Everyone, solve a physics problem, thank you

1) The volume of the bowl is 30cm3;

2) the mass of the bowl is 60g (3) the density of the bowl is 2g cm3 Solution: (1) from Figure A, the volume of water v1 = 500ml, from Figure C, the total volume of bowl and water v3 = 530ml, the volume of the bowl v = v3-v1 = 530ml-500ml = 30ml = 30cm3;

So the answer is: 30

2) As can be seen from Figure B, the total volume of water and the boiled water discharged from the bowl v2=560ml, the volume of the boiled water discharged from the bowl vrow=v2-v1=560ml-500ml=60ml, the mass of the boiled water discharged from the bowl m water = water v drain = 1g ml 60ml = 60g;

When the bowl floats on the water, the bowl is subject to the gravity g bowl = m bowl g under the vertical line, and the vertical upward buoyancy f floats m water g, which is in a state of equilibrium, and is obtained by the equilibrium condition: g bowl = f float, that is: m bowl g = m water g, so m bowl = m water = 60g;

So the answer is: 60

3) The density of the bowl Bowl = M bowl V = 60g * 30cm3 = 2g cm3 so the answer is: 2

Solution: Fill the bottle with boiling water, and hear the sound of the air column in the bottle vibrating, the structure and material of the air column remain unchanged, and the timbre remains the same

At the beginning, the drop of the water is large, the impact force is large, the air in the bottle vibrates greatly, the amplitude is large, the loudness is large, and the sound is loud The air column in the bottle is of large quality, large volume, difficult to vibrate, low frequency, low tone, and a low "knock, knock" sound is heard

When the water is about to be full, the drop of the water is small, the impact force is small, the vibration amplitude of the air in the bottle is small, the amplitude is small, the loudness is small, and the sound is small The air column in the bottle is small in quality, small in size, easy to vibrate, high in frequency, high in tone, and you can hear a sharp "squeak, squeak" sound

Therefore, choose D

Conservation of momentum problem.

The original question is incorrect, and according to the answer, the question should be corrected to read: "A sandbag of mass m is thrown at ship B from ship A at a rate of v relative to ship A".

Solution] The direction of the initial velocity v of the first ship is taken as the positive direction, and the velocity of the sandbag is taken as a negative value when the ground reference frame is uniformly selected.

Before the sandbag is thrown, the total momentum of the sandbag and the boat is mv

After the sandbag is thrown, the momentum of ship A is (m-m)v A, and the momentum of the sandbag is m(v A-v) According to the law of conservation of momentum, there is:

mv = (m-m)v a, m(va-v).

v A = (mv + mv) m

Taking the sandbag and the B boat as the research objects, according to the law of conservation of momentum, there are:

mv m(v a-v) (m+m)v b.

v B = (m + m )v [m(m+m)]= (m +mm + m -mm)v [m(m+m)]=[m(m+m)-m(m-m)]v [m(m+m)]=v- m(m-m)]v [m(m+m)] exactly matches the answer you provided, indicating that the original question is wrong.

First of all, the speed of throwing the sandbag is -v, pay attention to the rate is v, but the direction is opposite, otherwise it will not be able to throw the ship B.

Therefore, mv=-mv+(m-m)(vA), and vA=(mv+mv) (m-m).

And ship B mv+(-mv)=(m+m)(vB), we get vB=(mv-mv) (m+m).

What I have in mind is to use the conservation of momentum to solve the problem, and the result is different from yours.

When going up, f=mgsin+ilb, when going down, mgsin=ilb is the same because v is the same, so i is the same.

There is f = 2 mgsin i = mgsin lb = e r = lvb r and v = mgsinr (l 2b 2).

100W and 40W bulbs.

Because p=ui=u2r, i.e., pr=u2

If the two bulbs are connected in parallel.

That is, when the voltage is equal.

The higher the power, the lower the resistance.

When two bulbs are connected in series, the current passing through the two lamps is equal.

According to p=ui=i 2*r

The greater the resistance, the greater the power, according to the conclusion of 1, the resistance of 40W is large, so the two lights are connected in series, the power of 40W is large, and the brightness is large.

By: Impulse equals the change in momentum, ft=mv

Know: If m1=m2, v1>v2 then the impulse of force f1 on object a is greater This statement is true.

The maximum instantaneous power of force f1 is a double of the maximum instantaneous power of force f2, nor is it twice as much.

So the answer to this question is wrong.

Analysis: The maximum rise of b is required to find the speed when the rope is disconnected, then h=s+v 2 2g. Answer:

Kinetic energy theorem (4mgsin -mg) s=1 2 (4m+m) v 2, v 2 = 2gs 5 so h=s + s 5 = 6s 5

Isn't the rise height the same as the height of block B when the string breaks? Because the angle is 30 degrees. So the maximum height is 1 2s.

Second=210mms=

2: According to the law of uniform acceleration motion: from the beginning of rest, the ratio of the distance passed in an equal time interval is the ratio of the consecutive odd number 1:

9...It can be deduced according to the formula of s=1 2at 2, and you can also draw a speed time graph, and you can reason about it yourself, and the memory will be profound, don't be lazy. ab distance:

Distance between the last two points =: 97; It can be seen that point A is point 11, point B is point 12, followed by point 49 and point 50, respectively. There are 13 48 missing paper segments in the middle, for a total of 36 points;

3: In the same way, the distance between point 1 and point 2 is, according to the formula s=1 2at 2, a=1m s 2.

1.v= t1= v2= (points The third question is not difficult, you can write it yourself.

50\50\

100\10=10m/s

The average speed at non-uniform speed is related to the distance chosen.

50\50\

100\10=10m/s

The average speed during variable speed movement is related to the distance chosen.