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0

There can still be forces involved even if nothing is moving. If all the forces balance, then the only thing that will say is that there should be no net acceleration (which is a change in velocity, either through change in speed or change in direction… remember these are vectors). In your system, gravity is constant, but the acceleration of the tree is ...


0

To obtain the fall velocity we have: $$mgh=\frac{1}{2}mv^2$$ From this you can extract a value for the velocity, as a function of the altitude. Subsequently, we need to slow the tree down. If we have a distance to stop, we know $$\Delta s= \Delta v \cdot \Delta t$$ If we know the impulse, and the time available to stop, we can use: $$m\Delta v = F \Delta ...


2

Hint: You're far enough from the ground that the formula $mgh$, which is an approximation, doesn't really apply anymore. What is the true formula for gravitational potential energy?


0

Here is a general figure of an hard spheres collision drawn in the center of mass of the mass $m_2$ before the collision. The black dot is attached to this frame. To solve the problem, you need to observe Conservation of energy: $m_1v_1^2=m_1(v'_1)^2+m_2(v'_2)^2$. Conservation of momentum: $m_1\vec v_1=m_1\vec v'_1+m_2\vec v'_2$ Conservation of torque ...


1

It's just a matter of conservation of volume. You need to start by stating assumptions that are obvious, prima-facia, that anyone would agree with, such as the liquid being incompressible. Then you say, if one accepts these assumptions, then do like @User58220 said, assume the contrary and derive a contradiction. Suppose you have to explain it to someone, ...


2

Too long for a comment, so in an answer: Not all the springs are a function of $x_1$ and $x_2$, only spring $K_b$ is a function of both $x_1$ and $x_2$. Spring force is a function of how much a spring is stretched, e.g. how much difference er is between the beginning and the end of a spring, so $x_{begin}$-$x_{end}$ or $x_{top}$-$x_{bottom}$ for this case. ...


0

If you look at the example in Aschroft and Mermin for the body centered cubic case (which is on page 105 in the version of the book I have) you'll see that you should use the simple cubic reciprocal vector, not the face-centered cubic reciprocal lattice vector.


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This answer assumes you want to find the projections of a vector onto the polar basis at some point away from the origin. This is equivalent to rmhleo's advice and differs from that given by Kyle in the comments to the question which address a different (and simpler) problem. The notion of projection is the same. Given an arbitrary vector $\vec{v}$, ...


0

Projecting can be understood like decomposing a vector into the sum of vectors whose direction match that of the coordinate system versors. In polar coordinates the versors are: $\hat \rho$ oriented along the radius pointing outwards of the origin; and $\hat \phi$ which is a vector tangential to the circle formed by the counter clockwise rotation of $\rho$ ...


0

Draw a diagram, then you'll find out the answer. If $\mathbf T$ is the tension in the string, $\delta$ is the angle, $m$ is the mass of the medal and $\mathbf g$ is the gravitational acceleration then $$T\cos\delta=mg$$ and $$T\sin\delta=ma$$ where $a$ is the acceleration, then $$\tan\delta=\frac{a}{g}$$ $T, m,\cos\delta\neq 0$ which leads $$a=g\tan\delta$$ ...


1

Yes, the force $A$ exerts on $B$ has to be equal to the one $B$ exerts on $A$. And when $F$ vanishes the forces between $A$ and $B$ vanishes too. More specifically, if in the time $F$ is acting no movement occurs (i.e. $F$ does not overcome the total friction of $A$ and $B$) the only thing causing $A$ and $B$ to interact is $F$, so this action reaction ...


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You are almost done. If you simplify the express you found for $v_{avg}$ and compare it with the initial equation for $v_{avg}$, in which you have to substitute $v_i$ for $v(t_i)$ and $v_f$ for $v(t_f)$, you will see that they are equivalent. Another way of deriving this is by expressing $v(t)$ in terms of $v_i$ and $v_f$: $$ v(t) = v_i + \frac{v_f - ...


