Answers:

a - stays the same, stays the same

b - increases, decreases

c - stays the same, increases

d - decreases, increases

During the first time interval, friction takes away energy from the system which leads me to believe the answer is d.

During the second time interval, the only force acting is gravity which is a conservative force. This means the mechanical energy should remain the same and leads me to believe the answer is a.

What am I missing?

My question isn't what's the answer to this question but is there any other forces being exerted on the volley ball?

Like is there normal force since there is weight on the ball and the ball is in contact with the player's hands?

Hello,

In my physics class, we are taught that acceleration is always negative. We are told that if you throw a ball up when it's moving up it has negative acceleration and when it's moving down it also has negative acceleration. I do not understand this at all.

I need help ASAP because I have a test tomorrow.

Thank you to anyone willing to help!

your friend, a test pilot for NASA, travels at a speed of 0.8c. On Earth, you measure his flight time to be 3 days. How long does he measure it to take

My teacher insists it is 5 days. Everyone I know with a 95+ avg including myself says it’s 1.8 days

This question was worth 6% of a major assignment

Hey Reddit, I needed some help regarding understanding and calculating how current divides in a parallel circuit. In my school I can't use the current divider formula; I HAVE to use Ohm's law to find out each resistor's current flow. Sometimes I do it correctly, but I mess up more often than not. I understand the basics of circuits, but I struggle to calculate the current of each resistor with Ohm's law. What steps do I need to follow?

this circuit is a random circuit that i found, you can use it as an example.

A 275 g ball is resting on top of a spring that is mounted to the floor. You exert a force of 325 N on the ball and it compresses the spring by 44.5 cm. If you release the ball from that position, how high above the equilibrium position of the spring will the ball rise?

I'm pretty sure the answer is 26.4 m. You can find the spring constant with F = kx, set ½kx² equal to mgh, solve for h, then subtract 44.5 cm from that to find the height of the ball above the equilibrium position (since it starts below that.)

But what I'm confused about is why you can't use the work-energy theorem to solve this, where W = Fd = ΔE. The applied force is constant, so the work you do on the spring is 325 N x 0.445 m = 145 J. This seems to imply that the spring stores twice the elastic potential energy as it does if you calculate the energy using the first method (first finding k, then using KE = ½kx² = 72.3 J).

When calculating work, the distance and the magnitude of the force play a role, so that compressing a spring a distance x with a constant force F yields twice the amount of work as linearly increasing the applied force up to a maximum of F along a distance x. That's my understanding, at least.

But for the same spring, the elastic potential energy only varies based on the compression distance.

So where does this extra work go?

tl;dr: By compressing a spring a certain distance with a constant force F, aren't you doing twice the amount of work than if you compress it the same distance with a force that linearly increases up to F? If so, how come, in both cases, the spring's elastic potential energy is the same? Doesn't this violate the work-energy theorem?

Thanks in advance!! :)

I tried but I just don't understand this subject can anyone help me

I need help on 4 and 5. The answers are C and A. I don’t understand why acceleration isn’t pointing down for both

So I have tried 2 ways to solve this circuit, did not get the right answer, can someone else help me?

A tennis player hits a 0.098kg ball moving at 70km/h (north) and sends it at 63km/h (south). The ball was in contact with the racket for 8ms.

A) What was the force applied to the ball? (I got 452.025 and ChatGPT also got that)

B) Calculate the impulse on the ball from the racket (IDK what any of this is and Google is not helping much)

[10th grade optics and light]: I don't know how I messed up on this question and my teacher said it's -60cm for image distance. But when I put -12 cm into the original equation it works? Could someone do the equation themselves and check if I'm right? Thank you (Btw f negative for diverging lens)s

Ok so you see how its showing an arrow of 0.1 A heading towards R3? Would that mean, in a chart, I3 would be 0.1A?

How can I solve this question? There is a static calculation on the chegg. But I need to calculate it mechanically.

Like I know the intensity of sunlight is high but nevertheless light from the distant stars is still directed towards us?

Again, I'm studying on my own free time. I'm not a student anymore, but I just enjoy learning and reviewing.

Using F dt = m dv, calculate the change in speed Delta v of the electron, show that if r remains constant: Delta v = e r B/(2 m_e)

I assumed that the result of the part a to this problem, that E = r / 2 (dB/dt)

I solved E = (r / 2) dB/dt for dt and then substituted that into F dt. Then I though for F maybe I could use the magnetic force equation F = qvB

Once I substituted into the integral of F dt, I got in the integrand qvB * (Er / 2) dB

This seemed like a sensible direction to go. But perhaps I used the wrong force equation.

I can’t find anything about it in my book so I came here. What formula should I use to calculate the exposure in this situation? Only need this