For a physics problem, it is best to write a known solution

It is known that f=3*10 3n t1=10:00 t2=10:05 s=6km

Find: (1)t=?

2）v=？3）w=?

Solution: (1) t=t2-t1=5min

2)v=s/t=6000/300=20m/s3)w=f*s=3*10^3*6000=

A: The taxi journey time is 5min

The average speed at which taxis travel is 20m s

What the taxi did during this time.

First question: Because the boarding time is at 10:00 and the drop-off is at 10:05, the ride time is 5min

Second question: The distance is 6km and the time is 300s

So v=6000 300=20m s

The third question: w=f*s=3000*6000=180000000

Solution: t=10h5min-10h=5min=1 12h=5minv=s t=

w=fs=3*10^3n*6*10^3=

Hope it helps

Study hard, 、、、 such a simple problem every day, think about it yourself, think about it, 、、、 make it yourself, it's the best 、、、

Suppose that after hearing the alarm, when they run to the high ground of 3000 meters, the wave just catches them, and they happen to be safe, then: After the occurrence of **, the time when the wave reaches the high ground: t1 = (1500km + 3km) (500km h) = hours, which is 3 hours and seconds, and the time when the wave reaches the high ground is:

11:30 a.m. Time it takes for children to run to high ground: t2 = 3000m (100m min) = 30min So children must be alerted before 11:00 a.m.

The real topic: On December 26, 2004, at 8:30 a.m., a large tsunami occurred in the waters off India, causing a strong tsunami. **At the time of the incident, a group of children were playing on the beach in the Maldives, 1,500 km from the epicenter.

If they were alerted and immediately ran to high ground 3,000 meters from the shore, they would not be swept away by the waves. If so, at what time will they be alerted to the police before they can escape death?

The tsunami traveled up to 500 km h in the ocean, and the children fled to the high ground at a speed of 100 m min. ）

Suppose that after hearing the alarm, they ran to the 3000-meter high ground when the waves just caught them up, and they happened to be safe, then:

After the occurrence of **, the time when the wave reaches the high ground: t1 = (1500km + 3km) (500km h) = hours, or 3 hours and seconds, and the time when the wave reaches the high ground is: 11:30 minutes and seconds.

The time it takes for the children to run to the high ground: t2 = 3000m (100m min) = 30min

So the kids have to be alerted before 11:00 a.m.

This is a practice problem that is solved by multi-step calculations. First of all, according to the Celsius temperature regulations, the actual temperature represented by each small cell of this thermometer is calculated. According to the regulations of Celsius, this thermometer indicates a number of 4 in the ice-water mixture, and its actual temperature value is 0; The indication in boiling water at one standard atmosphere is 99, and its actual temperature is 100.

There are 95 bars between 4 and 99, which represents an actual temperature difference of 100. So the actual temperature represented by each cell is 20 19. Secondly, calculate the number of lattices at room temperature according to the conditions given by the question.

There are 21 slots between 4 and 25. Finally, the problem can be solved by multiplying the number of compartments of room temperature by the actual number of temperatures represented by each compartment.

Solution: (1) The actual temperature represented by each cell of this thermometer is: 100 (99-4) cells = 20 19 grids.

2) The number of grids at room temperature is (25 4) grids = 21 grids.

3) The room temperature is t= 20 19 grids * 21 grids =

It is known that the indicator is 4 in the ice-water mixture, 99 in boiling water, and 25 in the classroom

Seeking: The actual temperature of the classroom.

Solution: It is stipulated that at a standard atmospheric pressure, the temperature of boiling water is 100, and the temperature of the ice-water mixture is 0, so 99 on this thermometer is equivalent to the actual temperature 100, 4 is equivalent to the actual temperature 0, and 99 -4 = 95 on this thermometer is equivalent to the actual temperature 100 -0 = 100.

The 1 on this thermometer is equivalent to the actual temperature of 100 (99-4) = 20 19

When the thermometer measures the temperature in the classroom to be 25, the actual temperature is (25-4) 20 19 +0 =

A: The actual temperature of the classroom is:

25=100*1/4

Indicator 4 = (Actual) 0 Indicator 99 = (Actual) 100 Length 95 = (Actual) 100

The indication is 25 = 95-4 = 91

The actual temperature of the classroom is Celsius.

(1) The resistance of L1 and L2 when they emit light normally.

r1=（6＾2）/3=12 ω

r2=（6＾2）/6=6 ω

2) When the switch S1 is disconnected and S2 is closed, one bulb is shining normally, and the voltage of the other bulb should be low so as not to burn out. So the normal work is a bulb with a higher pressure drop, according to the partial pressure formula, this bulb should be the one with a larger resistance, that is, L1

L1 has a current rating of 3 6= A

Then the indication of the ammeter is a

The voltage drop across L2 is V

Then the voltmeter indicates 3 V

3) When the switch S2 is disconnected and S1 is closed, LL, L2 are short-circuited, and the luminous light is L3 with a power supply voltage of V

So the current flowing through L3 is 1 9= A

That is, the indication of the ammeter is A

The resistance of l3 is (9 2) 1=81

The square of the rated voltage of L3 is 4 81 = 324

L3 has a rated voltage of 18 V

(1) When emitting normally, according to p=u*i, u=i*r, you can push: p=u*u r, and then r=u*u p can be obtained;

Substituting the data can obtain:

r1=u1*u1 p2=12 ohms, r2=u2*u2 p2=6 ohms;

2) Seeing this, I would like to say that the title is not rigorous enough, it says that only one light shines normally, so the other one is just bright or dark? In this way, two situations arise, and there is no definite answer to this question.

The time required for one time t = 60 180 = 1 3 (s) and the time t for one time to get off the ground'=3/5

1 5=, this period includes two processes of ascent and descent, then the maximum height above the ground h= g(t'/2）²=

The work done by overcoming gravity is the absolute value of the negative work done by gravity, and the negative work done by gravity when rising from the ground is w=mgh=25(j).

If the negative work done by gravity in contact with the ground is not considered, the total work done to overcome gravity: 180*w The average power of work done to overcome gravity: p=180*w 60=75(w).