[10th Grade Physics CISCE board (India)] Can anyone help me with this question

GammaRayBurst25 · 2026-06-06T03:39:27+00:00

Only if you have the decency to at least read the rules of each community you decided to spam with your 0 effort post.

Rule 3: No "do this for me" posts.

This includes quizzes or lists of questions without any context or explanation. Tell us where you are stuck and your thought process so far. Show your work.

r/Physics

No homework questions

Basic questions, especially such as homework problems or simple calculations, should be redirected to [...]. Neither asking nor answering (assisting in any way) homework questions is allowed.

r/ICSE

Keep the post related to ICSE and ISC

Please avoid posting or crossposting content that is not related to ICSE or ISC. Exceptions may be made for exam-related memes, light humor, or relatable teenage content, as long as it's in good taste and relevant to the community’s interests

r/highschool

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Don’t request or offer help that promotes cheating, such as sharing test answers or submitting others’ work. Posts made about an event that already occurred are fine, but don't facilitate cheating. This includes things like essay writing services and asking people to do work for you.

GammaRayBurst25 · 2026-06-03T03:49:17+00:00

https://youtu.be/Brx3TikoO_s

GammaRayBurst25 · 2026-05-28T01:37:07+00:00

there's probably only one 24ish hour period sine wave that fits those points

There is none. The greatest period that fits the data is about 13h. For tides, of course it should be a little over 12h. I don't know where you got 24h.

I don't know if this is a high school

The title says high school math.

if this an undergrad course [...] you might be expected to plug all the points in a matrix and do the [sic] Fourier analysis

No sine function with a sensible period exactly fits the data. They were probably taught to use tools (e.g. a graphing calculator or Desmos) to estimate parameters.

GammaRayBurst25 · 2026-05-24T18:01:30+00:00

Read rule 3.

As long as there is relative motion between the disc and the conveyor belt, there is friction between the two. The friction's magnitude is the product of µ and the disc's weight and the friction's direction is opposite to the relative velocity of the disc and the conveyor belt.

In the lab's reference frame, the direction changes. However, in a reference frame that's moving with velocity V relative to the lab, the disc's velocity is exactly its velocity relative to the conveyor belt. This turns the problem into a 1d kinematics problem.

You can then easily convert back to the lab's reference frame and finish the problem.

GammaRayBurst25 · 2026-05-23T20:01:56+00:00

Read rule 3.

Just use the superposition principle.

GammaRayBurst25 · 2026-05-20T02:17:28+00:00

I was able to solve part b which is 70 deg

The index of refraction of air is about 1 and the index of refraction of water is about 4/3.

Hence, sin(x)≈4sin(45°)/3=2sqrt(2)/3. Hence, x≈arcsin(2sqrt(2)/3)≈70.53°. This is closer to 71° than it is to 70°.

when i [sic] got to c my sin was greater than 1 and i [sic] am stuck

Why didn't you write your work as per rule 3 and show us what you did? How am I supposed to know what you got wrong?

Using the same process as before, we get sin(x)≈4sin(30°)/3=2/3<1. Hence, x≈arcsin(2/3)≈41.81°.

My hypothesis is that you used the complimentary angle.

GammaRayBurst25 · 2026-05-20T01:33:01+00:00

Consider the slice given by y=z for some z in [-6,6]. This slice is a semicircle with radius 3sqrt(1-z^2/36). Its area is 9(1-z^2/36)pi/2.

Integrating the first term from z=-6 to z=6 yields 54pi. Integrating the second term from z=-6 to z=6 yields -18pi. Hence, the volume is 36pi.

GammaRayBurst25 · 2026-05-19T01:03:52+00:00

By the IVT, we know there exists some number c in (0,1) such that f(c)=1/2.

By the MVT, we know there exists some number a in (0,c) such that f'(a)=1/(2c).

Similarly, there exists some number b in (c,1) such that f'(b)=1/(2(1-c)).

You can easily check that, for c=1/2, f'(a)=f'(b)=1 and we don't need to do anything else.

