Hot answers tagged

4

Because what you feel as weight is actually the force you must exert to your surroundings not to collapse, or start falling anywhere. Weightless is just absence of this force. In fact, I can not imagine what should one feel in a free fall but weightless.


2

In your last question it is important as to what you mean by WRT the question. If you are trying to find the E-field due to a point charge using Gauss then to make the surface integration easier you choose a surface which has the following properties: the E-field direction is everywhere perpendicular to the surface the E-field has a constant ...


2

Reusing your car example: use the fact that acceleration is "change in velocity". This can be positive (acceleration in the usual/common sense), but everyone knows that velocity can also be decreasing. This is what phsicists call negative acceleration.


2

You feel weightless while free falling because no normal reaction force is present. While sitting on a chair you feel the normal reaction force that chair exerts on you which is equal to your weight and the force you exert on the chair.


2

The inertia matrix for a thin rectangular foil (laying along the xy plane) in body coordinates is $$ I_{body} = \begin{vmatrix} \frac{m}{12} b^2 & 0 & 0 \\ 0 & \frac{m}{12} a^2 & 0 \\ 0& 0 & \frac{m}{12}(a^2+b^2) \end{vmatrix} $$ where $a$ and $b$ are the side dimensions. The inertia matrix in world coordinates, while rotated by ...


2

You can calculate an average power per square metre, but it's going to be a somewhat meaningless figure because the Sun only shines for part of the day and the strength of the sunlight varies continuously throughout the day. But if you want to go ahead this is how to do it. Suppose the total power (not just the PAR) per square metre is $P$ (in watts per ...


2

Your mistake is: since $k$ is independent from the total length of the spring, the half long spring still has $k$ as force constant. We can define a constant $K$ to be the force per relative extension of the spring: $$ F = K \frac{x}{L} $$ where $x$ is the extension and $L$ is the original length. This constant $K$ is an intrinsic property of the ...


2

The work done by you when you apply a force on the ball does two things: It increases the gravitational potential energy of the ball up to the point of release. It increases the kinetic energy of the ball up to the point of release. This is you energy method. In terms of forces if you just lifted the ball and it had no kinetic energy at the end what force ...


2

You almost have it. The last step is $$ \Omega = \frac{\rho}{\rho_\text{c}} = \frac{\rho_0\,a^{-3}}{\rho_\text{c,0}}\frac{\rho_\text{c,0}}{\rho_\text{c}} = \Omega_m\,(1+z)^3\frac{H_0^2}{H^2}, $$ where we used the critical density $$ \rho_\text{c} = \frac{3H^2}{8\pi G},\qquad \rho_\text{c,0} = \frac{3H_0^2}{8\pi G}. $$ The result follows immediately from what ...


2

Solve for $F_b$ from the horizontal braking distance. Assume $F_b$ is constant, then during braking kinetic energy has been converted to friction work: $$F_b \Delta x = \frac12 mv^2$$ where $\Delta x=123\:\mathrm{ft}$ is the braking distance and $v=60.0\:\mathrm{miles/hour}$. I've not checked the rest of your work. You don't need to invoke friction ...


2

If it's falling only then you have $F_d=+Cv^2$, where up is the positive direction. You said there is a gravitational force $F_g=-mg$. Write a Newton's 2nd Law equation, set $a=\frac{dv}{dt}$, rearrange, to get dv/g(v) = dt, (I'll let you find $g(v)$) and integrate away. The $v$ integral is not trivial. Look it up in an integral table, if your teacher will ...


1

$$PV=1\times RT \space For \space O_2 $$ $$ P_2V=1\times R\times 2\times T \space for\space He2$$ Divide Both $\frac{PV}{P_2V}=\frac{T}{2T}$ $$\frac{P}{P_2}=1/2$$


1

At the instant you release the arrow, it begins to lose velocity due to gravitational acceleration, which can be represented by a vector pointing opposite to the arrow's direction of flight. The arrow loses about 9.8 m/sec of its velocity every second (this is the magnitude of gravitational acceleration at the Earth's surface). Solve for the time it would ...


1

How long does it take to stop? $v/a$ or roughly 2 seconds. How far does it go up in 2 seconds? ${1/2}at^2$ or 20m, roughly. You figure it out exactly.


