A physics problem in the second year of junior high school, is there anyone to help you do it?

Very simply, the temperature of the ice cubes taken out of the freezer compartment of the refrigerator is lower than O, and the water is released to cool down after being put into the water, and the ice absorbs heat and heats up; When the temperature of the water near the ice cube drops to o, the temperature of the ice is still lower than o; The ice cube continues to absorb heat, and the ice cube is nearby'c's water continues to exothermic and solidifies into ice. So the ice cubes will "stick" together.

The temperature of the ice cubes taken out of the freezer room of the refrigerator is lower than 0, and after being put in the water, the water is exothermic to cool down, and the ice absorbs heat to heat up; When the temperature of the water near the ice cube drops to 0, the temperature of the ice is still below 0; The ice cube continues to absorb heat, and the water near the ice cube 0 continues to release heat and solidifies into ice. Thus "sticking" several ice cubes together.

The temperature of the ice cubes taken out of the freezer compartment of the refrigerator is lower than 0 degrees Celsius, and the water is released to cool down after being put into the water, and the ice absorbs heat and heats up; When the temperature of the water near the ice cube drops to 0 degrees Celsius, the temperature of the ice is still below 0 degrees Celsius; The ice cubes continue to absorb heat, and the water near the ice cubes at 0 degrees Celsius continues to release heat and solidify into ice. So the ice cubes will"sticky" together.

This is easy to understand, because when the ice melts, the ice melts into water at zero degrees, and the ice melts into water because the heat from the water is transferred to the ice.

The problem is between ice and ice, when the ice is in contact with the ice, their temperature is still zero, so the small amount of water or the freshly melted ice that exists between the ice at zero will remain the same temperature as the ice between most of the ice, and it will also enter the state of ice and connect with the ice. So the water that is at the critical end will melt with the ice.

The temperature of the ice is low, below 0, it will absorb heat to the surrounding water, the water will release heat, and when the water temperature drops to 0, it will freeze, and the ice will freeze when the temperature drops to 0, and the ice will not melt at this time, so it will stick.

Because the ice cubes are lower than the water temperature, the ice cubes have to melt and absorb heat! If the temperature of the water next to it is high, it will provide heat back to the melting of the ice cubes and redistribute the heat! Just go back, the water between the two ice cubes froze.

It is in a state of solid-liquid coexistence.

Solution: T1 = S v steel = L 5200m s

t2 = s v air = l 340m s

t=t2-t1=l/340m/s-l/5200m/s=2sl≈

A: (i.e.) the length of this section of track is about.

I may have miscalculated, the answer is strangely ......

That, lz, is this a short-answer question? If yes, the physics short-answer questions are not allowed to list equations.

x=v steel*t=vair*(t+2); Solve this equation and get xThe distance is equal, the speed is equal by the time, the speed is known, the difference in time is 2, calculate the time first, and the distance can be found.

The electric energy consumed by the aluminum disc rotates 200 revolutions is w=200r 1800r The actual power of the electric water heater p=w t=( i1 r=1210w i2 r=1000w

Column equation, let the length of the rail x m

x 340 is the length of the rail divided by the speed of sound in the air to get the time he hears the sound in the air x 5200 is the length of the track divided by the speed of sound propagation of the track x 340 - x 5200=2 is the interval time x is the length of the rail

The length of the high-speed rail track is sRule.

He heard the sound in the air at s 340

The time when the sound coming from the railroad tracks is S 5200

s/340-s/5200=2

Find s=728m

Two sounds, one from the sound of the rails and one from the air.

Let the time of rail sound transmission be (t-2), then the air sound transmission is t, and the units above equation 5200*(t-2)=340*t (the distance is equal) are omitted. Please make it up by yourself.

I'm going to use equations to solve it.

Solution: Set the track length x, by"Xiao Gang heard two sounds at one end with an interval of 2s"There is x v track + 2s = x v empty.

The solution is approximately equal to 728m

Answer:.. After the problem is solved, the next step is to explain.

The time for sound waves to propagate on the track is X V rail, and Xiao Gang listened to this sound and only heard the sound from the air after two seconds, so the time for sound waves to travel in the air is "X V track" plus "2 seconds", which is also X V empty.

That's it.