0

A few clarifications first: I will assume that the bigger cylinder (radius $R$) is not moving, and I will be doing the problem in this frames. I will assume that both $\theta$ and $\phi$ are positive in the clockwise direction. This is against convention but it'll make a few negative signs disappear, and you many find it easier to think this way given how ...


1

There is one error in the derivation, if you want to have $v(t_i)=v_0$, you must have $$v(t) = v_0 + a(t-t_i)$$ You also have to use the fact that $v_f = v(t_f)$. Once you use all this, you should be able to divide out $t_f-t_i$ in the corrected version of your last line and get the result you seek.


-2

The question that has illogical parameters that are probably not correct. It is unclear to what force exactly they are referring here. Consider the following question: A mosquito and a truck collide, both have the same velocity, they collide and and it takes one second for them to stop. That clearly illustrates the foolishness of the question. It sets ...


3

...truck will experience larger force. But Newton's Third Law of motion says that 'to every action there is an equal and opposite reaction'. So the force experienced by the truck (M) should be same to that experienced by the car (m), but negative, isn't it? Third law states that momentum must be equal and opposite if both vehicles must come to a ...


0

The author of the question presumably expects you to notice these facts both have the same velocity they collide and stop The collision last for one second and to work out the total force acting on each vehicle from the change in their momentum. The answer given is arguable correct if the author is talking about the total force (i.e. not just the ...


1

They both experience the same force because of the impact, due to the Newton's third law, like you say. I think the question is not clear enough. If you assume there is no friction between the trucks and the ground, then you can use momentum considerations. I know this shouldn't be an answer, but I'm new and I can't post a comment, yet.


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Perhaps this can be done by the readings collected by doing simple archimedes displacement experiment. The volume of liquid overflow into the eureka can be considered as volume of object submerged since the water is initially full.


1

Assume the contrary, the volume of liquid displaced is greater than, or less than, the volume of the object. Then there is either a volume containing neither water nor object, or there is a volume occupied by both water and object.


1

You're just thinking 'upside down'. The 3 ohm resistor is just 1/4 the resistance of the 12 ohm resistor but, power is inversely proportional to the resistance. Thus, if you decrease the resistance by a factor of X, you increase the power by a factor of X. In this case, the resistance is decreased by a factor of 4 so the power is increased by a factor of ...


0

what is your logic behind setting V=1? you do not have enough info to assume that. What you do know is that the voltage is constant (same cell), so you can write: $P_{ini}*R_{ini}=P_{final}*R_{final}$ Knowing both R's gets you $P_{final}=4P_{ini}$


1

Original answer This question when applied to the Earth is purely academic. The easiest solution is the one posted by Johannes: One revolution per one hour and 24 minutes. Why go beyond that? The question is academic. It's not academic when applied to asteroids. There's an interesting effect, the Yarkovsky–O'Keefe–Radzievskii–Paddack effect, aka the YORP ...


2

How fast would a sphere need to rotate for a dust speck at its equator to achieve balance between gravitational attraction and centrifugal force? If you do the math (equating $G M m / R^2$ to $m \omega^2 R$ and using $M = \frac{4\pi}{3} \rho R^3$ as well as $\omega = 2\pi f$), it follows that the size of the sphere is entirely irrelevant and that only the ...


0

In this problem you need to consider the difference between the G.P.E. of the mass on the Earth's surface and the G.P.E. at this intermediate point. You haven't said how you have calculated this, but it seems likely that you have made the total gravitational field equal to zero at point P. This is an approximation. You should understand that if you really ...


0

This is an interesting question. Unfortunately it's impossible for everyone to be weightless at the same time because the spin effect would mostly effect people living on the equator and people on the poles would practically not notice it. Also if it's at the sea level then seas would be weightless at the surface as well and eventually that water would fly ...


