A cylinder of radius R, length L, and mass M is released from rest on a slope inclined at angle θ. It is oriented to roll straight down the slope. If the slope were frictionless, the cylinder would slide down the slope without rotating. What minimum coefficient of static friction is needed for the cylinder to roll down without slipping?
Answers
Answer:
[tex]\mu_s=\frac{1}{3}\tan \theta[/tex]
Explanation:
Let the minimum coefficient of static friction be [tex]\mu_s[/tex].
Given:
Mass of the cylinder = [tex]M[/tex]
Radius of the cylinder = [tex]R[/tex]
Length of the cylinder = [tex]L[/tex]
Angle of inclination = [tex]\theta[/tex]
Initial velocity of the cylinder (Released from rest) = 0
Since, the cylinder is translating and rolling down the incline, it has both translational and rotational motion. So, we need to consider the effect of moment of Inertia also.
We know that, for a rolling object, torque acting on it is given as the product of moment of inertia and its angular acceleration. So,
[tex]\tau =I\alpha[/tex]
Now, angular acceleration is given as:
[tex]\alpha = \frac{a}{R}\\Where, a\rightarrow \textrm{linear acceleration of the cylinder}[/tex]
Also, moment of inertia for a cylinder is given as:
[tex]I=\frac{MR^2}{2}[/tex]
Therefore, the torque acting on the cylinder can be rewritten as:
[tex]\tau = \frac{MR^2}{2}\times \frac{a}{R}=\frac{MRa}{2}------ 1[/tex]
Consider the free body diagram of the cylinder on the incline. The forces acting along the incline are [tex]mg\sin \theta\ and\ f[/tex]. The net force acting along the incline is given as:
[tex]F_{net}=Mg\sin \theta-f\\But,\ f=\mu_s N\\So, F_{net}=Mg\sin \theta -\mu_s N-------- 2[/tex]
Now, consider the forces acting perpendicular to the incline. As there is no motion in the perpendicular direction, net force is zero.
So, [tex]N=Mg\cos \theta[/tex]
Plugging in [tex]N=Mg\cos \theta[/tex] in equation (2), we get
[tex]F_{net}=Mg\sin \theta -\mu_s Mg\cos \theta\\F_{net}=Mg(\sin \theta-\mu_s \cos \theta)--------------3[/tex]
Now, as per Newton's second law,
[tex]F_{net}=Ma\\Mg(\sin \theta-\mu_s \cos \theta)=Ma\\\therefore a=g(\sin \theta-\mu_s \cos \theta)------4[/tex]
Now, torque acting on the cylinder is provided by the frictional force and is given as the product of frictional force and radius of the cylinder.
[tex]\tau=fR\\\frac{MRa}{2}=\mu_sMg\cos \theta\times R\\\\a=2\times \mu_sg\cos \theta\\\\But, a=g(\sin \theta-\mu_s \cos \theta)\\\\\therefore g(\sin \theta-\mu_s \cos \theta)=2\times \mu_sg\cos \theta\\\\\sin \theta-\mu_s \cos \theta=2\mu_s\cos \theta\\\\\sin \theta=2\mu_s\cos \theta+\mu_s\cos \theta\\\\\sin \theta=3\mu_s \cos \theta\\\\\mu_s=\frac{\sin \theta}{3\cos \theta}\\\\\mu_s=\frac{1}{3}\tan \theta............(\because \frac{\sin \theta}{\cos \theta}=\tan \theta)[/tex]
Therefore, the minimum coefficient of static friction needed for the cylinder to roll down without slipping is given as:
[tex]\mu_s=\frac{1}{3}\tan \theta[/tex]
The minimum coefficient of static friction that needed for cylinder to roll down without slipping is [tex]\mu_s= \frac{tan\theta}{3}[/tex]
Explanation:A cylinder of radius R, length L, and mass M is released from rest on a slope inclined at angle θ. It is oriented to roll straight down the slope. If the slope were frictionless, the cylinder would slide down the slope without rotating. What minimum coefficient of static friction is needed for the cylinder to roll down without slipping?
Given: radius R, length L, angle θ, and mass M
We need to calcuate the minimum static friction coefficient. It is useful so the cylinder will roll without slipping down the incline. The cylinder is also released from rest. As the cylinder is rolling, we have to consider the moment of inertia. Rolling of cylinder is happened due to the friction force
By applying Newton law of motion
[tex]F = M a\\\tau = I \alpha\\\tau = I \frac{a}{R} \\\tau = \frac{1}{2} M R^2 \frac{a}{R}[/tex]
From diagram
[tex]Mg sin\theta - f_{fr} = Ma\\f_{fr} = \mu_s N\\f_{fr} = \mu_s Mg cos \theta\\a = g sin \theta - \mu_s cos \theta[/tex]
Then also
[tex]\tau = f_{fr} R\\f_{fr} = \frac{Ma}{2} \\\mu_s Mg cos \theta = \frac{Mg (sin\theta - \mu_s cos \theta)}{2} \\\frac{3}{2} \mu_s cos\theta = \frac{sin\theta}{2}\\ \mu_s = \frac{tan\theta}{3}[/tex]
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Related Questions
Find the work done by the force F = xyi +(y-x)j over the straight line from (-1,1)to (3,-3). The amount of work done is ___?
Answers
Answer:
amount of work done is[tex] W = \frac{-4}{3}[/tex]
Explanation:
Formula for work done by force field
[tex]W = \int F. dr = \int_{t_o}^{t_1} f(r(t)) r'(t) dt[/tex]
where
r(t) is parametrization of line
as it is straight line so
[tex]r(t) = 1- t) r_o + tr_1 0 \leq t \leq 1[/tex]
thus,
r(t) = (1-t)(-1,1) + t(3,-3)
= (-1+t,1-t) + (3t - 3t)
= (-1+t +3t, 1-t-3t)
r(t) = (4t -1, 1- 4t)
r'(t) = (4,-4)
putting value in above integral
[tex]\int_{0}^{1} ((4t -1,1-4t)). (4,4) dt = \int_{0}^{1} (-(4t -1)^2 , 2-8t).(4,-4) dt[/tex]
[tex]= \int_{0}^{1} (-16 t^2 + 8t -1,2-8t) .(4,-4) dt[/tex]
[tex] =\int_{0}^{1} (4(-16t^2 +8t -1) -4(2-8t)) dt[/tex]
[tex]= 4[ -16 \frac{t^3}{3} + 16\frac{t^2}{2} - 3t]_{0}^{1}[/tex]
[tex]\int_{0}^{1} ((4t -1,1-4t)). (4,4) dt = \frac{-4}{3}[/tex]
To find the work done by the force F = xyi +(y-x)j over the straight line from (-1,1) to (3,-3), we compute the line integral of the dot product of the force and displacement vectors along the path. Performing the integrations using the limits set by the line's endpoints gives us the work done.
Explanation:The task involves physics concepts in vector calculus. Specifically, we're looking at the work done by a force vector over a path in two-dimensional space. Here, the force is described by the vector function F = xyi +(y-x)j. We need to calculate the work done by this force as it moves along a straight line path from (-1,1) to (3,-3).
