{-3})^2\times \dfrac{1.5}{0.125}\\\\\epsilon=8.48\times 10^{-5}\ V[/tex]
So, the average induced emf around the border of the circular region is [tex]8.48\times 10^{-5}\ V[/tex].
Answer:
[tex]84.8\times 10^{-6} V[/tex]
Explanation:
We are given that
Radius,r=1.5 mm=[tex]1.5\times 10^{-3} m[/tex]
[tex]1mm=10^{-3} m[/tex]
Initial magnetic field,[tex]B_0=0[/tex]
Final magnetic field,[tex]B=1.5 T[/tex]
Time,[tex]\Delta t=[/tex]125 ms=[tex]125\times 10^{-3} s[/tex]
[tex]1 ms=10^{-3}s [/tex]
We know that average induced emf
[tex]E=\frac{d\phi}{dt}=-\frac{d(BA)}{dt}=A\frac{dB}{dt}=A\frac{(B-B_0)}{dt}[/tex]
Substitute the values
[tex]E_{avg}=\pi (1.5\times 10^{-3})^2\times \frac{1.5-0}{125\times 10^{-3}}[/tex]
[tex]E_{avg}=84.8\times 10^{-6} V[/tex]
Frequently in physics, one makes simplifying approximations. A common one in electricity is the notion of infinite charged sheets. This approximation is useful when a problem deals with points whose distance from a finite charged sheet is small compared to the size of the sheet. In this problem, you will look at the electric field from two finite sheets and compare it to the results for infinite sheets to get a better idea of when this approximation is valid. What is the magnitude E of the electric field at the point on thex axis with x coordinate a/2?
Answer:
Explanation:
solution found below
The electric field at a midpoint between two finite charged sheets can be calculated using Gauss's Law. While the field strength would theoretically be zero if the sheets were infinite, because real sheets are finite, the field will not be exactly zero.
Explanation:This question concerns the physics concept of electric fields originating from two finite charged sheets. Given a point 'a/2' on the x-axis in-between these sheets, the magnitude E of the electric field can be calculated using Gauss's Law. The electric field E due to an infinite sheet of charge is given by σ/2ε₀ (σ is the charge density), which is independent of the distance from the sheet. Hence, for a point which is midway between two such sheets, and each sheet carries charge of opposite polarity, the field strength at that point will be zero. However, since real sheets will not be infinitely large, the field will not be exactly zero. The exact value will depend on the exact geometry of the problem and on the distance from the sheets to the point.
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In an alcohol-in-glass thermometer, the alcohol column has length 12.66 cm at 0.0 ∘C and length 22.49 cm at 100.0 ∘C. Part A What is the temperature if the column has length 18.77 cm ? Express your answer using four significant figures. T = nothing ∘C Request Answer Part B What is the temperature if the column has length 14.23 cm ? Express your answer using four significant figures.
Answer:
[tex]62.1566632757\ ^{\circ}C[/tex]
[tex]15.9715157681\ ^{\circ}C[/tex]
Explanation:
[tex]\Delta T[/tex] = Change in termperature
[tex]\Delta L[/tex] = Change in length
We have the relation
[tex]\dfrac{\Delta L}{\Delta T}=\dfrac{22.49-12.66}{100-0}=\dfrac{18.77-12.66}{t-0}\\\Rightarrow t=\dfrac{18.77-12.66}{0.0983}\\\Rightarrow t=62.1566632757\ ^{\circ}C[/tex]
The temperature is [tex]62.1566632757\ ^{\circ}C[/tex]
[tex]\dfrac{\Delta L}{\Delta T}=\dfrac{22.49-12.66}{100-0}=\dfrac{14.23-12.66}{t-0}\\\Rightarrow t=\dfrac{14.23-12.66}{0.0983}\\\Rightarrow t=15.9715157681\ ^{\circ}C[/tex]
The temperature is [tex]15.9715157681\ ^{\circ}C[/tex]
(A)
According to the question,
The change in length will be:
= [tex]l_2-l_1[/tex]
= [tex]22.49-12.66[/tex]
= [tex]9.83 \ cm[/tex]
The change per degree will be:
= [tex]\frac{Change \ in \ length}{Temperature}[/tex]
= [tex]\frac{9.83}{100}[/tex]
= [tex]0.0983 \ cm/deg[/tex]
Now,
The change in length,
= [tex]18.77-12.66[/tex]
= [tex]6.11 \ cm[/tex]
hence,
The temperature,
= [tex]\frac{6.11}{0.0983}[/tex]
= [tex]62.2^{\circ} C[/tex]
(B)
The change in length,
= [tex]14.23-12.66[/tex]
= [tex]1.57 \ cm[/tex]
hence,
The temperature will be:
= [tex]\frac{1.57}{0.0983}[/tex]
= [tex]15.98^{\circ} C[/tex]
Thus the above answers are correct.
