Physics calculation problems, physics calculation problems, I don t know how to do it!

In kinematics, you can choose the ground as the frame of reference, or you can choose an object that moves in a uniform straight line relative to the ground as the frame of reference. Here, we can choose the river moving at a uniform speed as a frame of reference (thinking that the river is immobile), so that the boat moves with the water and against the water is the same. That is, it can be considered that the boat is moving in still water.

When moving forward, something fell on the boat, just drifting in place, after finding it, turn back to look for it, the time to go back and forth is the same, now it is known: it takes half an hour to go back, of course, it is half an hour from lost to found, found at 6 o'clock in the afternoon, minus half an hour, the lost time is of course half past five!

Old question. When I was in junior high school, I did what looked like math...

You see, at the beginning, if the speed of the ship is 2, the current is 1, and the relative speed is 1, isn't it? That is, the actual speed of 1

Then, go along the water, the relative velocity of the water flow 1 is unchanged, and it is also 1, then the actual velocity is 2

That is, the speed of going down is twice as fast as going up, and it takes half an hour to catch up, that is, the dinghy has been gone for half an hour? Draw your own picture and see if that's the case, and you can see it by drawing it.

I can only tell you that the speed of the river and the speed of the ship relative to the water remain the same, which is useful for indicating that the speed of the ship and the speed of the water are not the same.

Answer: 40j, 75, oh.

Analysis: Calority: q i 2rt 2 2

Efficiency: W machine W total W machine (W machine Q) FS (FS+Q) 12 Resistance: motor work, w uit 12

Solution: U 4v

So the resistor voltage is: u' 9 4 5V

Resistance: ru'i 5 2 euros.

I don't know if I asked, but the circuit should be a motor and a resistor in series.

The useful work w of a DC motor is =gh=12*

The total work done by the current: w total = uit = 9 * 2 * 20 = 360 j heat generated by the current passing through the coil q1 = i square * rt = 2 * 2 * mechanical efficiency = w has w total.

The heat generated by the resistor r Q2 = W Total -w has -Q1 = 360-120-40 = 200J and then uses Q2 = i square RT to solve R

Mechanical output power = 12*

Motor input power = 9 * 2 = 18W

The heating power of the motor coil = 2*2*

Efficiency = 6 18 =

What is R?

1.Divide 160 trillion kilometers by 300,000 kilometers per second at the speed of light. The unit is seconds.

2,300000000m s divided by. Unit: meters.

3. Less 100 340 = 5 17 seconds.

Multiply by 20

Subtract the first 20 seconds (that's 5 times 20) and divide by 4

Question 1: Are the conditions insufficient?

The second question is solved like this.

The weight of the car rises at a constant speed, so the speed at h is still v=10m s, and t can be solved by the kinematic formula s=vt+1 2 at (where s=175m v=10m s a=10m s).

v is solved by 2as=v v (where s=175m v=10m s a=10m s).

If you do the calculations, you can do it yourself, and exercise and rent it.

The most primitive method:

s=vt+a/2t2

The drag is constant, so there is, the acceleration a(f=ma) is constant, according to a=(v-v0) t, t=(v-v0) a

Let the speed through one steel plate be v0, and the speed through three steel plates be v' According to the title:

3 v0 *v0/a+a/2*（v0/a）2=3 v’ *v’/a+a/2*（v’/a）2

The solution gives v'= v0 3 (open squared).

(1) Acceleration a=gsin37° m=gsin37°=10 (2) friction f= n= gcos37°

Acceleration: a'=(gsin37°-μgcos37°)/m=gsin37°-μgcos37°

2m/s²

Solution (1): The object is subjected to gravity, and the oblique support force on the object. One component of gravity is to be balanced with the supporting force, i.e. f = mgcos. Another component of gravity is the net force exerted on the block.

From f=ma, obtain: mgsin =ma. α=37°，g=10m/s²,a=6m/s²。

2): The object is subjected to gravity, and the oblique support force and friction force on the object. One component of gravity is to be balanced with the supporting force, i.e. f = mgcos. The other component of gravity minus friction is the net force experienced by the object. From f=ma, we get:

mgsinα-μmgcosα=。