Work done by a force in moving an object is given by the line integral of force · displacement. In mathematical terms, this definition reads as W = ∫F.dr. The computation requires us to take the dot product of the force and differential displacement vectors along the path.
The direction of displacement as we go from (-1,1) to (3,-3) is simply the path's tangent. Therefore, dr = dx i + dy j, where (dx,dy) is the differential displacement vector along the line. Since the line is straight, (dx, dy) = (3 - (-1), -3 - 1) = (4, -4) up to a constant. Then dot product becomes F · dr = (xy - (y-x))*(4).
Now, we take the line integral of this product over the path. The limits of integration, derived from the line's endpoints, are x = -1 to x = 3. We carry out the integration and simplify to get the work done by the force over the straight line. Note that the straight-line path simplifies the calculation. For paths with more complex shapes or forces that vary nonlinearly with displacement, such integrations may require specialist mathematical techniques.
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You have a glass ball with a radius of 2.00 mm and a density of 2500 kg/m3. You hold the ball so it is fully submerged, just below the surface, in a tall cylinder full of glycerin, and then release the ball from rest. Take the viscosity of glycerin to be 1.5 Pa s and the density of glycerin to be 1250 kg/m3. Use g = 10 N/kg = 10 m/s2. Also, note that the drag force on a ball moving through a fluid is: Fdrag = 6πηrv . (a) Note that initially the ball is at rest. Sketch (to scale) the free-body diagram of the ball just after it is released, while its velocity is negligible. (b) Calculate the magnitude of the ball’s initial acceleration. (c) Eventually, the ball reaches a terminal (constant) velocity. Sketch (to scale) the free-body diagram of the ball when it is moving at its terminal velocity. (d) Calculate the magnitude of the terminal velocity. (e) What is the magnitude of the ball’s acceleration, when the ball reaches terminal velocity? (f) Let’s say that the force of gravity acting on the ball is 4F, directed down. We can then express all the forces in terms of F.
Answers
Answer:
(a) check attachment
(b)5 m/s²
Explanation:
Given: radius = 2.00mm: density = 2500kg/m³: viscosity of glycerin = 1.5pa: decity of glycerin = 1250kg/m³: g = 10N/kg = 10m/s²: Fdrag = 6πnrv
(a) for answer check attachment.
(b) For the magnitude of the balls initial acceleration:
Initial net force(f) = mg - upthrust
= [tex]mg - (\frac{m}{p} )pg.g[/tex]
acceleration (a) = [tex]Acceleration(a)=\frac{f}{m}\\=g - (\frac{pg}{p})g\\=g(1-\frac{pg}{p} )\\=10(1-\frac{1250}{2500} )\\a=10(1-0.5)\\a=5 m/s²[/tex]
c.) fromthe force diagram in the attachment; when the ball attains terminal velocity the net force will be zero(0)
[tex]mg=6πnrv + upthrust[/tex]
d.) For the magnitude of terminal velocity:
[tex]mg=6πnrv + (\frac{m}{p})pg.g\\\\(\frac{4}{3}πr^{3} p)g=6πnrv +\frac{4}{3}πr^3pg.g\\\\V = \frac{2}{9}.\frac{(2*10^{-3})^{2}*(2500-1250)*10}{1.5}\\\\=0.79cm/s[/tex]
e.) when the ball reaches terminal velocity, the acceleration is zero (0)
A rectangular coil of 65 turns, dimensions 0.100 m by 0.200 m, and total resistance 10.0 ? rotates with angular speed 29.5 rad/s about the y axis in a region where a 1.00-T magnetic field is directed along the x axis. The time t = 0 is chosen to be at an instant when the plane of the coil is perpendicular to the direction of B with arrow.
(a) Calculate the maximum induced emf in the coil.
V
(b) Calculate the maximum rate of change of magnetic flux through the coil.
Wb/s
(c) Calculate the induced emf at t = 0.050 0 s.
V
(d) Calculate the torque exerted by the magnetic field on the coil at the instant when the emf is a maximum.
N
Answers
Answer:
Explanation:
N = 65
Area, A = 0.1 x 0.2 = 0.02 m^2
R = 10 ohm
ω = 29.5 rad/s
B = 1 T
(a) at t = 0
e = N x B x A x ω
e = 65 x 1 x 0.02 x 29.5
e = 38.35 V
(b) The maximum rate of change of magnetic flux is equal to the maximum value of induced emf.
Ф = 38.35 Wb/s
(c) e = NBAω Sinωt
e = 65 x 1 x 0.02 x 29.5 x Sin (29.5 x 0.05)
e = 38.174 V
(d) Maximum torque
τ = M B Sin 90
τ = N i A B
τ = N e A B / R
τ = 65 x 38.35 x 0.02 x 1 / 10
τ = 5 Nm
Sound waves are a longitudinal wave that have a speed of about 340 m/s when traveling through room temperature air. What is the wavelength of a sound wave that has a frequency of 6,191 Hz?
Answers
The wavelength of the wave is 0.055 m
Explanation:
The relationship between speed, frequency and wavelength of a wave is given by the wave equation:
[tex]v=f\lambda[/tex]
where
v is the speed
f is the frequency
[tex]\lambda[/tex] is the wavelength
For the sound wave in this problem we have
v = 340 m/s is the speed
f = 6,191 Hz is the frequency
Solving for [tex]\lambda[/tex], we find the wavelength:
[tex]\lambda=\frac{v}{f}=\frac{340}{6191}=0.055 m[/tex]
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A 10.0 g bullet is fired into a stationary 5.00 kg block of wood. The bullet lodges inside the block. The speed of the block plus bullet system immediately after the collision is measured at 0.600 m/s.
What was the original speed of the bullet?
Answers
Answer:
v = 300.6 m/s
Explanation:
given,
mass of bullet(m) = 10 g = 0.01 kg
mass of block of wood (M)= 5 Kg
speed of the block plus bullet after collision(V) = 0.6 m/s
speed of wood(u) = 0 m/s
original speed of the bullet(v) = ?
using conservation of momentum
m v + Mu = (M + m)V
0.01 x v + 5 x 0 = (5 + 0.01) x 0.6
0.01 x v = (5.01) x 0.6
[tex]v = \dfrac{3.006}{0.01}[/tex]
v = 300.6 m/s
the original speed of the bullet before collision is equal to v = 300.6 m/s
The original speed of the bullet can be determined using the law of conservation of momentum. Applying the formula and solving for the original bullet speed, the bullet was initially traveling at 3 m/s.
Explanation:This problem is an application of the law of conservation of momentum. The principle states that the total momentum before a collision is equal to the total momentum after the collision. In this case, before the collision, only the bullet has momentum as the block is stationary. After the collision, the momentum is shared between the block and the lodged bullet.
To find the original speed of the bullet, we can set up this equation representing the law of conservation of momentum: m1v1 (before collision) = (m1 + m2) v2 (after collision). Here, m1 = mass of the bullet, v1 = original speed of the bullet, m2 = mass of the block, v2 = final speed of the block and bullet together.
Substituting the given values: (10 g)x = (10 g + 5.00 kg)(0.600 m/s). Solving for x (original bullet speed) yields a value of x = 3 m/s.
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An antitank weapon fires a 3.00-kg rocket which acquires a speed of 50.0 m/s after traveling 90.0 cm down a launching tube. Assuming the rocket was accelerated uniformly, what is the average force that acted on it?
