A force F acts on a cart in motion on a frictionless surface to change its velocity. The initial velocity of the cart and final velocity of each cart are shown. You do not know how far or in which direction the cart traveled. Rank the energy required to change each cart's velocity from greatest to least.
Cart A changes from 5 m/s to 2 m/s and has a mass of 2 kg.
Cart B changes from 3 m/s to -3 m/s and has a mass of 3 kg.
Cart C changes from 5 m/s to 6 m/s and has a mass of 5 kg.
Cart D changes from -1 to 2 m/s and has a mass of 4 kg.
Does this problem use W = 1/2 mv^2 formula? Does mass matter since it's frictionless and is moving in the x direction?
]]>a) Derive an equation to find electric field on the surface of a big conductor plate.
b) What is the surface charge density (in C/m^2) on the above mentioned plate, assuming that is a conductor?
]]>This is my strategy:
The problem is that I cannot find the final pressure in the Bernoulli's equation. What I mean by that is that I have this:
P_{1} + 1/2ρv_{1}^{2} + ρgy_{1} = P_{2} + 1/2 ρv_{2}^{2}+ ρgy_{2}
Thanks! I appreciate any help that comes my way.
]]>I have a question for 2. a. Why is voltage across C0 equal to V0? I understand how this be the case if the resistor R1 was not there (since voltages are the same for components connected in parallel); however, since the resistor is connected to the capacitor in series, wouldn't the voltage V0 be split across C0 and R1 (which means C0 would have a voltage less than V0)?
Also, for 2. b - I originally thought that I2 would increase, because when the switch is first closed, the capacitor acts like a wire and allows current to flow freely, so the current will be split between both the R1 and R2 parallel branches. Then, after a long time, the capacitor acts like an open switch, so no current will flow through R1 and it will all flow through R2 - therefore, the current in R2 would increase. After watching this video, I understand Mr. Fullerton's reasoning as to why I2 remains constant, but I'm not sure what was wrong with my original thinking.
Thank you for any help!
]]>I'm confused about this question: "A proton is moving to the right at constant speed v and enters a region with uniform magnetic and electric fields. It continues to move in a straight line. The magnetic field is directed toward the top of the page. What is the direction of the electric field?"
Using the right hand rule, with fingers pointed to the right and curled fingers up, my thumb points out of the page, leading me to believe that would be the answer. The correct answer is actually that the electric field points into the page. Why is this?
Thank you!
]]>This question features a circuit with a resistor, an inductor, and a battery all connected in series. It then shows an increasing concave down graph of something vs. t - the graph started at the origin and had a horizontal asymptote at some positive value of x. The "something" could be:
A. The potential difference across the resistor
B. The potential difference across the inductor
and/or C. The current in the circuit.
I had reasoned that it would be all 3. When an inductor is first connected to a battery at time t=0, it doesn't allow electricity to flow, but as time goes on, it allows more and more electricity to flow until it is essentially acting like a wire. Therefore, current will increase, and voltage will also increase across the entire circuit. However, it turned out that B was incorrect, and I'm not sure why (I eventually received the answer to this question but not the explanation). Why is this the case?
Thank you!
]]>
I'm not understanding the worked out solutions, especially why the tensions were subtracted. Explanations would be appreciated
]]>In the explanation you talked about the pressure holding the lid, but you didn't include in the net force.
]]>Then, R24 (combined) would be 2R. R234 (combined) would be in paralled and be 2R/3. Then, the Req would be R1+R234 (R+2R/3) which is 5R/3. Then wouldn't the current be 3/5? Not .75.
]]>which unit is equivalent to a newton per kilogram
the answer is m/s^{2}
How
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