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A rectangular neoprene sheet has width W = 1.00 m and length L = 4.00 m. The two shorter edges are affixed to rigid steel bars that are used to stretch the sheet taut and horizontal. The force applied to either end of the sheet is F = 81.0 N. The sheet has a total mass M = 4.00 kg. The left edge of the sheet is wiggled vertically in a uniform sinusoidal motion with amplitude A = 10.0 cm and frequency f = 1.00 Hz. This sends waves spanning the width of the sheet rippling from left to right. The right side of the sheet moves upward and downward freely as these waves complete their traversal.
(A) Derive an expression for the velocity with which the waves move along the sheet. Express your answer in terms of the variables F, L, and M.
(B) What is the value of this speed for the specified choices of these parameters? Express your answer with the appropriate units.
Answer:
(a) [tex]\sqrt{FL/M}[/tex]
(b) [tex]9 ms^{-1}[/tex]
Explanation:
(a)
The speed of the wave depends on the type of the material. Here we have neoprene sheet so the speed of the wave will depend on the linear density of neoprene sheet. Linear Density is defined as Mass per unit length of the material (as materials of same type can have different thickness). The symbol used for the Linear Density is .
μ = Mass of the sheet / Length of the sheet
The speed of the wave in such material can be be found by using the equation:
[tex]\frac{1}{v^{2} } = {\frac{Linear Density}{Force} }[/tex] (where v is speed)
The equation can be rearranged:
v = [tex]\sqrt{Force/Linear Density}[/tex]
so,
v = sqrt(F/μ)
v = sqrt(F/ (M/L))
v = [tex]\sqrt{FL/M}[/tex] (answer)
(b)
Putting the values
F = 81 N
M = 4kg
L = 4m
v = [tex]\sqrt{FL/M}[/tex]
v = 9m/s
When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less damage. In one such accident, a 1750kg car traveling to the right at 1.60m/s collides with a 1450kg car going to the left at 1.10m/s . Measurements show that the heavier car's speed just after the collision was 0.260m/s in its original direction. You can ignore any road friction during the collision.
A:What was the speed of the lighter car just after the collision?
B:Calculate the change in the combined kinetic energy of the two-car system during this collision.
Answer:
Explanation:
We shall apply law of conservation of momentum .
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ , m₁ ,m₂ are masses of two bodies colliding with velocities u₁ and u₂ respectively . v₁ and v₂ are their velocities after collision.
m₁ = 1750 , m₂ = 1450 , u₁ = 1.6 m /s , u₂ = - 1.1 m /s , v₁ = .26 m /s
substituting the values
1750 x 1.6 + 1450 x - 1.1 = 1750 x .26 + 1450 v₂
2800 - 1595 = 455 + 1450v₂
1450v₂ = 750
v₂ = .517 m /s
B ) initial kinetic energy
= 1/2 x 1750 x 1.6²+ 1/2 x 1450 x -1.1²
= 2240+ 877.25
= 3117.25 J
final kinetic energy
= 1/2 x 1750 x .26²+ 1/2 x 1450 x .517²
= 59.15 + 193.78
= 252.93
loss of kinetic energy
= 3117.25 - 252.93
= 2864.32 J
Double the resistance of the resistor by changing it from 10 Ω to 20 Ω. What happens to the current flowing through the circuit?
Answer:
2 amps
Explanation:
you just divide
A spaceship from a friendly, extragalactic planet flies toward Earth at 0.203 0.203 times the speed of light and shines a powerful laser beam toward Earth to signal its approach. The emitted wavelength of the laser light is 691 nm . 691 nm. Find the light's observed wavelength on Earth.