Answers
Answer:
Force in the rocket will be 4166.64 N
Explanation:
We have given mass of the rocket m = 3 kg
Rocket acquires a speed of 50 m sec so final speed v = 50 m/sec
Initial speed u = 0 m/sec
Distance traveled s = 90 cm = 0.9 m
From third equation of motion we know that [tex]v^2=u^2+2as[/tex]
[tex]50^2=0^2+2\times a\times 0.9[/tex]
[tex]a=1388.88m/sec^2[/tex]
From newton's law we F = ma
So force will be [tex]F=1388.88\times 3=4166.64N[/tex]
So force in the rocket will be 4166.64 N
The average force that acted on the rocket can be calculated using Newton's second law, which states that force is equal to mass times acceleration. In this case, the mass of the rocket is 3.00 kg and the acceleration can be determined by dividing the change in velocity by the distance traveled. The average force that acted on the rocket is 166.8 Newtons.
Explanation:The average force that acted on the rocket can be calculated using Newton's second law, which states that force is equal to mass times acceleration. In this case, the mass of the rocket is 3.00 kg and the acceleration can be determined by dividing the change in velocity by the distance traveled. The change in velocity is 50.0 m/s and the distance traveled is 90.0 cm, which is equivalent to 0.90 m.
So, the acceleration of the rocket is 50.0 m/s divided by 0.90 m, which gives us 55.6 m/s². Now, we can use Newton's second law to calculate the force: F = m * a, where F is the force, m is the mass, and a is the acceleration.
Plugging in the values, we get F = 3.00 kg * 55.6 m/s² = 166.8 N. Therefore, the average force that acted on the rocket is 166.8 Newtons.
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Police lieutenants, examining the scene of an accident involving two cars, measure the skid marks of one of the cars, which nearly came to a stop before colliding, to be 78 m long. The coefficient of kinetic friction between rubber and the pavement is about 0.90. Estimate the initial speed of that car assuming a level road.
Answers
Answer:
37.11231 m/s
Explanation:
u = Initial velocity
v = Final velocity = 0
s = Displacement = 78 m
g = Acceleration due to gravity = 9.81 m/s²
[tex]\mu[/tex] = Coefficient of kinetic friction = 0.9
Acceleration is given by
[tex]a=-\frac{f}{m}\\\Rightarrow a=-\frac{\mu mg}{m}\\\Rightarrow a=-\mu g[/tex]
Equation of motion
[tex]v^2-u^2=2as\\\Rightarrow v^2-u^2=-2\mu gs\\\Rightarrow -u^2=2(-\mu g)s-v^2\\\Rightarrow u=\sqrt{v^2-2(-\mu g)s}\\\Rightarrow u=\sqrt{0^2-2\times (-0.9\times 9.81)\times 78}\\\Rightarrow u=37.11231\ m/s[/tex]
The initial speed of that car is 37.11231 m/s
For a very rough pipe wall the friction factor is constant at high Reynolds numbers. For a length L1 the pressure drop over the length is p1. If the length of the pipe is then doubled, what is the relation of the new pressure drop p2 to the original pressure drop p1 at the original mass flow rate?
Answers
Final answer:
The new pressure drop p2 will be equal to the original pressure drop p1 at the original mass flow rate.
Explanation:
The relation between the new pressure drop p2 and the original pressure drop p1 at the original mass flow rate can be determined using Poiseuille's equation. Poiseuille's equation states that the pressure drop p2 - p1 is equal to the product of the resistance R and the flow rate Q.
When the length of the pipe is doubled, the resistance R remains constant because the friction factor is constant at high Reynolds numbers. Therefore, the new pressure drop p2 will be equal to the original pressure drop p1.
A garden hose is attached to a water faucet on one end and a spray nozzle on the other end. The water faucet is turned on, but the nozzle is turned off so that no water flows through the hose. The hose lies horizontally on the ground, and a stream of water sprays vertically out of a small leak to a height of 0.78 m.
What is the pressure inside the hose?
Answers
To solve this problem it is necessary to use the concepts related to pressure depending on the depth (or height) in which the object is on that fluid. By definition this expression is given as
[tex]P = P_{atm} +\rho gh[/tex]
Where,
[tex]P_{atm} =[/tex] Atmospheric Pressure
[tex]\rho =[/tex]Density, water at this case
g = Gravity
h = Height
The equation basically tells us that under a reference pressure, which is terrestrial, as one of the three variables (gravity, density or height) increases the pressure exerted on the body. In this case density and gravity are constant variables. The only variable that changes in the frame of reference is the height.
Our values are given as
[tex]P_{atm} = 1.013*10^5Pa[/tex]
[tex]\rho = 1000Kg/m^3[/tex]
[tex]g = 9.8m/s^2[/tex]
[tex]h = 0.78m[/tex]
Replacing at the equation we have,
[tex]P = P_{atm} +\rho gh[/tex]
[tex]P = 1.013*10^5 +(1000)(9.8)(0.78)[/tex]
[tex]P = 108944Pa[/tex]
[tex]P = 0.1089Mpa[/tex]
Therefore the pressure inside the hose is 0.1089Mpa
A block of mass 11.8 kg slides from rest down a frictionless 35.6° incline and is stopped by a strong spring with k = 3.11 x 10^4 N/m. The block slides 2.17 m from the point of release to the point where it comes to rest against the spring. When the block comes to rest, how far has the spring been compressed?
Answers
Answer:
x = 6.58 m
Explanation:
For this exercise let's use the concept of conservation of mechanical energy. Let's look for energy at two points the highest and when the spring is compressed
Initial
Em₀ = U = m g and
Final
[tex]E_{mf}[/tex] = ke = ½ k x2
As there is no friction the mechanical energy is conserved
Em₀ = [tex]E_{mf}[/tex]
m g y = ½ k x²
Let's use trigonometry to face height
sin θ = y / L
y = L sin θ
x = √ 2mg (L synth) / k
x = √ (2 11.8 9.8 2.17 sin35.6 / 3.11 104)
x = 6.58 m
A 0.30-kg puck, initially at rest on a frictionless horizontal surface, is struck by a 0.20-kg puck that is initially moving along the x-axis with a velocity of 9.1 m/s. After the collision, the 0.20-kg puck has a speed of 5.5 m/s at an angle of θ = 53° to the positive x-axis. (a) Determine the velocity of the 0.30-kg puck after the collision, (b) Find the fraction of kinetic energy lost in the collision.
Answers
The problem presents a collision scenario where principles of momentum conservation and kinetic energy are applied to determine the final velocity of a striking puck and the fraction of kinetic energy lost during the collision.
Explanation:The question describes a collision scenario in physics involving two pucks, where conservation of momentum and kinetic energy principles are applied. The question can be resolved by dividing it into two sections: (a) Determination of the velocity of the 0.30-kg puck after the collision, and (b) Determination of the fraction of kinetic energy lost in the collision.
For part (a), we use the law of conservation of momentum, which states that the total linear momentum of an isolated system remains constant, regardless of whether the objects within the system are at rest or in motion. To find the final velocity of the 0.30-kg puck, we subtract the momentum vector of the 0.20-kg puck after the collision from the momentum vector before the collision. For part (b), we first find the initial and final total kinetic energy of the system. The fraction of kinetic energy lost is given by the difference between the initial and final kinetic energies, divided by the initial kinetic energy.