You're in the third year of junior high school, right?,I haven't seen this kind of question since I was in my first year of high school.,I miss it so much.。

Solution: It is known that t=2s

v=5200-340=4860m/s

s=vt2s*4680m/s

9360m A: This section of the railway is 9360m long.

Enough complete, I hope it helps you.........Naturally, it is also to be adopted by pulling, thank you.

Set up rails x meters.

x/340-x/5200=2

Time propagated in air - time propagated in steel = time difference is found x

Xiao Gang heard two sounds at one end.

Gravity of the tractor:

g = mg = 5000kg * 10n kg = 50000n tractor pressure on the ice:

f=g=50000n

The contact area between each track of the tractor and the ground:

s=25m*28*10^-2m=7m^2

The pressure of the tractor on the ice:

p1 = f1 2s = 50000n 2 * 7m 2 = because there is: p1 = so this tractor can pass through the ice.

The maximum pressure that the ice can withstand:

p2=p-p1=

The maximum weight of the object can be loaded:

f=ps=9440pa*7m^2=66080n

The mass of the tractor is 5000kg, the gravity is g = mg = 5000 *, the contact area between the two tracks and the ground is s =, so the pressure is p = g s = 49000, 98000-35000 = 63000pa, so it can withstand the external weight of f = 63000 *, so the mass that can be loaded is m = f g = 88200, and the maximum weight of the object can be 9000kg

According to p=f s, the maximum gravity f= of the river surface can be obtained to the power of 5 cattle, and according to f=mg, m=

Answer: Can pass up to m object = m-m drag =

Why isn't it the topic of the first year of junior high school?,Speechless、、、 handed over to other prawns、

1 All (1) Let the density of 60 degree liquor be wine, then wine = m alcohol + m water v alcohol + v water.

60cm3× / 100cm3==

2) If the volume of 60 degrees of wine is V, then the volume of 30 degrees of wine is (1000ml-V), according to the title: 60100 V + 30100 (1000ml-V) = 420ml

The volume of 60 degrees of liquor v = 400ml, then the volume of 30 degrees liquor is 600ml

What's the question! Say. will do some help, they are all like-minded people.

The time for a listener 20m away is 20 340 = 1 17 seconds.

The time of the listener in front of the radio at 200 km from the auditorium is 200000 3 * 10 8 = 1 1500 seconds < 1 17 seconds.

So, the listener in front of the radio, hear the report first.

t1=20/340

t2=200/3*10^5

The calculation shows that the listener in front of the radio hears the report first.

I guess it's a listener in front of the radio.

Hello friends:

Solution: The distance of the train is s=l train + l tunnel = 200m + 6700m = 6900m, the movement time of the train is t = , and the speed of the train is v =

s/t=6900m/414s

Okay?

Solution: (copy 1) the distance of the train S = L train + L tunnel attack = 200m + 6700m = 6900m, the movement time of the train bai.

dut=, the speed of the train.

zhiv=st=6900m414s≈

2) The time it takes for a person to run to the safe zone DAO t=sv=600m5m s=120s, and the length of the fuse l=v, the fuse t=

3) The speed of the car v = 36km h = 10m s, the time when the sound reaches the cliff t1 = sv = 875m 340m s;

In this time, the distance of the car is s1=v car t1=10m s, and the distance from the car to the cliff is s2=s-s1=;

Taking the car as a reference, the sound is relative to the velocity v of the car'= v + v sound = 340m s + 10m s = 350m s, the time when the sound reaches the car t2 = s2v =, the time when the driver hears the echo t = t1 + t2 =

Answer: (1) The speed of the train as it passes through the tunnel is; (2) The length of the fuse wire is at least 96cm; (3) It takes 5s for the driver to hear the echo of the horn.

Because the oil is lighter than water, when the left tube is full, the right tube is not yet full.

When the oil is full, the height of the oil in the pipe is x, and the height of the water in the left pipe is 30-x; And because the total length of the water in the pipe is 20*2=40, the height of the water in the right pipe is 40-(30-x)=x+10

Taking the junction of oil and water in the left pipe as the reference plane, the oil height is x, and the water height is (x+10)-(30-x)=2x-20, and it can be seen from the whole force analysis that only the gravity of oil and water is equal, the liquid level will be maintained, so there is, the solution is x=16, that is, you can pour 16cm high oil.

m=pv=2*16*.75=24g

This question has been answered before, so I don't have to waste my money to ask it again, I still hope it can help you.