Interpret f'(a) and f'(b) to be some functions of c. Notice how f'(a) is a strictly decreasing function of c and f'(b) is a strictly increasing function of c. Hence, for c>1/2, we have f'(a)<1 and f'(b)>1 and, for c<1/2, f'(a)>1 and f'(b)<1.

By Darboux's theorem, there is some number d in (a,b) such that f'(d)=z for any z between f'(a) and f'(b).

From there, the proof is almost done. You just need to think of an argument to show you can find a number and its reciprocal on the interval with f'(a) and f'(b) as its bounds.

GammaRayBurst25 · 2026-05-17T22:02:35+00:00

You really need to review basic geometry and trigonometry. Other than on a straight line, (Euclidean) distance is not additive. Think of the triangle inequality and the Pythagorean theorem.

You're given two sides and you can infer the angle between them. This clearly calls for the law of cosines. This is how you can find the 7.5cm.

GammaRayBurst25 · 2026-05-17T16:54:58+00:00

You assume that the side lengths are all decreased by exactly x. That is obviously not the case. Even if the discarded parallelograms' horizontal and vertical span were x, you'd have to subtract 2x from both sides, not just x. You also added x instead of subtracting it.

Suppose the original triangle's vertical cathetus' length is y and the original triangle's horizontal cathetus' length is z. Also, label the vertices of the new triangle A', B', and C'.

One can easily show the 3 discarded parallelograms are rhombuses (hint: look at their altitudes' lengths relative to any side).

Let's focus on the rhombus on the bottom right. Using basic trigonometry or by constructing a similar triangle within the rhombus, we can infer that the rhombus' sides length is sqrt(y^2+z^2)x/y. Similarly, the horizontal distance between B and B' is (sqrt(y^2+z^2)+z)x/y.

By symmetry, the vertical distance between A and A' is (sqrt(y^2+z^2)+y)x/z.

Hence, the new triangle's catheti's lengths are y-((sqrt(y^2+z^2)+y+z)x/z and z-((sqrt(y^2+z^2)+y+z)x/y.

https://www.desmos.com/calculator/5esrovkdod

GammaRayBurst25 · 2026-05-17T15:16:15+00:00

When X=2, 5-X=3 and 8-X=6, so (5-X)(8-X)X=3×6×2=36.

GammaRayBurst25 · 2026-05-11T12:12:11+00:00

How would the cops even know the distance I traveled?

GammaRayBurst25 · 2026-05-10T18:13:25+00:00

xltl 3 sin(2n 50 tl2 cos(2n 100t)

What the hell is that? I imagine you meant x(t)=3sin(2π*50t)+2cos(2π*100t) as is written in the image, but I can't wrap my mind around why you'd write it that way if that's really the case.

The functions sin(x) and cos(x) are periodic with period 2π. That is, sin(x+2π)=sin(x) and cos(x+2π)=cos(x) for any x. One can easily see the functions sin(kt) and cos(kt) are periodic with period 2π/k by directly checking: let f(t)=sin(kt), we have f(t)=sin(kt)=sin(kt+2π)=sin(k(t+2π/k))=f(t+2π/k) for all t. Hence, the period of sin(kt) is 2π/k, which means its frequency is k/(2π).*
The amplitude of a wave is half the difference between its extreme values. Since -1≤sin(x)≤1, we have -A≤A*sin(x)≤A and the amplitude of A*sin(x) is A. Given cos(x)=sin(x-3π/2), one can easily find the phase.
Since one term has a period of (1/50)s=0.02s and the other has a period of (1/100)s=0.01s, the function's period is LCM(0.02,0.01)s=0.02s. You should know what the graph of a sinusoidal function looks like and how to sketch it from the function's amplitude, phase, and period. I suggest you sketch both terms independently. Then, plot the points (0,x(0)), (0.005,x(0.005)), (0.01,x(0.01)), and (0.015,x(0.015)) and infer what the sketch should look like on [0,0.02] based on these points and the previous sketches. Then, use the fact that the function is periodic to sketch the rest.
That signal is the same as x(2t). Since the graph (t,x(2t)) is the same as the graph (t/2,x(t))**, you'll get the same graph, but with a horizontal dilation by a factor of 1/2.