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The principle of conservation of momentum and kinetic energy principles are employed to figure out the final velocity of the 0.30-kg puck and the fraction of kinetic energy lost in the collision. Initial and final velocities of the pucks, as well as their masses, are used in the process.
Explanation:To solve this problem, we need to apply the principle of conservation of momentum, which states that the total momentum of a system before a collision is equal to the total momentum after the collision. To find out the velocity of the 0.30-kg puck after the collision (a), the momentum equation (m₁v₁_initial + m₂v₂_initial = m₁v₁_final + m₂v₂_final) is applied. It is known that initial velocities of 0.30-kg and 0.20-kg pucks are 0 and 9.1 m/s respectively; the final velocity of the 0.20-kg puck is 5.5 m/s at 53° to the x-axis.
To calculate the fractional kinetic energy lost in the collision (b), firstly, the kinetic energy before and after the collision is calculated using the equation KE = 1/2mv₂. The fraction of kinetic energy lost is then (KE_initial - KE_final)/KE_initial.
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When the palmaris longus muscle in the forearm is flexed, the wrist moves back and forth. If the muscle generates a force of 45.5 N and it is acting with an effective lever arm of 2.45 cm, what is the torque that the muscle produces on the wrist?
Answers
Answer:
Torque on the rocket will be 1.11475 N -m
Explanation:
We have given that muscles generate a force of 45.5 N
So force F = 45.5 N
This force acts on the is acting on the effective lever arm of 2.45 cm
So length of the lever arm d = 2.45 cm = 0.0245 m
We have to find torque
We know that torque is given by [tex]\tau =F\times d=45.5\times 0.0245=1.11475N-m[/tex]
So torque on the rocket will be 1.11475 N -m
A transformer has 500 primary turns and 9.7 secondary turns.
(a) If Vp is 120 V (rms), what is Vs with an open circuit?
(b) If the secondary now has a resistive load of 13 Ω, what is the current in the primary and secondary?
Answers
Answer: a) 2.33V b) Ip = 0.0035A, Is = 0.179A
Explanation:
The relationship formula between the Voltage, Number of turns and current in a transformer is given as
Vp/Vs = Np/Ns = Is/Ip
Vp is voltage in the primary coil
Vs is voltage in the secondary coil
Ns is number of turns in the secondary coil
Np is number of turns in the primary coil
Is is current in the secondary coil
Ip is current in the primary coil
a) substituting the values in the formula
Vp/Vs = Np/Ns
120/Vs = 500/9.7
Vs = 120×9.7/500
Vs = 2.33V
b) since secondary now has a resistive load of 13 Ω, Using ohm's law to get current in the secondary
V=IR
Is = Vs/R
Is =2.33/13
Is = 0.179A
To get Ip, we will use the relationship Np/Ns = Is/Ip
500/9.7= 0.179/Ip
Ip = 9.7×0.179/500
Ip = 0.0035A
You can, in an emergency, start a manual transmission car by putting it in neutral, letting the car roll down a hill to pick up speed, then putting it in gear and quickly letting out the clutch. If the car needs to be moving at 3.5 m/s for this to work, how high a hill do you need? (You can ignore friction and drag)
Answers
Answer:
The hill should be not less than 0.625 m high
Explanation:
This problem can be solved by using the principle of conservation of mechanical energy. In the absence of friction, the total mechanical energy is conserved. That means that
[tex]E_m=U+K[/tex] is constant, being U the potential energy and K the kinetic energy
[tex]U=mgh[/tex]
[tex]K=\frac{mv^2}{2}[/tex]
When the car is in the top of the hill, its speed is 0, but its height h should be enough to produce the needed speed v down the hill.
The Kinetic energy is then, zero. When the car gets enough speed we assume it is achieved at ground level, so the potential energy runs out to zero but the Kinetic is at max. So the initial potential energy is transformed into kinetic energy.
[tex]mgh=\frac{mv^2}{2}[/tex]
We can solve for h:
[tex]h=\frac{v^2}{2g}=\frac{3.5^2}{2(9.8)}=0.625m[/tex]
The hill should be not less than 0.625 m high
The hill should have a height of 0.625 m and above.
What is Height?This is defined as the measurement of the vertical position of a body.
Total mechanical energy = Potential energy + kinetic energy.
Potential energy = mgh
Kinetic energy = 1/2mv²
We can infer that:
mgh = 1/2mv²
h = v² / 2g
= (3.5)² / 2(9.8)
= 0.625m.
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Medical cyclotrons need efficient sources of protons to inject into their center. In one kind of ion source, hydrogen atoms (i.e., protons with one orbiting electron) are fed into a chamber where there is a strong magnetic field. Electrons in this chamber are trapped in tight orbits, which greatly increases the chance that they will collide with a hydrogen atom and ionizes it. One such source uses a magnetic field of 70 mT, and the electrons' kinetic energy is 1.2 eV. If the electrons travel in a plane perpendicular to the field, what is the radius of their orbits?
Answers
Answer:
[tex]r=5.278\times 10^{-4}\ m[/tex]
Explanation:
Given that:
magnetic field intensity, [tex]B=0.07\ T[/tex]kinetic energy of electron, [tex]KE=1.2\ eV= 1.2\times 1.6\times 10^{-19}\ J= 1.92\times 10^{-19}\ J[/tex]we have mass of electron, [tex]m=9.1\times 10^{-31}\ kg[/tex]Now, form the mathematical expression of Kinetic Energy:
[tex]KE= \frac{1}{2} m.v^2[/tex]
[tex]1.92\times 10^{-19}=0.5\times 9.1\times 10^{-31}\times v^2[/tex]
[tex]v^2=4.2198\times 10^{11}[/tex]
[tex]v=6.496\times 10^6\ m.s^{-1}[/tex]
from the relation of magnetic and centripetal forces we have the radius as:
[tex]r=\frac{m.v}{q.B}[/tex]
[tex]r=\frac{9.1\times 10^{-31}\times 6.496\times 10^6 }{1.6\times 10^{-19}\times 0.07}[/tex]
[tex]r=5.278\times 10^{-4}\ m[/tex]
The radius of orbit of the cyclotrons in the given field is 5.287 x 10⁻⁵ m.
Force of the electron
The magnetic force and centripetal force of the electron is calculated as follows;
qvB = mv²/r
qB = mv/r
r = mv/qB
where;
m is mass of electronv is speed of the electronq is the magnitude of the chargeB is magnetic field strengthKinetic energy of the electronThe kinetic energy of the electron is given as;
K.E = ¹/₂mv²
mv² = 2K.E
v² = 2K.E/m
v² = (2 x 1.2 x 1.6 x 10⁻¹⁹)/(9.11 x 10⁻³¹)
v² = 4.215 x 10¹¹
v = √4.215 x 10¹¹
v = 6.5 x 10⁵ m/s
Radius of their orbitThe radius of their orbit is calculated as follows;
r = mv/qB
r = (9.11 x 10⁻³¹ x 6.5 x 10⁵) / (1.6 x 10⁻¹⁹ x 70 x 10⁻³)
r = 5.287 x 10⁻⁵ m
Thus, the radius of orbit of the cyclotrons in the given field is 5.287 x 10⁻⁵ m.
