Ratio Problem The problem of physical power in the third year of junior high school

Set the rod length to 3x, 5x

The weights were 5y, 6y

Then the work done by these two people is: A = 3x * 5y B = 5x * 6 y A power = 3 x * 5 y 9

B power = 5x*6y 10

A is better than B = the above is OK.

That's 5:9

For such ratio problems, the special value method can be used to solve the problem. The ratio of the length of poles A and B is 3:5, so that pole A is 3m long and pole B is 5m long.

The ratio of the weight of A and B is 5:6, and the weight of A is 5N, and B weighs 6 New. Then the work done by A overcoming gravity is w A = gh = 3m * 5n = 15j

The work done by B overcoming gravity is w B = g'h'=5m*6n=30j。then pA=w t=15 9;then p B = w'/t'=30 10=3, so pA pB=5 9

Power p = fv f A: f B = g A: g B = 5:6 v A: v B = (la t A) :(lb t t B) = 3 9:5 10 = 2:3

pA: pB = 10:18 = 5:9

The definition of power is to divide the work done by time, in this problem assuming that A's weight is 5x, then B's weight is 6x, the length of a rod is 3y, the length of B is 5y, A's power is 5x*3y 9, B's power is 6x*5y 10, and the two unknowns of x and y can be eliminated compared to get 5:9

This kind of problem is best solved by the special value method, which is to consider the rod length as 3 and 5, and the weight as 5 and 6, and then use the formulas w=gh and p=w t to solve the problem. Of course, you can also use the velocity formula to find the velocity ratio first, and then use p=fv to solve the problem.

Solution: The water Qi's ridge pump does at least w=pvgh=1000*, and the power of the pump is at least p=w t=4500000 and hypertonic 60w=75000w=75kw

1.No, the drop-off time is 9:05 and I have missed it.

2.Taxi speed: 12km, 10min=12km, (1 6h)=72kmh=20ms

3.Power p=fv=3x10 3 x 20=60kw

1. Late. Because it is 9:05 to get off the bus, the test starts at 9:00

2，v=s\t=

3,p=fv=3x10*

c w1 w2, p1w=gh, is the same object, g carries the same dust vertically; ab=bc, so h is the same, so w1=w2 is calculated

p=w t, free fall, the speed will be more and more argumentative, v=s t, s is the same, the speed of bc is fast, so bc is used less between brothers and friends. According to p=w t, it can be seen that in the case of the same work, the power is greater if the time is less, so please click on p1. If you don't know, ask questions.

Select cw=fs are the same p=fv The speed of the BC segment is greater than that of the AB segment.

C. The height of the barbell rising when lifting weights is about 2m, so the work of jerk can be obtained from the work finding formula, w stiff=m stiff gh=154kg 10n kg 2m=3080j, close to option a; Grab the work w grab = m grab gh = 128kg 10n kg 2m = 2560j, p grab = w grab t = 2560j, which is the closest to c.

This question examines students' participation and understanding of physical activities, and some students do not know that the height of the barbell rise when lifting weights is about 2m plus arm length, so they fall into a state of confusion. Therefore, when solving problems, I don't know where to analyze, and of course my thinking can't.

Solution: (1) The work done by the miner in 10 seconds, that is, the total work w total = fs rope = f 2h = 150 N2 4 meters = 1200 joules.

The power of the work done by the miner in 10 seconds, i.e., total power p total = w total t = 1200 joules 10 seconds = 120 watts.

2) The gravitational force of the object g = m g = 32 kg 10 N kg = 320 N.

The work done by the gravity of the object is useful work w useful = g h = 320 N 4 m = 1280 joules 1200 joules, that is, the useful work is actually greater than the total work, so your data is wrong, either the tensile force of 150 N is small, or the mass of the object is 32 kg.

Changed: Changed rally force from 150 to 180 N. Here's how to solve it:

Solution: (1) The work done by the miner in 10 seconds, that is, the total work w total = fs rope = f 2h = 180 N2 4 m = 1440 joules.

The power of the work done by the miner in 10 seconds, i.e., total power p total = w total t = 1200 joules 10 seconds = 144 watts.

2) The gravitational force of the object g = m g = 32 kg 10 N kg = 320 N.

The work done by overcoming the gravity of the object is useful work w useful = g object h = 320 Nn 4 m = 1280 joules.

If all friction is not taken into account, the work done to overcome the gravity of the moving pulley is the extra work w extra = g moving h

1440 1280) joules = g movement 4 meters.

The gravitational g-motion of the movable pulley = 40 N.

3) The mechanical efficiency of the pulley block = w useful w total = 1280 joules 1440 joules

Answer: The work done by the miner in 10s is 1440 joules, and the power he does is 144 watts; The weight of the movable pulley is 40 N ; The mechanical efficiency of the block pulley.

(1) There are two sections of rope that bear the weight of the object, i.e., n=2

The distance traveled by the end of the rope is s=2h=2 4,m=8mThe work done by the miner w=fs=150n8m=1200jPower p=wTotal t=1200J 10s=120w

2) G = mg = 32kg 10n kg = 320ng kinetic = 2f-g = 150n 2-320n = ?

3) w useful = gh = 320n 4m = 1280j = w useful w total = 1280j 1200j = ?

There is a problem with the data you gave, the weight is 320N, each section is at least 160N, and the number you give is 150N, the mechanical efficiency will be more than 100! Check it out again! The steps are no problem!

Solution: 1) Cause n=2

then s=nh=8m

Hence w=fs=......

p=w/t=……

2) by f=(g+go) n g=mg=......go=nf-g=......=？

3) W has = gh

wtotal = fs then =w has w total = ......

A: ......There is a numerical problem with this question - is the f-value 150N? - Wrong input, right? It should be 180n, right?

Otherwise, the weight of the movable pulley in 2) is negative.

3) Is it greater than 100%?

Is it possible? Relationships and formatting are no problem! Find out the numerical problem yourself!

The correct answer should be c, because the sum of the resistance of the electric wire and the resistance of the lamp must be greater than the resistance of the lamp. According to p=U2 r, when the voltage is constant, the resistance increases, and the power decreases. Therefore, the sum of the electrical power consumed by the wire and the lamp must be less than the rated power of the lamp.

Hope it helps.

It turned out to glow normally, but later the resistance of the wire was increased. Therefore, from the value r, it can be seen that the total power decreases (less than 60W). Whereas, the actual power consumed by the bulb is 55W, so the power consumed by the wire is less than 5W

It must be c No need to doubt it.

60W is the rated power not the total power, the actual voltage and current can be calculated from the actual power of 55W, and then the power consumed by the wire can be calculated.

When the sliding blade is closed at the A end, the lamp glows normally, and when the S is disconnected, the current representation decreases ,--

U RL+U 32=U RL+>U=16V Total S is still disconnected, the slider is connected to the B terminal, and the power of the lamp is 1/9 of the rated power---

16^2/rl=9*(16/(rl+20))^2-->rl=10ohm

The rated voltage of the lamp is U=16V

Rated power p=U2 rl=

The thick and thin segments are connected in series, and the current is the same. It is easy to break and is thin in the place where the resistance is large. So A: No, the voltage of the minutes is high.

B: True C: False, reversed.

D: False. Same.

The thick and thin segments are connected in series, and the current is the same. The resistance is large in the thin place, and the electric power is large when the current is the same.

A: No, the voltage with high resistance in the series circuit is high.

B: Yes, p=i 2*r The electrical power with large resistance C: False, when the length of the material is the same, the cross-sectional area is small and the resistance is large.

D: False. The current is equal everywhere in a series circuit.