Urgent: a physics calculation problem in senior one The buoyancy of a hydrogen balloon is 5N and it rises at a constant speed. Suddenly, an object with a mass of 100g originally tied to the balloon falls off from the balloon. If the air resistance is not considered and the buoyancy of the balloon remains unchanged, the acceleration of the balloon rising after the object falls off is calculated

Urgent: a physics calculation problem in senior one The buoyancy of a hydrogen balloon is 5N and it rises at a constant speed. Suddenly, an object with a mass of 100g originally tied to the balloon falls off from the balloon. If the air resistance is not considered and the buoyancy of the balloon remains unchanged, the acceleration of the balloon rising after the object falls off is calculated


2.5m/s^2
The initial velocity is uniform, which means that the force is balanced, Mg = f, buoyancy = 5N
M = 100g, f = f-mg = 5N - (mg mg) n = 1n
f=ma ,a=f/M-m=1/0.5-0.1=2.5m/s^2



A calculation problem of physical electric power
A lamp "6V 3W". B lamp "18V 18W" (1) connect a and B in series to a power supply, in order to make one of the lamps work normally, what is the power supply voltage? What is the power ratio of the two lamps? (2) connect a and B in parallel to a power supply, in order to make one of the lamps work normally, what is the power supply voltage? What is the power ratio of the two lamps?


First calculate the resistance, r = u ^ 2 / P ----- R1 = 12 Ω, R2 = 18 Ω
Two lamps are connected in series, and the voltage is proportional to the resistance. U1 / U2 = R1 / r2 = 2 / 3
Therefore, U1 = 6V, U2 = (3 / 2) U1 = 9V (a normally emits light)
The power supply voltage is u = U1 + U2 = 15V (if U2 = 18V, U1 = 9V, the lamp will burn out)
The power is also proportional to the resistance, P1 / P2 = 2 / 3
In parallel connection, the voltage is still 6V, and a still lights normally
The power of the two lamps is inversely proportional to the resistance, P1 / P2 = 3 / 2