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When light of wavelength 350 nm is incident on a metal surface, the stopping potential of the photoelectrons is measured to be 0.500 V. What is the work function of this metal? (c = 3.00 × 108 m/s, h = 6.626 × 10-34 J ∙ s, 1 eV = 1.60 × 10-19 J) 3.05 eV 4.12 eV 3.54 eV 0.500 eV
Answers
The work function of the metal can be calculated using Einstein's photoelectric equation and the given information about the wavelength of light, stopping potential, Planck's constant, and speed of light. The initial step is to calculate the energy of the photon; thereafter, calculate the kinetic energy of the photoelectron with the stopping potential, and finally find the work function by subtracting the kinetic energy from the photon energy.
Explanation:To calculate the work function of the metal, we need to first figure out the energy of the photon hitting the metal surface. This can be found using Einstein's photoelectric equation: E = hc/λ. Here, h is Planck's constant (6.626 x 10-34 J · s), c is the speed of light (3.00 x 108 m/s), and λ is the wavelength of the light (350 nm or 350 x 10-9 m).
After calculating the energy of the photon, we can determine the energy required to eject the photoelectrons, known as the stopping potential, which in this case is 0.500 V. Note this is the extra energy required to stop the photoelectron and is equal to the kinetic energy of the photoelectron. Hence, 0.500V = 0.500 eV because 1V = 1eV for a single electron with charge e. To convert this to joules, we use the conversion 1eV = 1.60 x 10-19J.
Finally, in order to find the work function, we use Einstein's photoelectric equation again which states that the energy of the incident photon is equal to the work function (φ) plus the kinetic energy of the photoelectron. That is, E = φ + KE. From this, we can rearrange the equation to solve for φ, the work function of the metal where: φ = E - KE.
By substituting the appropriate values into this equation, the work function of the metal can be found out. But remember, work function is a property of the metal only and is independent of the incident radiation. Also, it determines the threshold frequency or cutoff wavelength of the metal beyond which photoelectric effect will not occur.
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Two physical pendulums (not simple pendulums) are made from meter sticks that are suspended from the ceiling at one end. The sticks are uniform and are identical in all respects, except that one is made of wood (mass=0.36 kg) and the other of metal (mass=0.82 kg) They are set into oscillation and execute simple harmonic motion. Determine the period of (a) the wood pendulum and (b) the metal pendulum
Answers
Answer:
a) T = 0.579 s , b) T = 0.579 s
Explanation:
The great advantage of wave mechanics is that the general equations for different systems are the same, what changes are the physical parameters involved, the equation of motion is
x = A cos (wt +φ)
Where w is the angular velocity that is this case for being a solid body is
w = √ (mg d / I)
Where I is the moment of inertia and d the distance to the pivot point
The moment of inertia for a ruler hold one end is
I = 1/12 M L²
The lost is related to the frequency and is with the angular velocity
T = 1 / f
w = 2π f
w = 2π / T
T = 2π / w
T = 2π √ I / mgd
For our case
d = L
T = 2π √(1/12 M L²) / M g L)
T = 2π √(L/(12 g))
a) wood suppose it is one meter long (L = 1m)
T = 2π √ (1 / (12 9.8))
T = 0.579 s
b) metal length (L = 1m)
T = 2pi RA (1 / (12 9.8))
T = 0.579 s
The period does not depend on mass but on length
how does the gravitational force between two objects change if the distance between the objects is cut in half?
A. the gravitational force doubles
B. the gravitational force decreases by half
C. the gravitational force decreases by a factor of 4
D. the gravitational force increases by a factor of 4
Answers
D. the gravitational force increases by a factor of 4
Explanation:
The magnitude of the gravitational force between two objects is given by:
[tex]F=G\frac{m_1 m_2}{r^2}[/tex]
where
[tex]G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}[/tex] is the gravitational constant
m1, m2 are the masses of the two objects
r is the separation between them
Here let's call F the initial gravitational force between the two objects. Later, the distance between the objects is halved, so the new distance is:
[tex]r'=\frac{r}{2}[/tex]
Substituting into the equation, we find the new force:
[tex]F'=G\frac{m_1 m_2}{(r/2)^2}=4(G\frac{m_1 m_2}{r^2})=4F[/tex]
Therefore, the gravitational force increases by a factor of 4.
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Final answer:
The gravitational force between two objects doubles if the distance between them is cut in half.
Explanation:
The gravitational force between two objects is inversely proportional to the square of the distance between them. This means that if the distance between the objects is cut in half, the force will increase by a factor of 4.
For example, let's say the original distance is 10 units. If we cut it in half, the new distance would be 5 units. The force would increase by a factor of (10/5)² = 4, meaning the gravitational force would quadruple.
So the correct answer is, A. the gravitational force doubles.
In a CD player, a CD starts from rest and accelerates at a rate of . Suppose the CD has radius 1212 cm. At time 0.15 sec after it started spinning, what is the magnitude of the linear acceleration∣ for a point on its outer rim?
Answers
Answer:
[tex]a =29.54\ m/s^2[/tex]
Explanation:
given,
radius of CD player = 12 cm
assume rate of acceleration = 100 rad/s²
times = 0.15 s
now,
tangential acceleration
[tex]a_t = \alpha r[/tex]
[tex]a_t = 100 \times 0.12[/tex]
[tex]a_t = 12 m/s^2[/tex]
now using equation
v = v₀ + a_t x t
v =0+ 12 x 0.15
v = 1.8 m/s
now, radial acceleration
[tex]a_r = \dfrac{v^2}{r}[/tex]
[tex]a_r = \dfrac{1.8^2}{0.12}[/tex]
[tex]a_r =27\ m.s^2[/tex]
now
acceleration
[tex]a = \sqrt{a_r^2+a_t^2}[/tex]
[tex]a = \sqrt{27^2+12^2}[/tex]
[tex]a =29.54\ m/s^2[/tex]
A spring has a 12 cm length. When a 200-g mass is hung from the spring, it extends to 27 cm. The hanging mass were pulled downward a further 5 cm. What is the time-dependent function of the position, in centimeters, of the mass, assuming that the phase angle LaTeX: \phi=0ϕ=0?
LaTeX: x(t)=15\cos{(8.08\textrm{ }t)}x(t)=15cos(8.08 t), where the coefficient inside the trigonometric function has units of rad/s.
LaTeX: x(t)=5\cos{(8.08\textrm{ }t)}x(t)=5cos(8.08 t), where the coefficient inside the trigonometric function has units of rad/s.
LaTeX: x(t)=5\cos{(1.29\textrm{ }t)}x(t)=5cos(1.29 t)x(t)=5cos(1.29 t), where the coefficient inside the trigonometric function has units of rad/s.
LaTeX: x(t)=15\cos{(1.29\textrm{ }t)}x(t)=15cos(1.29 t)x(t)=15cos(1.29 t), where the coefficient inside the trigonometric function has units of rad/s.
Answers
Answer:
The equation of the time-dependent function of the position is [tex]x(t)=5\cos(8.08t)[/tex]
(b) is correct option.
