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

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Measure F at 3 (or more) different velocities and curve fit a parabola to the data.

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There are couple of methods that is widely used. incline plane method horizontal plane method There is a very good expriment suggested by Pellissippi State Community collage Below is the link that defines what and how to do it. Check out the below link ...

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You have a net force $$F = F_G - F_D$$ $$ma = -mg + C_0 v^2$$ Dividing both sides by m, letting $C_1 = C_0 / m$, $$a = -g + C_1 v^2$$ $$\frac{dv}{dt} = -g + C_1 v^2$$ $$\frac{dv}{-g+C_1 v^2} = dt$$ Now to integrate both sides, we need to use the limits $[0,t]$ for $dt$, $[v_0,v]$ for $dv$ (where $v_0$ is our initial velocity. $$\int_{v_0}^{v} ... 2 If it's falling only then you have F_d=+Cv^2, where up is the positive direction. You said there is a gravitational force F_g=-mg. Write a Newton's 2nd Law equation, set a=\frac{dv}{dt}, rearrange, to get dv/g(v) = dt, (I'll let you find g(v)) and integrate away. The v integral is not trivial. Look it up in an integral table, if your teacher will ... 2 Solve for F_b from the horizontal braking distance. Assume F_b is constant, then during braking kinetic energy has been converted to friction work:$$F_b \Delta x = \frac12 mv^2$$where \Delta x=123\:\mathrm{ft} is the braking distance and v=60.0\:\mathrm{miles/hour}. I've not checked the rest of your work. You don't need to invoke friction ... 1 You almost have it. The last step is$$ \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}. $$The result follows immediately from what ... 1 Hint : the block starts from rest 11 cm from equilibrium. You don't give it a push. That means that after one oscillation, it'll come back at the same place and never go further away than 11 cm. That makes the rest of the work become trivial. 1 Your mistake is: since k is independent from the total length of the spring, the half long spring still has k as force constant. We can define a constant K to be the force per relative extension of the spring:$$ F = K \frac{x}{L} $$where x is the extension and L is the original length. This constant K is an intrinsic property of the ... 1 You can calculate an average power per square metre, but it's going to be a somewhat meaningless figure because the Sun only shines for part of the day and the strength of the sunlight varies continuously throughout the day. But if you want to go ahead this is how to do it. Suppose the total power (not just the PAR) per square metre is P (in watts per ... 0 When a torque is applied to an object it begins to rotate with an acceleration inversely proportional to its moment of inertia. This relation can be thought of as Newton's Second Law for rotation. The moment of inertia is the rotational mass and the torque is rotational force. Angular motion obeys Newton's First Law. If no outside forces act on an object, ... 1 Your solutions are wrong. As$$ \frac{dq}{dt}=\frac{\partial H}{\partial p}\qquad \frac{dp}{dt}=-\frac{\partial H}{\partial q} $$you get$$ \frac{dq}{dt}=10\,p\qquad \frac{dp}{dt}=0 $$i.e.$$ q(t)=10\,p_{0}\,t+q_{0}\qquad p=p_{0} $$The q coordinate flows in time in straight lines, while the p coordinate doesn't change in time. So each phase ... 1 No, placing the spheres at he vertices of a cube is not the most efficient packing. The most efficient packing of spheres is in a face-centred-cubic arrangement. The volume taken up is 74%. Computing this value follows a similar procedure to what you used. https://en.wikipedia.org/wiki/Cubic_crystal_system 0 \displaystyle \nabla \times \frac{dE}{dt} is equivalent to \displaystyle \frac{d}{dt} \left( \nabla \times E \right) This can be shown by breaking down the curl. For example, \displaystyle \frac{d}{dx} \frac{d}{dt} E = \frac{d}{dt} \frac{d}{dx} E since x is independent of t (Newtonian speeds) Take the curl of one of the existing equations (same ... 0 Hey I had the same problem, but the key is that you have to take care about the following thing, if you write$$ \mathcal{M}^\dagger = v^\dagger \gamma^5 (\gamma^0)^\dagger u = - v^\dagger \gamma^0 \gamma^5 u = -\bar{v} \gamma^5 u $$so taking the conjugate of a spinor don't give you automatically the "barred version" The same actually happens when you take ... 1 Here, in the case of your question, the brass rod and the steel rod are connected in parallel combination in between the hot reservoir and the cold reservoir. So, the thermal resistance will be given by, 1/Rp = (k1A1+k2A2)/l..... And, we know that, Q/t = dT/Rp = (k1A1+k2A2)*dT/l. You have been given the values of Q, t, k1, k2, A1, and dT. So, you can easily ... 1 Your system has 2 degrees of freedom, but using x_1 and x_2 will not be helpful in determining the effective spring rate. To get the spring rate you need the extension x of the connection point with mass M and the tilt angle \theta. Do the substitution:$$ \begin{align} x_1 & = x - a \theta \\ x_2 & = x + b \theta \end{align} $$The ... 1 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 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. 0 The "should have been proof" relies in the extensions of the springs being the same. If you put a=b that condition is not satisfied. Find a spot where the two extensions are the same and you will get "should have been proof" answer. 0 Forget all ideas about vectors and velocity and relate it to a car which is travelling at a speed of 70 mph slowing down to a speed of 50 mph. So the idea is negative means slowing down and positive means speeding up. This would be a reasonable start and all the subtleties can come later. 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 ... 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. 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. 0 6.4A (amps) is 6.4C/s (Coulombs/second). Do you know how many Coulombs pass in 15s? Each electron has a charge of 1.60217662 × 10^-19 coulombs. Equally, 6.24150913 × 10^18 electrons make 1 Coulomb (1/1.60217662 × 10^-19). Do you know how many electrons it would take to make the same Coulombs as above? 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} $$where a and b are the side dimensions. The inertia matrix in world coordinates, while rotated by ... 1 First of all, as this is a homework question, I can't tell you the complete solution. Choose P as the origin of coordinate system and resolve the forces into x and y component. And as the body is in equilibrium, the net force is zero. So you get these two relations (when the net force on x and y component equated to zero.)$$G\cos\theta=H\cos\phi\\ ...

