What is Net: Definition and 1000 Discussions

A diagram is also provided, which looks like this: I'm not sure what is correct and after doing some digging online I still haven't been able to come to a consensus. I'm currently stuck between one of two possibilities: positive or zero. I'm pretty sure work on A (Wa) and work on B (Wb) are...

I started by summing the forces and torques to get: - ma = mg-T - I*alpha=Tr I then used a=alpha*r and I=Mr^2 to combine the equations and solved for angular acceleration equals 81.75rad/s^2. Plugging this back into a torque equation I got that the net torque is 1.04Nm. However, the problem...

The mathematical representation of the net external force on a system(obtained from Newton's second law) is ##\vec F_{net} = \frac {d\vec P}{dt}##, which is the rate of change of linear momentum of the system. If we substitute ##\vec P = m\vec v## into the formula for force and differentiate, we...

This is how I interpreted the problem, a) The net torque about point A is zero. This is because the forces F1 and F2 are equal and opposite, and they act at the same distance from point A. Therefore, they produce torques that cancel each other out.. The force F3 doesn’t does not produce any...

I have tried finding the tension of the string through the equation Net Force = T - mg = ma but I am struggling with coming up with the correct acceleration. Is my net force equation correct?

m1 = 4 kg, m2 = 12 kg, m3 = 8 kg. k = 327 N/m for all three blocks. The elevator accelerates upwards at 3.8 m/s^2. Net force of block one would be equal to force applied by top spring minus weight of system, since top spring is holding all 3 blocks. F1 = 4*3.8= Fs,top - Wsystem = Fs,top -...

I'm a little confused because my teacher used Bill's 500J of work for the kinetic energy equation and I don't understand why. I used the net work, so 300J, to find the speed and I'm not sure why that's wrong. Wouldn't friction make the wagon move slower than if there was no friction? So why...

The net torque about an axis through point A is given by, If I take the axis of rotation perpendicular to the paper and the solution I arrive would be the following below Net torque = 30 cos45 x 1.5 - 10 cos30X 3 = 5.829Nm ( counterclockwise) But the book gives an answer...

Net torque = 2gx20xcos30 - 2gx35xcos20 - 2gx15xcos60 + 2gx30xcos80 = -357Nm Have I done it right?

If there is no force being applied can there still be a net force? For example, supposed you apply a force F forward on an object on a frictionless plane, the moment you stop applying the force, the object may still be accelerating towards its final velocity but there is no applied force. So if...

In a system, the net internal torque should be 0. If we have two fly wheels, one spinning with angular velocity w, and the other at rest and the flywheel at rest is dropped onto the other flywheel, the two fly wheels reach the same angular velocity due to friction between the two wheels. I am...

We know the net force on the charged particle in the uniform electric field pointing up is mg - qE. To get acceleration, divide the net force by mass to get g - qE/m Plug into kinematic equation and get velocity by itself and substitute$$\sqrt{h(2g - \frac{q \sigma}{\epsilon_o m})}$$

Find the net force of A as shown in the diagram.

1 example to represent the question in semantic net. https://lh4.googleusercontent.com/Gu5Cd7Xha3IDmtFfCHA7P_GCsQc9HyN9gt3a_M_2g1v_U3g8dHQyZkM537SZvQyhum0DyaaTGrmvVLBZWNq5z2tRFK_2fqg7y739u79WcDN31ipqtNzFH7CL95AlQgAMZOE7IPr9 Second question-: every student loves to party. Represent this in...

Hey Everyone, I am working on making a simulation of a rock climber on a wall in 3D space. The end result is to help visualize the free body diagram of the climber and show how moving your limbs moves your center of mass and changes the forces on your hands and feet. I've been struggling to...

Q1a) - My current wrong answer is <-5.9e7, -3.3e8, -2.17e8> I used the Fnet = Gm1m2 / d^2 <unit vector> But i keep getting a dif answer each time Q2a) - I thought i could find net force and then divide it by the mass, and multiply it by the time interval. However I got the answer <-4.5, 4.5...

in a cours of electrostatic when we have a positive charge and we bring another one (also postitive)we have to do work and apply a force that equals the force of repultion over the distance which seems weird because if we do that the net force will be equal to 0 and the charge will not move can...