1

Think of $\lvert \psi \rangle$ as being written $a \lvert n \rangle + b \lvert n{+}1 \rangle$ -- it is just a linear combination of $\lvert n \rangle$ and $\lvert n{+}1 \rangle$ with (possibly complex) coefficients $a$ and $b$. Converting from a ket to a bra (i.e., finding the dual) distributes over addition and scalar multiplication, and it ...


2

You calculated the specific potential energy at that distance. You were asked to calculate the potential energy needed to reach that point. You did two things wrong in that calculation. You forgot to multiply by 930 kg and you forgot to use the given condition "that the potential energy of a 930 kg object on the Earth's surface is -58.7GJ". What you need ...


2

Hint: Divide the $n$-sided regular polygon into $2n$ right-angled triangles. Calculate the moment of inertia for a triangle with the appropriate dimensions and appropriate mass rotating about the appropriate point, which will require a much easier integration than would calculating the moment of inertia for the whole polygon at once. Then multiply the ...


1

There's an error in that the type of pipe for each of the two fundamental frequencies as described in your comment don't match the problem description. The pipe with a fundamental frequency of 440Hz is open-closed, and the pipe with a fundamental frequency of 660Hz is open-open. You actually said "closed-closed", which isn't even an option, but even if ...


0

Once you have the Lagrangian (1) (there is a mistake in your equation, look at the book), just express the momemtum as $p = \partial L / \partial \dot{q} = \dot{q}/\omega$. Insert this expression in the lagrangian expression, and you have the result.


1

The angle between $a_1$ and $a_2$ should be $2 \pi/3$, not $\pi/3$. Your equation is then $$\left(\frac23 a\right)^2 + \left(\frac13 a\right)^2 + \left(\frac12 c\right)^2 + 2\left(\frac23 \frac13 a^2\cdot \cos(\frac{2\pi}{3})\right)=a^2.$$ This gives $\frac49 a^2 + \frac19 a^2 + \frac14 c^2 - \frac29 a^2 = a^2$ which implies $c^2/a^2 = 8/3.$


1

When the magnet is rotating and the coils are stationary, you have an AC generator unless you play tricks with brushes. This is because the net flux change in the coil after a complete revolution is necessarily zero - and thus the net voltage must be zero too. The sinusoidal waveform in your second sketch is therefore what you expect to see in a brushless ...


3

Yes is correct. The rotation of the armature commutes the connection to the circuit so current in the circuit flows in the same direction regardless of the direction in the armature. If the armature is fixed, it will only couple to the circuit in a unique way, and the circuit will experience the current in the same sense as the armature, which is the way you ...


2

The differential equation is not complicated and has an analytical solution. Using $F=ma$ you get $m\frac{dv}{dt}=-a\cdot e^{b\cdot v}$, move both dt and the exponential to the opposite side and you get the solution for $v$ after integration of a simple exponential. Repeat using $v(t)=dx/dt$ to get the distance


0

There is only one force to consider, the friction force. So you use Newton's 2nd law to write the equation of motion, remembering the relationship between velocity and acceleration. The rest is just solving that differential equation for the velocity, which can be done numerically, or by looking it up in a table, or with math software. Once you have that ...


2

"I found out that F applied to mC must equal to the tension exerted on mA" is wrong F=(m_a+m_b+m_c)*a T=m_a*a T_x=m_b * a T_y=m_b *g T^2=T_x^2+T_y^2 Replace and get F=[(m_a+m_b+m_c)*m_b*g]/sqrt(m_a^2-m_b^2)


0

Hint 1: Everything accelerates (the velocity increases) downwards by 9.81 m/s, every second - on earth. Hint 2: The relationship between height and velocity can be calculated using kinetic and potential energy equations: 0.5*m*v^2 = m*g*h m = mass, v = velocity on the ground, g = gravity (9.81) and h = initial height. Since you know the velocity ...


0

To be honest I'm having a hard time understanding your question. Do I get you right, that the wind velocity is given at the positions $(x,y,z)$ by $(vx,vy,vz)=(az,bz^2,c)$? Which of $az$, $bz$ and $c$ are constants or do they all depend on any of the variables $(x,y,z)$? I also don't get what you should calculate "How much wind" is not specific. Do you ...