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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 ...

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you never feel your weight (you feel weightless ) due to no force reminding your brain of your weight. In case while you stand on earth the earth's crust reminds you of your weight by providing a normal reaction force.

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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 ...

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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.

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Because you never feel your own weight! I'm sitting on this chair to write this, and I can feel the push of the chair up against my body. That is what I think of as my weight - but it is not - it is the push of the chair against my body in response to my weight! If the chair disappeared I'd be in free fall, and I still wouldn't be able to feel my weight, ...

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The reason you feel weightless is because there is no force pushing against you, since you are not in contact with anything. Gravity is pulling equally on all the particles in your body. This creates a sensation where no forces are acting on you and you feel weightless. It would be the same sensation as if you were floating in space.

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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.

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The result you described says that the projection of antisymmetric rank 2 $SO(4)$ tensors onto self-dual and anti-self-dual subspaces commutes with the action of $SO(4)$. This just implies that the space of antisymmetric rank 2 tensors of $SO(4)$ is reducible. To show that the self-dual and anti-self-dual subspaces are themselves irreducible, think of ...

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The action of a massless scalar field is given by: \begin{eqnarray} S(\phi)&=&\int{\cal{L}}dt\\ &=&\int d^{4}x \sqrt{-g}\left(g^{\mu\nu}\phi_{,\mu}\phi_{,\nu}\right) \end{eqnarray} Now choosing a tetrad, i.e., a basis of one form at each spacetime point $\{e^{a}=e^{a}_{\mu}dx^{\mu}\}$ we can rewrite the action as: \begin{eqnarray} ...

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The vapor pressure of your water column at 50 degrees C is approximately 100 Torr, which is 1/7.6 of one atmosphere. The pressure in the water column is given by the equation $P = \rho g h$ , where $\rho$ is the density of the water, g is the local acceleration due to gravity, and h is the height of the water column above a stated point. The water will ...

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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 ...

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I am giving the solutions of original task (to get the speed at point B). I am not sure if the questions are necessary to perform the task. If you are sure the path taken does not matter (and I will assume that per your statement). So, let us consider a straight line path. Vertical component of F overcomes gravity and causes vertical move. Only horizontal ...

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Does the electric field at a point on the surface change? Yes, it change. Where distance is decreased Electric field increase. Does the total flux through the Gaussian surface change? No, it remains same. Consider a spherical gaussian surface of radius $2m$ and a charge $q$ inside it. Let this charge emit $10$ field lines. Then, clearly, all the field ...

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In such circuit see that all resistance at rim have same potential, since no resistance reduced it. At rim all have same potential as that at $a$. At center all have same potential as that at $b$ So, all have same potential difference $V_{a} - V_{b}$ Same potential difference means that they are connected in parallel.

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They aren't. Frictional force down an incline is the coefficient of kinetic or static friction multiplied by mass and the acceleration of gravity and the sine of the angle of elevation. Meanwhile, the component of weight down the incline is all of the above without the coefficient of kinetic or static friction.

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You have identified the problem as the factor $2$ between your calculation and the book. You have read the problem correctly, so it seems the book is wrong.

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You are going about it the wrong way. You need to find $\omega$ first - think about the definition of shm. A very useful relationship when trying to solve this sort of problem is $v^2=\omega^2(A^2-x^2)$. Having used that relationship you can sort out the form of the equation for $x$ and the phase.

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If you did not have resistor R in the circuit the voltmeter would always give the same reading - the voltage of the battery. A thermistor changes its resistance when a temperature changes. You are not given a resistance meter. All you have is a battery, a resistor R and a voltmeter. The circuit as set up is called a potential divider which means that the ...

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Using the first $2$ statements or conditions, apply the principle of conservation of energy and you will be able to calculate the resistive force offered by the target material. Next use the second condition and again apply the principle of conservation of energy to get the required answers, mainly the velocity of the bullet after penetrating the target.

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Potential at center due to +ve sphere is not correct. What you had found is when cavity is at center. However potential due to $-\rho$ is correct. First consider no cavity Potential at center of sphere due to uniformly charged complete sphere $V = 3kq/2a$ Now, potential due to positive charged sphere $cavity$ at center. $$V_{1}= \frac{4\rho π ... 2 Always, always, always start problems like this by drawing a diagram: This make it obvious why cos and sin are used as they are. 1 Not sure about your question. Here are some suggestions: In thermodynamics we define internal energy as U = U(S, V) (number of particles is constant). Then we write it's differential:$$dU = \frac{\partial U}{\partial S}dS + \frac{\partial U}{\partial V} dV Now we recall 1st law of thermodynamics $dU = \delta Q + dW$ together with 2nd (or it's ...

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It says at the start of the solution the fraction of incident particles scattered through an angle greater than $\theta$ is given by . . . . So that formula is being applied. It might well be that the formula was derived by first finding the fraction scattered from $0^\circ$ up to and including $\theta$ and then that fraction taken away from $1$ to ...

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