Recent content by Istiak

The equation on #20 were saying pressures are same. But in #16 I got that pressure times volume are the same. If we consider pressure to be same then Volume should be same. That's what was confusing me that moment. The fire increases the pressure of the balloon if we consider volume to be...

How to change username? I had visited "MY PF" but can't find out anything. There's no edit option of username.

##\rho=\frac{P}{RT}## ##\dfrac{\rho_1}{\rho_2}=\frac{T_2}{T_1}## ##\dfrac{P_1RT_2}{RT_1P_2}=\frac{T_2}{T_1}## ##\dfrac{P_1}{P_2}=1## ?

don't know. But I may try, at the end I will consider their temperature is same. So I will start with ##PV=nRT## ##\frac{P_1V_1}{P_2V_2}=1##? Considered for a single mole ##n=1##.

I think that depends. Like, if outside pressure is higher than it shrinks and inflates for less pressure outside

Are you talking about ##(\dfrac{V_1}{V_2})^\gamma=\dfrac{P_2}{P_1}##?

Consider a hot-air balloon with fixed volume VB = 1.1 m3. The mass of the balloonenvelope, whose volume is to be neglected in comparison to VB, is mH = 0.187 kg. The balloon shall be started, where the external air temperature is ϑ1 = 20 oC and the normal external air pressure is po = 1.013 ⋅...

The problem said that, "density of first object is ##n## times of second object". Hope it's clearer more now

No. ##n## is just an integer.

##\rho_1=n\rho_2##.

first object and 2nd object. e.g. ##\rho_1## density of first object.

I just found here(https://byjus.com/physics/relation-between-density-and-temperature/#:~:text=Density and Temperature Relationship 1 When density increases,,reduces. 4 When the temperature decrease, density increases.) that P=##rho##RT. So they just took ##\rho=\frac{n}{V}##...

Since they have same charges (not magnitude) so they distract. They simply stop as further as they move, at infinity? E=0 at infinity. dunno 🤔 ##U=\int \vec F \cdot d\vec r## ?

Generally, energy is ##U=9\times 10^{9} \times \frac{5\times 10^{-6}30\times 10^{-6}}{2+(10+20)\times 10^{-2}}=0.5869 J## <br/> After touching, they have charges ##q_1 and q_2 = 35\mu C-q_1## ##\frac{q_1}{10}=\frac{35\mu C-q_1}{20}## I was wondering where 1/10 and 1/20 coefficients come...

According to an Olympiad contestant (reached ipho), the answer is ##I=2nr##. (asked for hint) And he have made sure that the ##r=\frac{mv}{eB}##.