Physics questions, master 5, physics questions, master progress

15 words. With a triangle resemblance comes out. Definitely 15 words.

The key to analyzing this issue is:

Who bears the pressure and weight of the mercury column in the glass tube?

The analysis process is as follows:

Take a point A on the mercury surface of the mercury tank container, and a point B at the same height as A in the glass tube. According to the communicator principle, pb = pa = atmospheric pressure. It is easy to understand that as long as the pressure at b and a is not equal, a flow will be generated, and eventually equilibrium will be reached.

Since the pressure at B and the same pressure at A are equal to the atmospheric pressure, this means that the result of having a column of mercury above B and no column of mercury above A is the same, that is, the pressure and weight of the column of mercury above B are not borne by the tank. So who bears the weight of the mercury column? I was undertaken by an iron frame.

This is easy to do, as the atmospheric pressure increases, the height of the mercury column increases, and the corresponding mercury in the mercury tank decreases, so the weight of the mercury tank decreases. The spring scale becomes smaller.

You mentioned the buoyancy problem, which is negligible, because the air pressure increases, and the upward and downward pressure of the mercury tank increases at the same time. If you have to consider the buoyancy problem, it is also very simple: the air density increases due to the increase of air pressure, and the weight of the air discharged under the same volume of the mercury tank increases, so the buoyancy becomes larger and the spring scale becomes smaller.

Considering that the mercury in the mercury tank decreases, the buoyancy decreases as follows: if you lose a basket of iron lumps and pick up a basket of cotton, the result is still a decrease.

The first whistle is from the tunnel mouth S, the speed of the train V car.

s=(v car t1 + v sound t1) 2

That is, s = (3V car + 340 * 3) 2 - 2s = 3V car + 1020 - the second whistle is the distance from the tunnel mouth S.

Then the S distance + V car T car = S

That is, S distance + 4V car = S ——

and S distance = (V car t2 + V sound t2) 2 - 2s distance = 2V car + 680 - simultaneous three-formula solution.

s=v-car=

Distance: (340*3) 2

Speed: ((340*(3-2)) 2) 4

The explanation of the second question is that the first three-second echo and the second two-second echo are one second apart, that is, the speed of sound goes back and forth in one second, so divide by two to calculate the distance of one second on a business trip, and finally divide by 4 seconds to get the speed.

2s1 = (v car + v car) t1

2s2 = (v car + v car) t2

S1-S2 = V car t

Among them, S1 is the distance from the train to the tunnel mouth when the first whistle is sounded, and S2 is the distance from the train to the tunnel mouth when the second whistle is sounded, t=4 seconds, the answer is the same as the teacher, and the students should pay attention to drawing the location map of the train when doing the question.

This is not a simple physics problem, nor is it a simple electrical mechanics engineering complex, which must be combined with the needs of the engineering site.

For example, the speed of picking up is required.

50 degrees is basically not labor-saving, directly calculated by 8 tons, anyway, it also has a safety margin.

If the lifting speed is required in meters per second, the power is 8000kg * gravitational acceleration * speed.

It is about equal to 40kw, considering the safety factor, you have to choose 60 80kw motor.

The approximate formula is weight * 10 * speed.

Weight by ton. 10 is the acceleration due to gravity.

Speed in seconds.

Finally, there is the kw number.

If it's meters.

Then only 8*10* is directly kw, optional 20 25kw Note that this is only a rough algorithm, you need to design the reducer, as well as mechanical efficiency concerns.

One. 1) Volume = 20 * 10 * 6 = 1200 cm 3 Mass = Density = 1800 1200 =

2) Gravity = mg =

Base area s1=20*10= p1=900pas2=20*6= = p2=1500pa

s3=10*6= = p3=3000pa

3) The maximum storage is 4500 900 = 5 pieces.

Two. A 5 laps B 4 laps The time is the same then v A: v B = s A: s B = 5: 4 A speed 5000 24 * 60 1000 =

B speed = B time = 5km 10 = i.e. 30 minutes.

Yijia arrived at the finish line 6 minutes late.

3 g takes 10 n per kilogram to indicate that it is calculated according to the standard air pressure.

Air density. The air thickness is taken as 3000 meters.

Then the volume of air = (6370,000+3000,000) 3)-6370,000 3)4 3

10^18 m^3

m 3 mass = kg

The words 2500r kWh on the meter indicate that when you turn 2500 times, you will consume 1 kWh (i.e. 1 kWh of electricity), so the amount of electricity consumed in these 2 minutes is 5 2500 = kWh 7200 joules.

The power of a monitor is about p (e total t) 20 [7200 (2*60)] 20 3 watts.

The enlightenment of this question in terms of power saving is: everyone is responsible for saving electricity, and the power supply should be closed thoroughly.

The first void: w=5r 2500rkw h=the transformation of kw h and j designed here)——

Second empty: p=w t=7200j 120s=60wpo=p 20=3w

The third empty: the monitor still consumes power in the standby state, so the monitor should be turned off when the computer is not in use.

(1)1000x3600x5/2500

2) The first empty answer is divided by 120 and then divided by 20

3) It is best to turn off electrical appliances when not in use, as standby will also consume power.

Its significance is: when the actual voltage U = rated voltage Uo = 220V, the electric power PO (the work of the current in 1s) is 40W

Connected to a home circuit, the current through the lamp i=po uo=40 220=(2 11)a=

The filament resistance r=u i=220 (2 11)=1210 1° electricity can be used for it to emit light normally.

General lighting Rated voltage 220 volts Rated power 40

When the switch S1 is placed at the B terminal and S2 is disconnected, the two resistors in the circuit are connected in series, the total resistance is the largest, and the heating power is the smallest, which should be a low gear.

When the electric oven is working in the middle gear, there should be only one resistor working in the circuit, that is, S1 is connected to B, S2 is closed, and only R1 is working at this time.

So the power consumption is 220 2 watts in p.

When the electric oven is working at a high level, it should be the least total resistance in the circuit, that is, S1 is connected to A, S2 is closed, and the two resistors are connected in parallel.

The total resistance is r1*r2 (r1+r2) ohms, so the electrical energy consumed is w (u 2 r r) * t (220 2 joule kWh.