Physics master advanced on 2 questions One physics problem master advanced don t give me the answer

The temperature does not decrease, and the water does not exothermic and solidifies into ice.

The ignition point of the paper is higher than the boiling point of water, and the heat of the paper is absorbed by the water, so the water boils, and the temperature of the paper does not reach its ignition point.

As the temperature inside the thermos remains the same, the water solidifies when condensed, and the mass decreases, so the ice mass increases (although the original ice may have a melting point, but the overall water mass decreases and the ice increases).

When the temperature of the ice-water mixture is 0 and the external temperature is also 0, the water cannot be exothermic and solidified, and the ice cannot absorb heat and melt, so the quality of water and ice remains unchanged.

The fifth question is changed to a place of minus 5 degrees Celsius, the water is cold and exothermic solidifying, the ice temperature decreases, and eventually the water will solidify, and the room temperature may rise, but the amount of change is very small. (These 5 questions are very good, you need to figure them all out.) )

The landlord has learned the torque? The mass of the strips is evenly distributed, and the center of gravity is at the midpoint. Let the support force of A to the wooden strip be fa, the supporting force of B to the wooden strip is fb, and the gravitational force is g.

At the beginning, fa=fb=1 2g, b has a leftward dynamic friction fb x to the wooden strip, a has a directional static friction to the wooden strip, and the maximum static friction is fa x =, so the actual static friction is , and the direction is to the right. The wooden strips are still.

The following is a discussion of the future state of exercise. When B moves to the distance A is S, the wooden strip does not rotate, A is selected as the reference point, and the force moment of the wooden strip is balanced g x 1 2 l - FB x S =0, and the vertical force is balanced g = fa + fb. The solution of the solution is FB=L 2S·G, Fa=(2S-L) 2S·G, and when the dynamic friction of B reaches the maximum static friction of A, the wooden strip will begin to slide.

In this case, there is (2s-l) 2s·g =l 2s·g x can be solved to get s= 3 4 l, fb=2g 3, fa=g 3At this time, a will suddenly change into a dynamic friction force with a magnitude of (2s-l) 2s·g, and b will be a static friction force with a magnitude of 1 15g. After that, the wooden bar starts to slide to the left.

Let the sliding h stop after a long time, and the dynamic friction provided by A is equal to the static friction provided by B. Or choose point A as a reference, and balance the force moment of the wooden strip g x (l 2 - h) -fb x(3l 4 - h) = 0 vertical force balance g = fa + fb .Horizontal force equilibrium fa x - fb =0

Synonomic solution yields FB=G3, FA=2G3, H=3L8

Later, and so on.,It's too late.,Go to bed.,The landlord has time to continue by himself.。

1.In order to ensure that car B does not collide with car A, the speed of the two cars is the same, and the time is t

There is 16-3t=16-4(, i.e., t=2s, so the common velocity v=16-3x2=10ms

Car A brakes, the acceleration is a1 = 3m s, the forward distance is x1, there is 2 (-3) x1 = 10 -16

x1=26m

Car B advanced x2= due to the driver's reaction time

Car B brakes A2 = 4M s, forward distance x2', there are 2 (-4) x 2 = 10 -16

That is, x2'=In order to ensure that car B does not collide with car A, the distance between the vehicle should be maintained at least (x2+x2')-x1=

2.The motorcycle accelerates to a = 028m s for 3min, and the speed will reach m=at= =, in the process of the motorcycle catching up with the car, it accelerates to the maximum speed m first, and then chases the car at this maximum speed m and sets the time required to accelerate to the maximum speed m t0, then the time to catch up with the maximum speed m is t-t0

The acceleration section for the motorcycle is: m=at0

From the equal displacement of the motorcycle and the car: at0 2+ m(t-t0) = t+s0

Solution: a =

1. A braking distance s1 = -v * v a1 = 256 3 B braking distance s2 = -v * v a2 + 8 = 256 4 + 8 A B distance should be greater than s2-s1 = calculate it yourself minutes The car leaves the original position of the motorcycle distance s1 = 1000 + 20 * 180 = 4600

The motorcycle accelerates for 4600 meters in 3 minutes, t = 180 seconds, the first stage of the motorcycle accelerates: s2 = (1 2) ATT motorcycle in the second stage of constant speed: s3 = 30 * (180-t) s1 = s2 + s3

4600=(1/2)att+30*（180-t）at=30

Solution: t=160 3; a=9/16

Explanation, there is a slight problem with the title, it should be "At least how much acceleration does the motorcycle start?" ”

Otherwise, there will be an infinite number of answers.

Because the motorcycle can accelerate to 30 meters and seconds with an acceleration greater than a = 9 16, and then quickly catch up with the car.

1. Use the formula: s=vt-1 2at 2 to calculate S A and S B, and then subtract S A with S B to be the distance between the cars, remember that the time of S B is less than that of S A, and the time of S A should be calculated t=v A

2. Use the distance to calculate equally

Let the acceleration be a

1000 + 20 * 180 "a*180 2 2 calculate the minimum value of a, and you can calculate the rest by yourself."

The first question can be solved by drawing pictures. The ordinate is the velocity, and the abscissa is the time. As shown in the figure, because the brakes of the car behind are relatively strong, if the distance is long, then the distance between the two cars will start to widen again after the T1 moment.

Therefore, the minimum distance is at T1, and the distance between the two cars is exactly zero.

This distance is the area of the triangle above.

Find the velocity at t1 and set it to v, then (16-v) 3=(16-v) 4 + v=10

Triangle area s = 1 2 * 16-v) = m The second problem you can still solve with drawing. It's a little more troublesome, and you have to learn to draw inferences. The answer given upstairs is not correct, because the speed of the motorcycle is limited, and it will not be a uniform acceleration all the time.

The answer is, I don't know if it's right or not, you can calculate it yourself.

One. When the velocity is equal, they do not collide without colliding. That is, 16-4*(t-t1)=16-3t

t=2Distance from 16t-1 2a ( II. Column at1 = 30 vt2 + 1 2 at1 2 = 4600 t1 + t2 = 180 a=

Such an easy question requires a master... Look it up yourself. The basic formula can be solved...

The extension of the current in the right direction passes through point p, where b is zero.

The upward current is a semi-infinite current-carrying wire, b = 0 i (4 a).

So, fill in the two blanks separately:

b = 0 i (4 a), orientation: perpendicular to the paper face.

This is a physics outline. The physics outline is a lot of hard to write improvements. Most you can go to the internet to raise this. Go translate it and he'll give it to you. The answer is very clear.

Ask the experts to help solve the physics questions. The physics questions are too difficult, please teacher. Help to answer! Thank you, teacher, for your trouble.

1. The maximum work done by the water flow per hour on the turbine.

w=gh=pvgh=

2. W electricity = 40% * w =

The generator can reach the maximum power p=w t=

The water surface drops 60 per hour

Work 60 per hour

The maximum power of the generator is W

200 days turns into seconds multiplied by megawatts to get *** joules.

Divide the joules by the calorific value of anthracite to get kilograms.

1. The maximum work done by the water flow per hour on the turbine.

w=gh=pvgh=

2. W electricity = 40% * w =

The generator can reach the maximum power p=w t=