0

Since this is a homework question, I will point you in the right direction rather than do your work for you. Reading the question carefully, I believe you are trying to solve the following situation: There is nothing ambiguous here. You have two "red" charges (identical, 1.8 uC), and need to determine the value of the "green" charge such that the ...


0

Doesnt the system collapse into ζ1 given we know this is the state at time t=0? True, but after the measurement at $t_0$ the system follows the unitary dynamics given by $H$. As you can easily check, $\zeta_1$ is not an eigenstate of the Hamiltonian, iff $E_1\neq E_2$. Hence the system will not remain in the state $\zeta_1$ during time-evolution. If ...


0

Intuitively, the force will be applied to the total mass of the blocks and cause an acceleration (in addition to the one of gravity) of $\frac F{m_1+m_2}$, so the total acceleration will be $\frac {F-m_1g}{m_1+m_2}$ In your solution you cannot replace $F$ with $m_2a$ because $F$ is not the whole force on $m_2$ It is correct to say $F-T=m_2a$. You ...


0

Mixing units is always a recipe for trouble. As you know, angular acceleration follows similar laws as linear acceleration. Specifically, the angle $\theta$ at time $t$ is given by $$\theta(t) = \theta(0) + \omega(0) t + \frac12\alpha t^2\tag1$$ which is the analog of $$x(t) = x(0) + v_0 t + \frac12 a t^2$$ In your case, the angular velocity is ...


4

Using cylindrical coordinates with the origin at the center and the $\phi = 0$ direction 'down' (the OP says the image should be rotated CCW 90 degrees), the electric field appears be have only a radial component with a sign change for $\phi = \frac{-\pi}{2}$ and $\phi = \frac{\pi}{2}$ $$\vec E = E(\rho,\phi)\hat\rho $$ $$E(\rho,\phi) = ...


1

I have came up with this: Charges are the sources of the electric field. So, whatever the point that field lines are "created" or "destroyed", must be a charge. Then, if there are a charge, then must be on the center. Calculating the electric flux: $$ \phi = \iint_S\ \mathbf E\cdot d\mathbf s = \frac{Q}{\epsilon_0} $$ Let's pick a sphere as gaussian ...


1

what is the energy of a photon? how much has that energy changed after interacting with the electron? what is the formula for the kinetic ennergy of an electron? if the electron was not moving, and all of the change of the photon energy changed th eelectron's kinetic energy, then what is the speed of the electron? divide the answer from 4. by c.


1

If the drag force is being modeled as a linear function of velocity $(\vec{F}_D=-b\vec{v})$, then the problem is straightforward. The vertical force balance for a falling droplet is $$\Sigma F_y=mg-bv=m\dot{v},$$ which gives the following differential equation for the velocity: $$\boxed{\dot{v}+\frac{b}{m}v=g}.$$ In the limiting case of the maximum ...


1

My interpretation of the question; specifically the "lost seconds per day" A pendulum is used to regulate a clock; in its original location, the clock keeps perfect time, as judged against some standard: WWV time signal, chronometer, etc... The clock is packed up, transported to a new location, set up and set running. The clock is set to the correct ...


0

The acceleration is in the opposite direction from the initial velocity. If you are measuring positive upwards (by taking the initial velocity to be $+16.1$) the acceleration of gravity is negative. Your second answer is correct, but has one place too much accuracy as your data only goes to one place behind the decimal point.


1

This is my interpretation: If the pendulum has a period $\tau$ (2 seconds in this case) then the number of oscillations per day,$N$, is the number of seconds in a day divided by $\tau$: $$ N = \frac{86400}{\tau} $$ If the pendulum loses $T$ seconds per day (20 seconds in this case) then the number of seconds lost per oscillation, $\Delta\tau$, is: $$ ...



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