Answer:
567.321nm
Explanation:
See attached handwritten document for more details
Answer:
The observed wavelenght on earth will be 5.51x10^-7 m
Explanation:
Speed of light c = 3x10^8 m/s
Ship speed u = 0.203 x 3x10^8 = 60900000 = 6.09x10^7 m/s
Wavelenght of laser w = 691x10^-9 m
Observed wavelenght w' = w(1 - u/c)
u/c = 0.203
w' = 691x10^-9(1 - 0.203)
w' = 5.51x10^-7 m
Point charges 1 mC and −2 mC are located at (3, 2, −1) and (−1, −1, 4), respectively. Calculate the electric force on a 10 nC charge located at (0, 3, 1) and the electric field intensity at that point.
Answer:
See attached handwritten document for answer
Explanation:
Answer:
Explanation:
a) You can compute the force by using the expression:
[tex]F=k\frac{q_1q_2}{[(x-x_o)^2+(y-y_o)^2+(z-z_o)^2]^{\frac{1}{2}}}[/tex]
where k=8.98*10^9Nm^2/C^2 and q1, q2 are the charges. By replacing for the forces you obtain:
[tex]F_T=k[\frac{(1*10^{-3}C)(10*10^{-9}C)}{[(3-0)^2+(2-3)^2+(-1-1)^2]}}][(3-0)\hat{i}+(2-3)\hat{j}+(-1-1)\hat{k}]\\\\ \ \ \ \ +k[\frac{(-2*10^{-3}C)(10*10^{-9}C)}{[(-1-0)^2+(-1-3)^2+(4-1)^2]}}][(-1-0)\hat{i}+(-1-3)\hat{j}+(4-1)\hat{k}]\\\\F_T=6.41*10^{-3}N[3i-j-2k]-6.9*10^{-3}N[-1i-4j+3k]\\\\=0.026N\hat{i}-0.034N\hat{j}-0.033\hat{k}[/tex]
b)
[tex]E=k[\frac{1*10^{-3}C}{[(3-0)^2+(2-3)^2+(-1-1)^2]}]+k[\frac{(-2*10^{-3}C)}{[(-1-0)^2+(-1-3)^2+(4-1)^2]}]\\\\E=641428.5N/C+690769.23N/C=1332197.73N/C[/tex]
[tex]E=k[\frac{1*10^{-3}C}{[(3-0)^2+(2-3)^2+(-1-1)^2]}][3i-2j-2k]+\\\\k[\frac{(-2*10^{-3}C)}{[(-1-0)^2+(-1-3)^2+(4-1)^2]}][-1i-4j+3k]\\\\E=641428.5N/C[3i-2j-2k]-690769.23N/C[-1i-4j+3k]=2615054.7i+1480219.92j-4973626.23k\\\\|E|=\sqrt{(E_x)^2+(E_y)^2+(E_z)^2}=5810896.56N/C[/tex]
where we you have used that E=kq/r^2
A centrifuge in a forensics laboratory rotates at an angular speed of 3,700 rev/min. When switched off, it rotates 46.0 times before coming to rest. Find the constant angular acceleration of the centrifuge (in rad/s2). Consider the direction of the initial angular velocity to be the positive direction, and include the appropriate sign in your result.
Answer:
Explanation:
Given,
initial angular speed, ω = 3,700 rev/min
= [tex]3700\times \dfrac{2\pi}{60}=387.27\ rad/s[/tex]
final angular speed = 0 rad/s
Number of time it rotates= 46 times
angular displacement, θ = 2π x 46 = 92 π
Angular acceleration
[tex]\alpha = \dfrac{\omega_f^2 - \omega^2}{2\theta}[/tex]
[tex]\alpha = \dfrac{0 - 387.27^2}{2\times 92\ pi}[/tex]
[tex]\alpha = -259.28 rad/s^2[/tex]
A standard door into a house rotates about a vertical axis through one side, as defined by the door's hinges. A uniform magnetic field is parallel to the ground and perpendicular to the axis. Through what angle must the door rotate so that the magnetic flux that passes through it decreases from its maximum value to 1/4 of its maximum value?
Answer:
75.5degrees
Explanation:
Magnitude of magnetic flux= BA
If rotated through an angle= BAcos theta
= (0.25)BA=BAcos theta
= costheta= 0.25
= theta= cos^-1 0.25
=75.5degrees
A wheel moves in the xy plane in such a way that the location of its center is given by the equations xo = 12t3 and yo = R = 2, where xo and yo are measured in feet and t is measured in seconds. The angular displacement of a radial line measured from a vertical reference line is θ = 8t4, where θ is measured in radians. Determine the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s.
