Prawns come to see the physics questions in the third year of junior high Thank you

When the person crosses the Q point and continues to fall, the rope has an upward pull force on the person, and the pull force is getting bigger and bigger. When the pulling force is less than the gravitational force, the person still accelerates the decline and the kinetic energy increases; When the pulling force is greater than the gravitational force, the person decelerates and descends, and the kinetic energy decreases Select d !!If you are interested, look at the 2001 Jiangsu competition

Of course, the kinetic energy of people is getting smaller and smaller. Because when falling from Q, the kinetic energy of the person is continuously converted into the elastic potential energy of the rubber band, and the lowest point is that the kinetic energy of the person is all converted into the elastic potential energy, and the kinetic energy is 0, and then the elastic potential energy of the rubber band is converted into kinetic energy in the upward process.

Yes, we have this topic in our exercise.

Same. Because at this time, kinetic energy turns into potential energy.

Obviously c. Kinetic energy 1 2*m*v2, m remains the same, v keeps getting smaller (because the resultant force keeps going up), so the kinetic energy keeps getting smaller.

c.According to the law of conservation of energy, the rubber rope is straightened at point Q, then the rubber rope is deformed and elongated after point Q, then part of the kinetic energy and part of the gravitational potential energy are converted into elastic potential energy, so the kinetic energy decreases and the speed slows down.

In addition, from a medical point of view, if it accelerates after Q, it may cause cerebral hemorrhage and limb ischemia when you stop exercising.

After the point is passed, the kinetic energy of the person becomes the elastic potential energy of the rubber rope, and the rear is basically all changed to the elastic potential energy, so the kinetic energy of the person is gradually reduced.

But we're not going to learn the energy yet, we'll talk about it in about two weeks, and we'll tell you ...... then

C, for sure.

After passing the Q point, there is more elastic potential energy as resistance.

c because after emphasizing the q point.

(1) Knowing that the "detective hound" is driving in a straight line at a uniform speed, in a balanced state, the traction force and resistance in the horizontal direction are a pair of balanced forces, and the size of the traction force is obtained accordingly;

2) Knowing the distance traveled, find the work done by the traction force of the car according to W=FS; Knowing the speed of driving at a constant speed, use the formula p= = =fv to find the work power of the traction force of the car;

3) The car is driving at high speed and is subject to greater air resistance, and the answer is analyzed from this aspect Answer: Solution: (1) The "detective hound" is driving in a straight line at a uniform speed, and the car is in a balanced state, and the traction and resistance are a pair of balanced forces, Traction:

f=f=2×105n；

2) The work done by the traction of the car:

W=fs=2 105N2000m=4 108J,(3) Reduce resistance.

1) Because of the constant velocity, f drag = f traction = 2 10 to the 5th power n2) w = fs = 2 10 5 to the power n * 2000m = 4 * 10 8 power p = fv = 2 10 5 power n * 400m s = 8 * 10 7 power 3 to reduce the resistance.

The horizontal traction force of the car is 2 10 5 power n The work done by the traction force of the car w = fs = 2 10 5n 2000m = 4 10 8j

P=FV=2 10 5N 400m S=8 10 7W The appearance of the car is designed to be streamlined, the air flow rate above the car is fast and the pressure is small, the air flow rate below is slow and the pressure is strong, and the pressure difference is generated up and down, giving the car an upward lift. Reducing the impact of dead weight on speed is conducive to increasing vehicle speed.

Reduce 50g after spill water

That's a 50ml reduction

300+m=500

The weight of the alloy is 200g

So the density is 4000kg m3

After removing the alloy, the total mass is 350-300 = 50g less, indicating that the volume of the alloy is equal to the volume of 50 grams of water.

The volume of 50 grams of water is: 50 grams 1 gram cm = 50 centimeters So the alloy volume is 50 centimeters

The mass of the alloy = 500 grams - 300 grams = 200 grams.

So the density of the alloy is: 200 grams 50 centimeters = 4 grams centimeters

When they meet in situ, there is.

t=2 r n1 v1=2 r n2 v2 thus has: n1 n2=v1 v2=10 4=5 2 that is, A 5n circles, B 2n circles will meet, the shortest time is n=1, that is, A 5 circles, B 2 circles, and the time used at this time is:

t=2 r2 v2=2 100 2 4=100 seconds.

Find the circumference of the circle by 200

Then, the velocity of v A and V B is 10 m per second, and V B is 4 m per second and in the same direction at the same time, so B is stationary relative to A, and A's speed is 6 m per second, so only the time required for A to circle the runway at a speed of 6 m per second is required.

Let the shortest time be t, then the distance traveled by A and B at the time of meeting is 10t and 4t respectively, because they meet in the same place, so 10t and 4t are integer multiples of the circumference of 200, so the minimum value of t is 100

v A*t=2*pie*r*n vB*t=2*pie*r*n The two formulas are compared, and we get: vA, vB=n n, i.e., 5 2=n n. Let n=5, n=2, calculable t=100

Solution: 1. As can be seen from the figure, two ropes bear heavy objects, so the speed of the goods moving v = 1 2xv people =

So the distance that the weight moves is the object = v object xt =

The frictional force f=m of the heavy object.

The useful work done by man is the work done to overcome the orbital friction w useful = fxs material = 600x25 = 15000j

2. Because it is moving at a constant speed, the pulling force of man f pull=

3. Because the mechanical efficiency of the pulley block is =75%, the work done by people w people = w useful = 1500

The power of human pulling force p=w human t=20000 100=200w

1. The gravitational force on the object is: g=mg=600kg 10nkg=6000n;

Frictional force experienced by the object: f=;

The distance traveled by a person in 100s is s person = 100s the distance the object moves s object = 1 2 * s person =

W has =fs=600N, 25m=15000J2, by =w has w total=fs fs'=fs fns=f nf get f=f n to find the required tensile force:

f=f/nη=600n/(2×75%)=400n；

3. The power of the pulling force is: P total = FV = 400N

Known: g 6000n f machine efficiency = 75% human distance s 100* cargo distance h 1 2*s 25m

Seek: w has f p

Solution: w has fh 600n*25m=15000jw total w has 15000j f=w total s 20000j 50m 400n

p=w t =20000j 100s=200w A: .

Very good to ask for.

1: The friction of the goods is 600N, and the work is w=fs=600*.

2: The pulling force of man is f pull = (f object 2) = (600 2). (for mechanical efficiency).

3: Power p=w t=400*.

Guaranteed to be correct.

Solution: , the distance traveled by the cargo s=(

w=f*s=600*25=15000j

Human = f 2 = 300n

3.The power of the cargo p=w 100=150w

p person = p *75

So p person = 200w

I don't know if it's right or not, and I'm not very sure, because it's a junior high school topic, so there are some things that I haven't considered.