Here is the link to the question. https://www.khanacademy.org/science/physics/forces-Newtons-laws/Newtons-laws-of-motion/a/what-is-Newtons-second-law What is the magnitude of the force F1 ? What is the magnitude of the force F2 ? Here is my drawing I made At this step I am lost. ## ay...

The slinky is designed to fully contract in 1 second. During this one second, the mass is weightless and move up at constant speed of 1m/s. After 1 second the mass gain 1m height in potential energy. Am I missing something?

Here is a diagram of my interpretation of the problem: Where I'm thinking that the engine originally takes heat from ##T_h## to ##T_l##, in which case ## \frac { Q_{h} } { T_{h} } = \frac { Q_{l} } { T_{l} } ## and ## W_{out} = Q_{in} - Q_{out} = Q_h \left( 1 - \frac {T_l} {T_h} \right) ##...

Hi! Given three voltages as follows; Q1 = 1C,Q2 = 1C,Q3 = 2C The distance a is 1m and b = 2m a) Find the values of the forces that are acting on Q2 I did that like this; $$ F_{12} = \frac{Q1*Q2}{4\pi\epsilon r^2} $$ $$ F_{32} = \frac{Q1*Q3}{4\pi\epsilon r^2} $$ The results are ...

I think both spools will land about at the spot x (there aren't any horizontal forces causing them to land away from the x). Also, I think student 2 might be the closest to being correct, but I'm not sure if they're entirely correct; isn't it possible that force doesn't only go to translational...

I don't understand why the answer is B. Here's my thinking: Since it wasn't mentioned that there's any other force aside from the drag force, then Fnet = Drag force I know that Fnet = ma, and since they have different masses, Fnet must be different for both balls? Acceleration cannot be...

Hey! 😊 You participate in the following game : You toss a fair coin until heads falls, but no more than three times. You have to pay $1$ euro for each throw. If your head falls, you win $3$ euros. The random variable $X$ describes your net profit (profit minus stake). Give the values that $X$...

The net Electric field(inside the dielectric): $$E_{net} = \frac{1}{4\pi \varepsilon_0 \varepsilon_r} \frac{q}{r^2}$$ $$\vec E_{net} = \vec E_{applied} - \vec p$$ where p is the polarization vector. let charge ##q_{-}## be present on the inner surface of dielectric and ##q_{+}## on the outer...

Hello, My question relates to gamma spectroscopy. I understand how the net peak area is calculated for any photopeak. Fortunately, gamma-spec software (e.g., Genie-2000 from Canberra) provides Net peak area and associated uncertainty (for Cs-137 661.7 keV peak, as an example). My question: are...

I'm not sure of my answer but it is e(downwards). I arrive at this answer since gravity always exists so no matter what happens it always pulls an object downwards. But I am not sure. Thank you -sun1108

Just to be clear, this isn't a homework problem. it is an example problem found on page 68 of the text "Matter and Interactions" 4th edition. The solution is given in the book, but I'm having difficulty following their reasoning. according to the book the net force is not constant, therefore we...

I tried just calculating the force with Coulomb's law, then calculating the forces for each vector individually and adding, but I got it wrong both ways

In general, it's your assets minus your liabilities. However, how detailed do you go into things you materially own for which there is an unknown/fluctuating market value for - things like clothing, furniture, gadgets, etc. (or do you even include these) - or are invested in (which can also...

Here's what I did: ∮ B * dl ＝μ0 ＊ I ∮ AR * 2π*R ＝μ0 ＊ I ∮ 2π*AR^2 / μ0 = I ∮ 2π*AR^3 / 3μ0 = I Where did I do wrong?