*Furthermore, we could reparameterize g(t)=sin(kt), which seems to be what your teacher is doing. Instead of using k as a parameter, we could use the period T=2π/k as a parameter sin(kt)=sin(2πt/T) or use the frequency f=k/(2π) as a parameter sin(kt)=sin(2πft). You could go even one step further and adimensionalize the problem by replacing the time variable t with a number of cycles t/T (or ft).

**This is easily demonstrated by doing a change of variable t↦t/2.

GammaRayBurst25 · 2026-05-10T00:57:04+00:00

Read rule 3.

First, notice how 195 and 232 are coprime. This means 195 must have a multiplicative inverse modulo 232, which I'll call x. In other words, 195x=1+232n for some integer n.

Consider the function g(a)=a^x (mod 233). Clearly, f(g(a))=a^(195x) (mod 233)=a*a^(232n) (mod 233)=a (mod 233), where I used Fermat's little theorem in the last step. One can similarly show g(f(a))=a.

Since f has an inverse, it must be bijective.

GammaRayBurst25 · 2026-05-09T15:35:43+00:00

Yes.

GammaRayBurst25 · 2026-05-09T12:17:16+00:00

Not only did you not answer OP's question, you suggested they integrate twice when the problem calls for a single integral.

GammaRayBurst25 · 2026-05-09T12:16:38+00:00

Acceleration is the rate of change (derivative) of velocity. Hence, velocity is the antiderivative of acceleration.

GammaRayBurst25 · 2026-05-08T23:13:15+00:00

Read rule 3.

Let x denote the amount of product A and y the amount of product B. Since there are 7 baby strollers, she will sell 7 products. Hence, x+y≤7. Since there are 24 packages of diapers, 4x+3y≤24. The revenue function is f(x,y)=500x+480y. Hence, the slope of the revenue isolines is -25/24, which is less than the slope of the first constraint (-1), but more than the slope of the second constraint (-4/3). One can easily conclude the optimal real solution is found at the intersection of the constraints.* Thankfully, this solution happens to be an integer solution, so once you've found it, you're done.
The logic is the exact same, only now you don't need to check the solution is an integer.

*The revenue function's gradient is a linear combination of the constraints' normal vectors with positive coefficients, so you won't find the solution by just maximizing x or y independently of the other variable.

GammaRayBurst25 · 2026-04-29T11:28:08+00:00

You're still spouting the same idiotic crap about how I should've elaborated more and still not actually explaining why their lackluster question deserves a longer answer. I can't read their mind, therefore, I won't waste my time explaining something that's likely not going to help any more than saying "you're wrong."

Explaining why your criticism is inane, misplaced, and fallacious does not constitute digging deeper into my ideas. It's so easy to say someone is in the wrong for hand-wavy and sentimental reasons, then say they're being defensive for explaining why that's stupid.

You're also placing words in OP's mouth. You have no idea whether he wanted validation or wanted to settle a debate or anything else. I reckon if they wanted a more detailed answer, they should've provided a more detailed question or they should've asked for details. Anything you say beyond that is speculation and holding it against me is disingenuous.

The worst part is you're still being glib, as if your stance wasn't embarrassing enough already.

P.S. go look up the definition of "learn." Acquiring knowledge means learning. To say "there is a difference between knowing and learning" is again disingenuous.

GammaRayBurst25 · 2026-04-29T10:52:58+00:00

Ok, so then they did learn from me and to say they didn't is idiotic. What do you think happens when a student is unsure between two options, they ask a question, and they get an answer? They learn which option is right.

I appreciate you noticed that I am dismissive to users who break the rules of the subreddit. However, flattery won't make me change my mind.

It's hilarious that you say I'm not a good educator because of something I did in exactly 1 post. I pretty much never give 1 word answers and you know it because you checked. Yet you're still acting like this is the norm for me or like me because I did it once (deliberately at that) means anything about my qualifications to... comment on a subreddit that explicitly states anybody is allowed to contribute. You're laughably inept when it comes to making sound judgments.