Answer:
the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s is [tex]P = 104.04 \hat{i} -314.432 \hat{j}[/tex]
Explanation:
The free-body diagram below shows the interpretation of the question; from the diagram , the wheel that is rolling in a clockwise directio will have two velocities at point P;
the peripheral velocity that is directed downward [tex](-V_y)[/tex] along the y-axisthe linear velocity [tex](V_x)[/tex] that is directed along the x-axisNow;
[tex]V_x = \frac{d}{dt}(12t^3+2) = 36 t^2[/tex]
[tex]V_x = 36(1.7)^2\\\\V_x = 104.04\ ft/s[/tex]
Also,
[tex]-V_y = R* \omega[/tex]
where [tex]\omega[/tex](angular velocity) = [tex]\frac{d\theta}{dt} = \frac{d}{dt}(8t^4)[/tex]
[tex]-V_y = 2*32t^3)\\\\\\-V_y = 2*32(1.7^3)\\\\-V_y = 314.432 \ ft/s[/tex]
∴ the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s is [tex]P = 104.04 \hat{i} -314.432 \hat{j}[/tex]
In 1958, Meselson and Stahl conducted an experiment to determine which of the three proposed methods of DNA replication was correct. Identify the three proposed models for DNA replication. Conservative Semiconservative Dispersive Answer Bank The Meselson and Stahl experiment starts with E. coli containing 15 N / 15 N labeled DNA grown in 14 N media. Which result did Meselson and Stahl observe by sedimentation equilibrium centrifugation to provide strong evidence for the semiconservative model of DNA replication? Both the first and second generation have both 15 N / 15 N DNA and 14 N / 14 N DNA. No hybrid 15 N / 14 N DNA was observed. The first generation has hybrid 15 N / 14 N DNA and the second generation has both hybrid 15 N / 14 N DNA and 14 N / 14 N DNA. No 15 N / 15 N DNA was observed. The first generation has hybrid 15 N / 14 N DNA and the second generation has hybrid 15 N / 14 N DNA. No 15 N / 15 N DNA nor 14 N / 14 N DNA was observed.
Answer:
Explanation:
The original has hybrid 15N/14N DNA, and the second generation has both hybrid 15N/14N DNA and 14N/14N DNA. No 15N/15N DNA was observed. In this experiment:
Nitrogen is a significant component of DNA. 14N is the most bounteous isotope of nitrogen, however, DNA with the heavier yet non-radioactive and 15N isotope is likewise practical.
E. coli was developed for several generations in a medium containing NH4Cl with 15N. When DNA is extracted from these cells and centrifuged on a salt density gradient, the DNA separates at which its density equals to the salt arrangement. The DNA of the cells developed in 15N medium had a higher density than cells developed in typical 14N medium. After that, E. coli cells with just 15N in their DNA were transferred to a 14N medium.
DNA was removed and compared to pure 14N DNA and 15N DNA. Immediately after only one replication, the DNA was found to have an intermediate density. Since conservative replication would result in equal measures of DNA of the higher and lower densities yet no DNA of an intermediate density, conservative replication was eliminated. Moreso, this result was consistent with both semi-conservative and dispersive replication. Semi conservative replication would result in double-stranded DNA with one strand of 15N DNA, and one of 14N DNA, while dispersive replication would result in double-stranded DNA with the two strands having mixtures of 15N and 14N DNA, either of which would have appeared as DNA of an intermediate density.
The DNA from cells after two replications had been completed and found to comprise of equal measures of DNA with two different densities, one corresponding to the intermediate density of DNA of cells developed for just a single division in 14N medium, the other corresponding to DNA from cells developed completely in 14N medium. This was inconsistent with dispersive replication, which would have resulted in a single density, lower than the intermediate density of the one-generation cells, yet at the same time higher than cells become distinctly in 14N DNA medium, as the first 15N DNA would have been part evenly among all DNA strands. The result was steady with the semi-conservative replication hypothesis. The semi conservative hypothesis calculates that each molecule after replication will contain one old and one new strand. The dispersive model suggests that each strand of each new molecule will possess a mixture of old and new DNA.
Two disks are mounted (like a merry-go-round) on low-friction bearings on the same axle and can be brought together so that they couple and rotate as one unit. The first disk, with rotational inertia 3.76 kg·m2 about its central axis, is set spinning counterclockwise (which may be taken as the positive direction) at 436 rev/min. The second disk, with rotational inertia 9.20 kg·m2 about its central axis, is set spinning counterclockwise at 953 rev/min. They then couple together. (a) What is their angular speed after coupling? If instead the second disk is set spinning clockwise at 953 rev/min, what are their (b) angular velocity (using the correct sign for direction) and (c) direction of rotation after they couple together?