Hi, I was working on the problem below: Question: An optical fibre transmission system uses a step-index multimode optical fibre which has a core refractive index of 1.49 and a cladding refractive index of 1.48. The fibre is also subject to material dispersion which is a function of wavelength...

Can someone help me understand what I did wrong when solving this problem?

In the solving portion of the textbook, the reasoning of the author in solving this problem is that the net work done on the system is zero because the net force of the system is zero. So my question is how was the system in equilibrium (net force=0)? My thinking is that since it is stated...

I am reviewing this problem from my textbook and am having some trouble with it. When solving for ΣTorque I chose the hinge to be the axis of rotation. I understand that from this point if the board is being held up and it were to fall after removing support, mg (the center of mass of the board)...

I'm tunneling into my corp's webclient. Every time it wants to do something, it overrides my spotify in my local (untunneled) browser - resulting in my music stuttering constantly). I'm using 'Global Protect' to tunnel into vdi.[mycorp]/portal/webclient/... Though it might solve the problem, I...

In AB, dU=0 (Isothermal) Hence dQ=dW W1= RT ln(2V/V) = RT ln(2) = 800R*ln(2) Q1=W1 = 800R*ln(2) Q2= Cp(300-800)=-500Cp = -500*2.5R Q3 = Cv(800-300) = 500Cv = 500*1.5R So Q= 800R*ln(2) -500*2.5R + 500*1.5R = 800R*ln(2) -500R Hence net heat absorbed is 800R*ln(2) -500R But. Suppose I find...

I realized that the tension in the rod is not uniform and found it to be ##T = Wx/L## I found this by splitting the rod into two sections one of length ##x## and the other of length ##L-X## where x is the length from the base of the hanging rod To find the total elongation in the rod I...

So what I did was find each of the forces the masses had on m1 using the above equation. From m2 I found 19.975 in the negative i hat and for m4 i found 29.96 in the positive k hat direction using the above equation. For m3 I used pythagorean theorem to calculate r, which was 3.25 (so r^2 was...

So 0=q1q3/r+q2q3/(13.6-r) 0=-8.5/r-3.63/13.6+3.63/r 0.2669=3.63/r-8.5/r r=-18.2465, but the answer is supposed to be 8.24

w= 0.50N, Fn= 7.8N, angle= 29 Fnet= Fn*fd* costheta = 7.8*0.50*cos(29 = 3.41 is my answer correct

I understand that single integrals over a function can be interpreted as net change. Net change of the quantity between the bounds of the integration. But I am trying hard to understand if double integration can also be regarded as net change? That is, the net change in volume when the two input...

Hi All, Please see attached photo of the question. It is asking for net force and acceleration. Taking the forces acting upwards and downwards on the parachutist as vectors: for A the net force would be 800-300=500; B would be 0 and C; would that be 800-1500=-700? And if so would that imply...

The current direction is as follows I think so much and do the right hand rule i get 0 at the center, but not sure why the answer is non zero. I have shown the directions of the magnetic fields, i have not shown the magnitudes of equal length but they all are equal. Why the answer is non zero...

I draw the graph like this: For (b), I divided each force vector to e from p1 and p2 as x and y parts. I computed them and got Fx=-4.608*10^(-15)N Fy=-2.52*10^(-15)N However, I am not sure whether I did it correctly or not... I appreciate every help from all of you! Thank you!

The electric field caused by the surface distribution on a point ##a## meters far from it is$$E(a)=\frac{kQ}{(R+a)^2}$$from which I get$$Q=\frac{(R+a)^2E(a)}{k}=\frac{(R)^2E(0)}{k}=\frac{(0.705)^2\times867}{8.99\times10^9}\approx4.79\times10^{-8}$$and I take its negative because the direction of...

Suppose we have placed a cube in field which varies linearly with z axis so electric field magnitude on coordinates of face ABCD is clearly more than face EFGH and we know area of both faces are equal, So if we calculate flux then it would be non zero but it contradicts with the fact that...

2500 N [south] + 3200 N [south] = 5700 N [south] 5700 N [south] - 6000 N [north] = 300[north]