Explanation:
Given that,
Length = 12 cm
Mass = 200 g
Extend distance = 27 cm
Distance = 5 cm
Phase angle =0°
We need to calculate the spring constant
Using formula of restoring force
[tex]F=kx[/tex]
[tex]mg=kx[/tex]
[tex]k=\dfrac{mg}{x}[/tex]
[tex]k=\dfrac{200\times10^{-3}\times9.8}{(27-12)\times10^{2}}[/tex]
[tex]k=13.06\ N/m[/tex]
We need to calculate the time period
Using formula of time period
[tex]T=2\pi\sqrt{\dfrac{m}{k}}[/tex]
Put the value into the formula
[tex]T=2\pi\sqrt{\dfrac{0.2}{13.6}}[/tex]
[tex]T=0.777\ sec[/tex]
At t = 0, the maximum displacement was 5 cm
So, The equation of the time-dependent function of the position
[tex]x(t)=A\cos(\omega t)[/tex]
Put the value into the formula
[tex]x(t)=5\cos(2\pi\times f\times t)[/tex]
[tex]x(t)=5\cos(2\pi\times\dfrac{1}{T}\times t)[/tex]
[tex]x(t)=5\cos(2\pi\times\dfrac{1}{0.777}\times t)[/tex]
[tex]x(t)=5\cos(8.08t)[/tex]
Hence, The equation of the time-dependent function of the position is [tex]x(t)=5\cos(8.08t)[/tex]
A piece of curved glass has a radius of curvature of r = 10.8 m and is used to form Newton's rings, as in the drawing. Not counting the dark spot at the center of the pattern, there are one hundred dark fringes, the last one being at the outer edge of the curved piece of glass. The light being used has a wavelength of 652 nm in vacuum. What is the radius R of the outermost dark ring in the pattern? (Hint: Note that r is much greater than R, and you may assume that tan(θ) = θ for small angles, where θ must be expressed in radians.)
Answers
Answer:
Radius of the outer most dark fringe is 2.65 cm
Solution:
As per the question:
Radius of curvature of the glass, r = 10.8 m
No. of dark fringes, n = 100
Wavelength of light, [tex]\lambda = 652\ nm = 652\times 10^{- 9}\ m[/tex]
Now,
To calculate the radius R of the outermost ring:
Radius of the dark fringe of nth order is given by:
[tex]R^{2} = nr\lambda = 100\times 10.8\times 652\times 10^{- 9} = 7.042\times 10^{- 4}[/tex]
[tex]R = \sqrt{7.042\times 10^{- 4}} = 0.0265\ m = 2.65\ cm[/tex]
You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics, you ask one of the archers to shoot an arrow parallel to the ground. Unfortunately the archer stands on an elevated platform of unknown height. However, you find the arrow stuck in the ground 67.0 m away, making a 3.00 ∘ angle with the ground.How fast was the arrow shot?
Answers
To develop this problem it is necessary to apply the kinematic equations that describe displacement, velocity and clarification.
By definition we know that velocity is defined as the change of position due to time, therefore
[tex]V = \frac{d}{t}[/tex]
Where,
d = Distance
t = Time
Speed can also be expressed in vector form through its components [tex]V_x[/tex] and [tex]V_y[/tex]
In the case of the horizontal component X, we have to
[tex]V_x = \frac{d}{t}[/tex]
Here d means the horizontal displacement, then
[tex]t = \frac{d}{V_x}[/tex]
[tex]t = \frac{67}{V_x}[/tex]
At the same time we have that the vertical component of the velocity is
[tex]V_y = gt[/tex]
Here,
g = Gravity
Therefore using the relation previously found we have that
[tex]V_y = g \frac{67}{V_x}[/tex]
The relationship between the two velocities and the angle can be expressed through the Tangent, therefore
[tex]tan\theta = \frac{V_y}{V_x}[/tex]
[tex]tan \theta = \frac{g \frac{67}{V_x} }{V_x}[/tex]
[tex]tan 3 = \frac{9.8\frac{67}{V_x} }{V_x}[/tex]
[tex]tan 3 = \frac{9.8*67}{V_x^2}[/tex]
[tex]V_x^2 = \frac{9.8*67}{tan 3}[/tex]
[tex]V_x= \sqrt{ \frac{9.8*67}{tan 3}}[/tex]
[tex]V_x = 111.93m/s \hat{i}[/tex]
This is the horizontal component, we could also find the vertical speed and the value of the total speed with the information given,
Then [tex]V_y,[/tex]
[tex]V_y = g \frac{67}{V_x}[/tex]
[tex]V_y = 9.8*\frac{67}{111.93}[/tex]
[tex]V_y = 5.866m/s\hat{j}[/tex]
[tex]|\vec{V}| = \sqrt{111.93^2+5.866^2}[/tex]
[tex]|\vec{V}| = 112.084m/s[/tex]
Final answer:
To calculate the initial speed of the arrow in a projectile motion problem, one needs to find the time of flight using the horizontal range and the final impact angle and then solve for the initial horizontal and vertical velocity components.
Explanation:
To determine the initial speed of the arrow shot from the bow, we'll use the principle of projectile motion. First, we need to find the time of flight. The arrow makes a 3.00° angle with the ground upon impact, and since it was shot parallel to the ground, it maintains a constant horizontal velocity throughout its flight. So, we can calculate the initial horizontal speed using this final impact angle.
The horizontal range (R) is given by R = v0x * t, where v0x is the initial horizontal velocity and t is the time of flight. The arrow landed 67.0 m away, so R is 67.0 m.
To find t, we will use the vertical motion equation under constant acceleration due to gravity (g = 9.8 m/s2). We know the final vertical velocity component (vfy) can be related to the initial vertical velocity component (v0y = 0 since it was shot parallel to the ground) by vfy = v0y + g*t. We can also calculate vfy using the impact angle: vfy = tan(impact angle) * v0x. Combining these, we can solve for t and subsequently for the initial speed v0.
A bungee jumper of mass m = 65 kg jumps from a bridge.
A bungee cord with unstretched length L is attached to her leg.
After falling a distance L, the jumper is moving at speed v = 26 m/s and the bungee cord begins stretching and acts like an ideal spring with spring constant k.
Neglect air resistance and the mass of the bungee cord.
Which choice best represents the unstretched length of the bungee cord, L?A. 5.9 m
B. 26 m
C. 34 m
D. 65 m
E. None of these
Answers
To solve the problem it is necessary to apply energy conservation.
By definition we know that kinetic energy is equal to potential energy, therefore
PE = KE
[tex]mgh = \frac{1}{2}mv^2[/tex]
Where,
m = mass
g = gravitaty constat
v = velocity
h = height
Re-arrange to find h,
[tex]h=\frac{v^2}{2g}[/tex]
Replacing with our values
[tex]h=\frac{26^2}{2*9.8}[/tex]
[tex]h = 34.498\approx 34.5m[/tex]
Therefore the correct answer is C.
Unpolarized light of intensity 800 W/m2 is incident on two ideal polarizing sheets that are placed with their transmission axes perpendicular to each other. An additional polarizing sheet is then placed between the two, with its transmission axis oriented at 30∘ to that of the first. What is the intensity of the light passing through the stack of polarizing sheets? (Express your answer to two significant figures.)