Answer:
A) ω = 13.38 rev/s
B) ω = 9.167 rev/s
C) In clockwise direction
Explanation:
We are given;
Rotational Inertia of first disk; I_1 = 3.76 kg·m²
Angular velocity of first disk; ω_1 = 436 rev/min = 7.267 rev/s
Rotational Inertia of second disk; I_2 = 9.2 kg·m²
Angular velocity of second disk; ω_2 = 953 rev/min = 15.883 rev/s
Total rotational inertia is;
I_total = I_1 + I_2
I_total = 3.76 + 9.2 = 12.96 kg·m²
Now; total angular momentum will be;
L_total = L_1 + L_2
Where L_1 is angular momentum of first disk and L_2 is angular momentum of second disk.
Thus;
I_total•ω = I_1•ω_1 + I_2•ω_2
Plugging relevant values in, we can find their angular speed after coupling which is ω.
Thus;
12.96ω = (3.76 x 7.267) + (9.2 x 15.883)
12.96ω = 173.44752
ω = 173.44752/12.96
ω = 13.38 rev/s
B) since second disk is now spinning clockwise, thus;
I_total•ω = I_1•ω_1 - I_2•ω_2
12.96ω = (3.76 x 7.267) - (9.2 x 15.883)
12.96ω = -118.8
ω = -118.8/12.96
ω = -9.167 rev/s
The negative sign tells us that it is clockwise.
So we would say ω = 9.167 rev/s in clockwise direction
One consequence of turbulence is mixing. Different layers of fluid flow cross over one another very easily, and get blended together. This is another kind of "transport", which lets atoms which might have started out all in one place get uniformly mixed around. Would you expect turbulent mixing to happen most easily in:
A. waterB. motor oilC. airD. honey
Answer:
Turbulence mixing will happen mostly in air
Explanation:
A 2.13-kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 5.00 N/m. The object is displaced 3.54 m to the right from its equilibrium position and then released, initiating simple harmonic motion. (a) What is the force (magnitude and direction) acting on the object 3.50 s after it is released
Answer:
17.54N in -x direction.
Explanation:
Amplitude (A) = 3.54m
Force constant (k) = 5N/m
Mass (m) = 2.13kg
Angular frequency ω = √(k/m)
ω = √(5/2.13)
ω = 1.53 rad/s
The force acting on the object F(t) = ?
F(t) = -mAω²cos(ωt)
F(t) = -2.13 * 3.54 * (1.53)² * cos (1.53 * 3.50)
F(t) = -17.65 * cos (5.355)
F(t) = -17.57N
The force is 17.57 in -x direction
Which of the following is NOT a cause for the intervention of antitrust authorities?
a.
Companies abuse their market power by acquiring new firms that allow for the increase of prices above a level that would occur in a competitive market.
b.
The merger between two companies allows for several customer options and substantial competition within the industry.
c.
An acquisition that creates too much consolidation and increases the potential for future abuse of market power.
d.
A company cuts prices when a new competitor enters the industry to force the competitor out of business.
e.
Dominant companies use their market power to crush potential competitors.
Answer:
b.
The merger between two companies allows for several customer options and substantial competition within the industry.
Explanation:
Antitrust authorities will only come in if the merger was to remove competition and reduce consumer options.
Final answer:
Antitrust authorities intervene when firms engage in practices that reduce competition. These practices include abusive use of market power, anticompetitive mergers and acquisitions, and restrictive practices. A merger enhancing competition does NOT warrant such intervention. Thus, option B is correct.
Explanation:
The intervention of antitrust authorities is generally warranted when actions are taken by firms that reduce competition and harm the economic ideal of a competitive market. These actions may include abusive use of market power, mergers and acquisitions that significantly increase market concentration and reduce competition, and various restrictive practices that limit competition, such as tie-in sales, bundling, and predatory pricing.
However, not all actions that affect competition trigger antitrust intervention. For example, if the merger between two companies allows for several customer options and substantial competition within the industry, that would not be a cause for concern for antitrust authorities. In this case, the merger does not lead to abusive market power or reduce competition, and therefore, option b 'The merger between two companies allows for several customer options and substantial competition within the industry' would be the correct answer as it is NOT a cause for intervention by antitrust authorities.