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Physics applications, on the sinking and floating of objects (urgent,) 1. The displacement of the ship is 1000t. If the density of the river is 1.0 * 10 cubic kg / m3, what is the buoyancy of the ship? What is the volume of the ship discharging the river? (g = 10 / kg) 2. The volume of the hydrogen balloon is 4.5 cubic meters, and the mass of the shell is 4kg. In order to know that the density of air is 1.29kg/cubic meter, and the density of hydrogen is 0.09kg/cubic meter, how much force is used to pull the hydrogen balloon? OK, I'm adding points
one
ρ=m/v
Row v = (1000 * 1000) kg / 1.0 * 10 ^ 3kg / m ^ 3 = 1000m ^ 3
F floating = ρ liquid GV discharge
F = 1.0 * 10 ^ 3kg / m ^ 3 * 10N / kg * 1000m ^ 3 = 1.0 * 10 ^ 7n
two
F = ρ air GV hydrogen balloon
F = 56.889n
The downward force of hydrogen itself
F downward = ρ hydrogen balloon * V hydrogen balloon + m spherical shell * g = 39.605n
Required pulling force: F pull + F down = f float
Fra = 17.284n
The first question is mainly about the application of the formula, paying attention to the understanding of the formula
The second question is very rare in our senior high school entrance examination at that time, but it is also a test of buoyancy. When we do it, we should draw a picture to understand it. We should also pay attention to the fact that the density of hydrogen is smaller than that of air. In the original liquid buoyancy formula, f floating = ρ liquid GV row, we regard the air density as ρ liquid, that is to say, all relatively large densities as ρ liquid, combined with the original liquid buoyancy formula. In this way, we can get the buoyancy
This kind of problem should be done with one eye open and one eye closed. You should refuel well! Learning physics is great and glorious. I hope you can become Einstein of China in the future
The sinking and floating of objects
The density of a and B objects with equal volume is 0.8 × 10 ^ 3kg / m ^ 3 and 1.2 × 10 ^ 3kg / m ^ 3 respectively. When they are put into water at the same time, the buoyancy ratio after they are still is (); if a and B objects are equal, the buoyancy ratio after they are still in water is ()
In the activity of exploring "which factors are related to the floating or sinking of objects", the students immerse the fresh radish in the water. After letting go, all the radishes float until they float on the water, as shown in the figure. In order to make the floating radish sink, Xiao Ming immerses the fresh radish in the water in the activity of an iron nail. After letting go, all the radishes float until they float on the water, As shown in the picture. In order to make the floating radish sink, Xiao Ming inserts all the nails into the radish, and Xiao Hua uses the same nails, but only half of them are inserted into the radish. He submerges the radish again and lets go. As a result, both radishes sink to the bottom of the water. From the angle of the force on the object, they must insert the nails into the radish to change the force___ The magnitude of the force, so that buoyancy___ (less than / greater than / equal to) gravity; from the way they explore____ (Xiaoming / Xiaohua) is more reasonable because:
When the volume is equal, 4:5 and the mass is equal, 6:5
The law of projector imaging
Positive and negative. Right and left? Make it clear
The slide projector? Is it the same?
The projector is a convex lens
According to the imaging law of convex lens, the inverted real image is obtained
Not only up and down, but also left and right
If the equation 1 / 2 (x + 2) = 3 / 2x-1 has the same solution as the equation K (2x-1) = KX + 7 of X, the value of K is obtained
Solving the first equation, we get x = 2. Because the two equations have the same solution, we substitute x = 2 into the second equation K (2 * 2-1) = k * 2 + 7, k = 7
The polar coordinates of 12 points m are (2, - π / 6). Why are the polar coordinates of the symmetric points of the line passing through the pole and perpendicular to the polar axis (2,7 π / 6),
How do I get 120 degrees?
I mean, isn't the Title Vertical? Isn't it 90 ° vertical? How do you get 120 degrees?
The polar coordinates of M are (2, - π / 6) and (2,11 π / 6)
In your words, their intersection angle is 120 degrees, 11 π / 6-4 π / 6 = 7 π / 6,
Therefore, the polar coordinates of the symmetric point of the line passing through the pole and perpendicular to the polar axis are (2,7 π / 6),
As shown in the figure, in △ ABC, if de ⊥ BC intersects AC at F, the extension of Ba intersects e, and AE = AF, is △ ABC an isosceles triangle? Please give reasons
The reasons are as follows: in RT △ BDE, ∠ B + ∠ e = 90 °, in RT △ FDC, ∠ C + ∠ DFC = 90 °, ∵ AE = AF, ∵ e = ∠ AFE, ∵ AFE and ∠ DFC are opposite vertex angles, ∵ AFE = ∠ DFC,
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