Answers
Answer:
75 W/m²
Explanation:
[tex]I_0[/tex] = Unpolarized light intensity = 800 W/m²
[tex]\theta[/tex] = Angle between filters = 30°
Intensity of light after passing through first polarizer
[tex]I=\frac{I_0}{2}\\\Rightarrow I=\frac{800}{2}\\\Rightarrow I=400\ W/m^2[/tex]
Intensity of light after passing through second polarizer
[tex]I_1=Icos^2\theta\\\Rightarrow I_1=400\times cos^2(30)\\\Rightarrow I_1=300\ W/m^2[/tex]
Intensity of light after passing through third polarizer
[tex]I_2=I_1cos^2\theta\\\Rightarrow I_2=300\times cos^2(90-30)\\\Rightarrow I_1=75\ W/m^2[/tex]
The intensity of the light passing through the stack of polarizing sheets is 75 W/m²
The intensity of the light passing through three polarizing sheets, with the first and the third being perpendicular and the second making an angle of 30 degrees with the first, is zero, due to the rules of light polarization.
Explanation:Unpolarized light is composed of many rays having random polarization directions. When it passes through a polarizing filter, it decreases its intensity by a factor of 2, thus the intensity after the first polarizer is 800 W/m^2 /2 = 400 W/m^2. According to Malus's Law, the intensity of polarized light after passing through a second polarizing filter is given as I = Io cos² θ, where Io is the incident intensity (from the first polarizer) and θ is the angle between the direction of polarization and the axis of the filter.
The second polarizer is making an angle of 30 degrees with the first polarizer. Therefore, the intensity after the second polarizer is I = 400 W/m^2 * cos²(30) ≈ 346 W/m^2. The third polarizer is perpendicular to the first polarizer (or, equivalently, it makes an angle of 90 degrees with the second polarizer), so the final intensity is I = 346 W/m^2 * cos²(90), which is 0 because the cosine of 90 degrees is zero. So, no light (intensity=0) passes through the stack of these polarizing sheets.
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A graduated cylinder is half full of mercury and half full of water. Assume the height of the cylinder is 0.25 m and don't forget about atmospheric pressure. 1) Calculate the pressure at the bottom of a graduated cylinder. (Express your answer to two significant figures.)
Answers
To develop this problem it is necessary to apply the concepts related to the definition of absolute, manometric and atmospheric pressure.
The pressure would be defined as
[tex]P = P_0 + P_w +P_m[/tex]
Where,
[tex]P_0 =[/tex] Standard atmosphere pressure
[tex]P_W =[/tex]Pressure of Water
[tex]P_m =[/tex]Pressure of mercury
Therefore we have that
[tex]P = P_0 \rho_w g \frac{h}{2} + \rho_m g\frac{h}{2}[/tex]
Here g is the gravity, \rho the density at normal conditions and h the height of the cylinder.
Replacing with our values we have that,
[tex]P = 1.013*10^5+10^3*9.81*\frac{0.25}{2}+13.6*10^3*9.81*\frac{0.25}{2}[/tex]
[tex]P = 1.21*10^5Pa \approx 0.121Mpa[/tex]
Therefore the pressure at the Bottom of a graduated cylinder is 0.121Mpa
The pressure at the bottom of a 0.25 m-graduated cylinder half full of mercury and half full of water is 1.2 × 10⁵ Pa.
A graduated cylinder is half full of mercury and half full of water. If the height of the cylinder is 0.25 m, the height of each column of liquid is:
[tex]\frac{0.25m}{2} = 0.13 m[/tex]
We can calculate the pressure at the bottom of the cylinder (P) using the following expression.
[tex]P = P_{atm} + P_{Hg} + P_w[/tex]
where,
[tex]P_{atm}[/tex]: atmospheric pressure (101,325 Pa)[tex]P_{Hg}[/tex]: pressure exerted by Hg[tex]P_w[/tex]: pressure exerted by the waterWe can calculate the pressure exerted by each liquid (P) as:
[tex]P = \rho \times g \times h[/tex]
where,
ρ: density of the liquidg: gravity (9.81 m/s²)h: height of the liquid column (0.13 m)The pressure exerted by the column of Hg is:
[tex]P_{Hg} = \frac{13.6 \times 10^{3} kg}{m^{3} } \times \frac{9.81m}{s^{2} } \times 0.13 m = 17 \times 10^{3} Pa[/tex]
The pressure exerted by the column of water is:
[tex]P_{w} = \frac{1.00 \times 10^{3} kg}{m^{3} } \times \frac{9.81m}{s^{2} } \times 0.13 m = 1.3 \times 10^{3} Pa[/tex]
The pressure at the bottom of the cylinder is:
[tex]P = P_{atm} + P_{Hg} + P_w = 101325 Pa + 17 \times 10^{3} Pa + 1.3 \times 10^{3} Pa = 1.2 \times 10^{5} Pa[/tex]
The pressure at the bottom of a 0.25 m-graduated cylinder half full of mercury and half full of water is 1.2 × 10⁵ Pa.
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Calculate the angular momentum, in kg.m^2/s, of an ice skater spinning at 6.00 rev/s given his moment of inertia is 0.420 kg · m2. kg · m2/s (b) He reduces his rate of spin (his angular velocity) by extending his arms and increasing his moment of inertia. Find the value of his moment of inertia, in kg · m2, if his angular velocity drops to 1.95 rev/s. kg · m2 (c) Suppose instead he keeps his arms in and allows friction with the ice to slow him to 3.00 rev/s. What average torque, in N m, was exerted if this takes 23.0 seconds? (Indicate the direction with the sign of your answer. Assume that the skater's rotation is in the positive direction.) N.m
Answers
The angular momentum of the ice skater spinning at 6.00 rev/s is 15.82 kg · m²/s. When the skater reduces their rate of spin to 1.95 rev/s, the moment of inertia is 1.29 kg · m². If the skater keeps their arms in and allows friction with the ice to slow them to 3.00 rev/s in 23.0 seconds, the average torque exerted is -0.343 N · m (indicating the direction of the torque).
Explanation:(a) The angular momentum of the ice skater can be calculated by multiplying the angular velocity (in radians per second) by the moment of inertia (in kg · m²). First, we need to convert the angular velocity from rev/s to rad/s. Since 1 rev = 2π rad, the angular velocity in rad/s is 6.00 rev/s * 2π rad/rev = 37.70 rad/s. Now, we can calculate the angular momentum: Angular momentum = angular velocity * moment of inertia = 37.70 rad/s * 0.420 kg · m² = 15.82 kg · m²/s.
(b) To find the new moment of inertia when the angular velocity decreases to 1.95 rev/s, we can rearrange the formula for angular momentum to solve for moment of inertia. Angular momentum = angular velocity * moment of inertia. Rearranging the equation: Moment of inertia = angular momentum / angular velocity = 15.82 kg · m²/s / (1.95 rev/s * 2π rad/rev) = 1.29 kg · m².