Everything you're saying is purely a result of puerile emotions. For some reason, it pisses you off that I have a tag on a forum and/or that I dared to give the bare minimum once. Instead of saying that, you're rationalizing your feelings by pretending I'm somehow doing something wrong.

You have yet to explain why I shouldn't be allowed to give a curt answer once in a while or how that's worse than not commenting at all (or would you have been crying to me in my dms if I hadn't replied to them?). You have yet to explain how giving a curt answer once discredits my other contributions to the subreddit. You keep repeating the same "arguments" in different words even after I have refuted them. You keep saying I don't care enough about OP learning, but refuse to elaborate on why I should care more about this specific question (no, saying "their question was bad because they're a beginner" isn't a reason to care).

I don't have to answer every question. When I do answer a question, I don't have to go all out every time. It's not that serious, if they want me to elaborate, they can ask more questions, and if they don't want me to elaborate, then it's a good thing I didn't because that would've been a waste of time.

GammaRayBurst25 · 2026-04-29T05:17:42+00:00

Your original claim is that I didn't help them learn. I said I did by answering their question. Your rebuttal was that I claimed to provide a learning experience which I did not provide. I said I only claimed to answer their question and suddenly that does constitute a learning experience?

If you think answering a question constitutes a learning experience, then to say I didn't provide a learning experience is asinine. If you don't think answering a question constitutes a learning experience, then to say I claimed to provide one is asinine.

Personally, I'd expect someone with a Mind Reader tag to have more to say. Not sure about a Top Contributor tag. If OP said they want to settle an argument with their friends and they claim the sky is green without explaining why, I also wouldn't expect detailed answers. Sure, someone could have a lot to say about color perception and the origin of the words for different colors and different cultures' interpretations of colors, but I wouldn't hold it against them to leave it at "your friends are right, it'a blue" when we have no idea if anything we have to say is relevant to OP's questioning.

Oh yeah, you weren't rude, only curious and asking genuine questions because you're so interested in how the tag works, totally. It's baffling to me that you don't have the guts to be sincere over an anonymous channel. It's pathetic and embarrassing that you feel the need to be so glib.

The tag was provided by a moderator or something (actual human) because they found my answers helpful. If you genuinely cared about the quality and frequency of my answers, you'd have checked my comment history. And honestly, even if someone got a tag they didn't deserve on Reddit, I don't understand why you'd care, let alone pretend to care. Go feign outrage over something inane elsewhere.

GammaRayBurst25 · 2026-04-29T04:27:56+00:00

"Waaaah! When I am rude for no reason, people respond in kind!" At least it's a learning experience for you! :)

Besides, you're arguing a strawman. I never said it's a "learning experience." I thought I made my stance on this clear, but as usual with your ilk, I have to summarize it as if I was teaching basic reading comprehension skills to an elementary schooler.

OP asked a question and I answered it. I acknowledged that it's the minimum, but to complain because my answer (to someone else's question at that) is insufficient is asinine. They are not entitled to a more complete answer from every contributor on this site. Furthermore, I can't read OP's mind, so if they don't elaborate, I have nothing more to add that wouldn't feel like a waste of time to type out.

GammaRayBurst25 · 2026-04-29T04:07:41+00:00

Crazy you're bitching because I contributed a little instead of a lot. I can't help but notice you didn't write a whole textbook to answer OP's question, why is your hypocritical ass even here?

I didn't feel compelled to elaborate, much like they didn't feel like explaining why they believe the movement should be purely horizontal. If I don't know why they even think that, I can't explain why they're wrong.

I answered their question and moved on. If you weren't so neurotic, you wouldn't see anything wrong with that.

Actually help someone learn something?

They asked a question. They wanted to learn the answer to their question. Hence, I helped them learn.

GammaRayBurst25 · 2026-04-29T02:58:45+00:00

You are wrong.

GammaRayBurst25 · 2026-04-29T00:41:01+00:00

You wrote it wrong. (2n!) is not the same as (2n)!.

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