1

There are some very useful elementary equations that describe basic motion with constant acceleration, and these are: $$v=u+at,$$ $$v^2=u^2+2as,$$ $$v=ut+\frac{1}{2}at^2,$$ where $u$ is initial speed, $v$ is final speed, $s$ is displacement (how far the object has moved) and $a$ is acceleration. You must now think about your problem to determine which ...


1

You have to approach this from an energy standpoint. You start with a certain amount of energy, in this case gravitational potential energy, which is dependent on the initial height ($h=4\ \text{ft}$). At the end, you have a final energy, in the form of kinetic energy, which is given by $\frac{1}{2}mv^2$. With no other energy terms involved (meaning you ...


1

I left a hint yesterday and since no one has followed up on it, I'll provide the answer. Knowing that the radius of the neutron is about $10^{-15} m$(1 fm), one can write (using the uncertainty principle)$$\Delta x \Delta p >\frac {\hbar} {2}$$ Now using the fact that for a bound particle the uncertainty of momentum is approximately the RMS momentum ...


1

If $\vec{A}$ is timelike then we can find an inertial frame in which: $$ \vec{A} = (a, 0, 0, 0) $$ Can you take it from here?


1

Suppose you have an infinite thin sheet of charge with a certain charge density. By symmetry we know the electric field is perpendicular to the sheet. You can calculate the field using this method: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesht.html $E = \dfrac{\sigma}{2\epsilon_0}$ This field is constant everywhere. If we have a second sheet ...


1

I'm sorry but it's a easy problem just using the ladder operator.You need to let x applied on the wavefunction one by one and the rest is easy. Applying the following property which can easily derived from the ladder operator (or just using the wavefunction,either way is ok,the specific procedure can be found in Griffiths' book): $\displaystyle ...


1

Hint : the block starts from rest 11 cm from equilibrium. You don't give it a push. That means that after one oscillation, it'll come back at the same place and never go further away than 11 cm. That makes the rest of the work become trivial.


1

Positive means speeding up, negative means slowing down. Now this is assuming you are traveling in the positive direction but through an axial change you could always guarantee this. I think this would be a good starting point for a child.


1

It is true that two electrons can't have identical quantum numbers, but spin itself is a quantum number. That is, the state with quantum numbers 111 can hold two electrons: one of spin up and one of spin down. So when you are finding the ground state, for example, find the four lowest energy eigenstates (ignoring spin), and the ground state of eight ...


1

You don't feel gravity pulling on you, because gravity doesn't actually pull on you or anything else. Gravity bends spacetime. What this means is that your own perceived frame of reference (a system of coordinates to measure space and time), which is all nice and straight, does not match up with the actual shape of spacetime. So even though you think you are ...


1

First of all, as this is a homework question, I can't tell you the complete solution. Choose P as the origin of coordinate system and resolve the forces into x and y component. And as the body is in equilibrium, the net force is zero. So you get these two relations (when the net force on x and y component equated to zero.) $$G\cos\theta=H\cos\phi\\ ...


1

I would explain that we can feel acceleration in a car. As our speed increases we are pushed back into the car seats. This is positive acceleration. As we slow down we are pulled back by the edge of the seat/seat bell. If the child is happy that acceleration is a change of speed then they should be able to tell with their eyes closed if the ...


1

There should be a negative sign in Equation 2. The exercise is a very simple one in substitution and does not require any sign cancelling.


1

It is a method that is generally used for conservative unidimensional problems (problems with only one degree of freedom, here your angle $\theta$ or cartesian coordinate $x$). You'll notice that it is equivalent to using Newton's second law in this case : let us write the total energy $E = \frac{1}{2} m v^2 + V(x)$, $V$ being potential energy. The problem ...


1

Your system has 2 degrees of freedom, but using $x_1$ and $x_2$ will not be helpful in determining the effective spring rate. To get the spring rate you need the extension $x$ of the connection point with mass $M$ and the tilt angle $\theta$. Do the substitution: $$ \begin{align} x_1 & = x - a \theta \\ x_2 & = x + b \theta \end{align} $$ The ...


1

Here, in the case of your question, the brass rod and the steel rod are connected in parallel combination in between the hot reservoir and the cold reservoir. So, the thermal resistance will be given by, 1/Rp = (k1A1+k2A2)/l..... And, we know that, Q/t = dT/Rp = (k1A1+k2A2)*dT/l. You have been given the values of Q, t, k1, k2, A1, and dT. So, you can easily ...



Only top voted, non community-wiki answers of a minimum length are eligible