(c) To calculate the average torque exerted when the skater slows to 3.00 rev/s, we can use the equation torque = change in angular momentum / time. First, we need to calculate the change in angular momentum by subtracting the initial angular momentum from the final angular momentum. The initial angular momentum is 6.00 rev/s * 2π rad/rev * 0.420 kg · m² = 15.82 kg · m²/s, and the final angular momentum is 3.00 rev/s * 2π rad/rev * 0.420 kg · m² = 7.91 kg · m²/s. The change in angular momentum is 7.91 kg · m²/s - 15.82 kg · m²/s = -7.91 kg · m²/s. Finally, we can calculate the average torque: Torque = change in angular momentum / time = -7.91 kg · m²/s / 23.0 s = -0.343 N · m (negative sign indicates the direction of the torque).
(a) Calculate the number of free electrons per cubic meter for gold assuming that there are 1.5 free electrons per gold atom. The electrical conductivity of and density of gold are 4.3 X 107 (O-m)-1 and 19.32 g/cm3, respectively.
(b) Now compute the electron mobility for Au.
Answers
Answer:
Part A:
[tex]n=8.85*10^{28}m^{-3}[/tex]
Part B:
[tex]Electron Mobility=3.03*10^{-3} m^2/V[/tex]
Explanation:
Part A:
To calculate the number of free electrons n we use the following formula::
n=1.5N-Au
Where N-Au is number of gold atoms per cubic meter
[tex]N-Au=\frac{Density*Avogadro Number}{atomic weight}[/tex]
[tex]Density = 19.32g/cm^3[/tex]
[tex]Avogadro Number=6.02*10^{23} atoms/mol[/tex]
[tex]Atomic weight=196.97g/mol[/tex]
So:
[tex]n=1.5*\frac{Density*Avogadro Number}{atomic weight}[/tex]
[tex]n=1.5*\frac{19.32*6.02*10^{23}}{196.97}[/tex]
[tex]n=8.85*10^{28}m^{-3}[/tex]
Part B:
[tex]Electron Mobility=\frac{Elec-conductivity}{n * charge on electron}[/tex]
n is calculated above which is 8.85*10^{28}m^{-3}
Charge on electron=1.602*10^{-19}
Elec- Conductivity= 4.3*10^{7}
[tex]Electron Mobility=\frac{4.3*10^{7}}{ 8.85*10^{28} * 1.602*10^{-19}}[/tex]
[tex]Electron Mobility=3.03*10^{-3} m^2/V[/tex]
Some climbing ropes are designed to noticeably stretch under a load to lessen the forces in a fall. A weight is attached to 10 m length of climbing rope, which then stretches by 20 cm. Now, this single rope is replaced by a doubled rope--two pieces of rope next to each other. How much does the doubled rope stretch?
b. A 10 m length of climbing rope is supporting a climber, and stretches by 60 cm. When the climber is supported by a 20 m length of rope, by how much does the rope stretch?
Answers
When a doubled rope, i.e. two pieces of rope used together, is used, it stretches half as much as a single rope would under the same load. Correspondingly, a rope's stretch is directly proportional to its length. So, it will stretch twice as much when its length is doubled.
Explanation:The stretchiness of a rope under a load is a property called elasticity. When there are two ropes, they share the load between them. Therefore, each carries only half the load. Hence, a doubled rope would stretch half as much as a single rope when supporting the same weight. In this case, the single rope stretches by 20 cm under a load. Thus, the doubled rope would stretch by 10 cm.
Similarly, the amount a rope stretches is proportional to its length. If a 10 m rope stretches 60 cm under a certain load, a 20 m rope -- twice as long -- would stretch twice as much under the same load, or 120 cm.
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In the context of elasticity in physics, a doubled rope that halves each individual rope's load would stretch half as much. Therefore, if a 10m rope stretches by 20cm, a doubled rope would stretch by 10cm. If a 10m rope stretches by 60cm, a 20m rope under the same load would stretch twice as much, or 120cm.
Explanation:These questions refer to the concepts of strain and stress in physics, specifically within the context of elasticity. Strain refers to the deformation of a body due to the applied stress, in this case, the length of rope extending under the weight of the climber.
a. If a 10m climbing rope stretches by 20cm under a certain load, its strain is 0.02 (20cm/10m). When the load is shared by two ropes, each rope bears half of the load. Hence, each rope is likely to stretch half as much. So, the doubled rope would stretch by 10cm.
b. Stretching is directly proportional to the length of the rope. Therefore, if a 10m rope stretches by 60cm under a load, a 20m rope under the same load would stretch twice as much, which is 120cm or 1.2m.
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On which of the following does the moment of inertia of an object depend?Check all that apply.linear speedlinear accelerationangular speedangular accelerationtotal massshape and density of the objectlocation of the axis of rotation
Answers
Explanation:
Moment of Inertia in physics is the measure of the rotational inertia of an object. It is nothing but the opposition that the body offers to having its speed of rotation about an axis changed by application some external torque.
I= MR^2
M= mass of object R= radius of rotation.
Therefore, MOI depends upon
a)total mass
b)shape and density of the object
c)location of the axis of rotation
The moment of inertia of an object depends on its total mass, the distribution of this mass in relation to the rotation axis (shape and density), and the location of the rotation axis. The higher the moment of inertia, the greater the object's resistance to changes in rotational rate. The parallel axis theorem provides a tool to calculate the moment of inertia for different rotation axes.
Explanation:The moment of inertia of an object depends on certain factors. Specifically, its total mass, the shape and density of the object, and the location of the axis of rotation. The moment of inertia is defined as the measure of the object's resistance to changes in its rotation rate. The moment of inertia varies depending on how mass is distributed in relation to the axis of rotation.
For instance, an object with more mass distributed farther from its axis of rotation will have a higher moment of inertia. This is seen in a hollow cylinder versus a solid cylinder of the same mass and size. The hollow cylinder has a greater moment of inertia because more of its mass is distributed at a greater distance from its axis of rotation.
A principle known as the parallel axis theorem can be used to find the moment of inertia about a new axis of rotation once it is known as a parallel axis. This theorem involves the center of mass and the distance from the initial axis to the parallel axis.
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A loop of wire enclosing an area of 0.19 m2 has a magnetic field passing through its plane at an angle to the plane. The component of the field perpendicular to the plane is 0.5 T and the component parallel to the plane is 0.8 T. What is the magnetic flux through this coil? Round the final answer to three decimal places.
Answers
To solve this exercise it is necessary to apply the concepts related to Magnetic Flow which is defined through the Gaus Law of Magnetism as the measure of the total magnetic field that passes through a given area.
Vectorially it can be defined as
[tex]\Phi = \vec{A}\vec{B}[/tex]
Where,
A = Area
B = Magnetic Field
A scalar mode can also be expressed as
[tex]\Phi = A*B Cos\theta[/tex]
Where,
[tex]\theta[/tex] is the angle between B and A, at this case the direction are perpendicular then
Our values are given as,
[tex]A = 0.19m^2[/tex]
[tex]B_1 = 0.5T[/tex]
[tex]B_2 = 0.8T[/tex]
The magnetic field component that interests us is the perpendicular therefore
[tex]\Phi = 0.19* 0.5cos(90)[/tex]
[tex]\Phi = 0.095Tm^2[/tex]
Therefore the magnetic flux through this coil is 0.095[tex]Tm^2[/tex]