Solve the physics problems in the second year of junior high school, and give more points for the go

Solution: Known: The pressure on the road surface when the truck is full of goods:

f=g=mg

2 103kg+10 103kg) 10n kg=force area: s=6 200 10-4m2=

Pressure on the road surface: p card = f s = =

P card P accept.

will damage the road surface

Therefore, when the car is driving on the road, it will cause harm to the road

12 10 10) (power of 6 8 power of 10, so there is danger.)

The maximum pressure that a certain horizontal highway pavement can withstand is 8 105pa There is a truck, the weight is 2t, the load is 6t, there are 6 wheels, the contact area between each wheel and the road surface is 200cm2, and the volume of the compartment is that the car is filled with sediment The density of the known sediment is kg m3 (g=10n kg) Find:

Is the car overloaded?

Explain whether the car is harmful to the road surface by calculation

It is known that the distance from Huanglong to Zhangwan is about 24km, if the car is filled with sediment and drives from Huanglong to Zhangwan at a constant speed of 90kw, how much work does the car do and how much time does it take? (The resistance of the car is times the weight of the car) (1) Find the volume of the compartment (the volume of sediment at most), know the density of sediment, use the density formula to find the load capacity of the car, and then compare it with the load of 6t to judge whether it is overloaded;

2) The pressure of the car on the ground is equal to the sum of the gravity of the car and the sediment, according to the formula p=fs

The pressure on the ground can be obtained, and then compared with the maximum pressure that the horizontal highway pavement can bear, it can be judged whether it is harmful to the highway pavement;

3) find the resistance of the car according to the question, the car is in a state of equilibrium when driving at a constant speed, find the traction force according to the two-force balance condition, find the work done by the car according to w=fs, and find the time required according to w=pt Answer: Solution: The volume of sediment is equal to the volume of the compartment, that is:

v = , the cargo capacity of the car is equal to the mass of the sand inside the car:

m Sediment = Sediment v = so this car is overloaded

The pressure of this car on the road:

f = g = (m car + m sediment) g = (2000kg + 10080kg) 10n kg =, whether the car is pressure on the road:

p=f s = 200 10-4m2 =, so it will be harmful to the road

Because the car is driving at a constant speed, the traction and resistance are a balancing force, so the traction of the car:

f=f=car+mSediment)g=,The work done by the car:

t=w p = 90000w ≈1288s．

Answer: The car is overloaded;

The car will cause harm to the road surface;

The work done by the car is about 1288s

Look at the picture, and if you can't understand it, you can't find x, I have nothing to say.

GJ is the width of the river, EP is the height of the tree, AB is the height of the tree, and PL is the reflection of the tree.

From the knowledge of geometric similarity:

hg ef=gi fi=gi (gi+fg)= so gi=cd ak=jd jk=jd (jd+dk)= so jd=ef kl=fi ki=(fg+gi) (kd+dj+ji)=so ji=3

So gj=gi+ij+jd=

Hope it works for you!

p=f s=450 [ pa=15000 pa When calculating the pressure, *2 is required because it is two feet

2.His gravitational force does not change f=450n

p=f/s=450/[ pa

At this time, it is one leg, so don't *2

Solve the cow. p=450 (negative quadratic m2)=......

2.The indication is 45kg, p=450 (negative quadrature m)=......

The second asked. I don't know, but I think it should be the same!

If you need to know the density of wood and copper, you can check the density table, which is set to a, b

Bronze statue volume: 56 a

Quantity of copper required: (56 a)*b

How much copper? Does that refer to the quality of copper? Yes, you only need to find the volume of the bronze statue first, which is equal to the mass of the wood divided by the density of the wood, and then the density of the copper and then multiplied by the volume.

The density of copper and wood, you should have a table in your textbook to refer to).

First, divide the mass of the model by the density of wood to obtain the volume of the model. This volume is also the volume of the bronze statue, and the copper density can be multiplied by this volume to obtain the required mass.

First of all, we have to calculate the volume of the model Since the wood and copper are the same size, so they have the same volume, and then calculate the mass of the copper needed v mold = m mold wood density v copper = v mold m copper = v copper * copper density It's as simple as that.

The density of wood is p1

The density of copper p2 volume is the same because of that.

Calculate the volume first.

Let the bronze statue volume = v

then v=56 wood density.

The weight of copper required is v * copper density = 56 * copper density and density of wood.

This question is not also provided for the density of the wood, and if it is provided, it can be based on.

Mass = Density * Volume.

Calculate the mass of copper.

1. When only S2 is closed, the ammeter, R1 and R2 are connected in series, let the current in the circuit at this time be i1, and the indication of the ammeter is, then i1=. The voltage applied to both ends of r1 is u1 = i1 * r1 =. The voltage applied to both ends of R2 is U2=I1*R2=.

The power supply voltage u=u1+u2=6v+.

2. When only S2 is disconnected, R1 and R2 are connected in parallel, and the ammeter is on the main road, then according to the nature of the parallel circuit, the voltages at both ends of R1 and R2 are equal, which is equal to the power supply voltage. At this time, it is known that the current through R2 is I2=1A, then the voltage applied to both ends of R2 is the power supply voltage U=I2*R2=1A*R2.

In summary, we can get the equation 6V+, which can be calculated as R2=12 and U=1A*R2=1A*12=12V.

r1 and r2 in series, u (12 + r2) =

Disconnect only S2 (equivalent to R1 and R2 in parallel, ammeter on the dry road. ), the current through R2 is 1A, U R2 = 1A(2).

Solution (1) and (2) yields u=12v r2=12

When S2 is closed, the total supply voltage is u=

When S2 is disconnected, the voltage across the resistor is equal to the supply voltage, so the voltage of R2 = 1A x R2 = U

The simultaneous equation can be solved to u=12v r2=12 ohms.

r1 and r2 in series: u=i*(r1+r2).

R1 and R2 in parallel: U=I2*R2

r1 i i2 is known to solve the equation yourself.

(12* then r2=12 euros u=12*.)

Do you see it ?? Think for yourself why you're doing it.

Solution: Known: The pressure on the road surface when the truck is full of goods:

f=g=mg

2 103kg+10 103kg) 10n kg=force area: s=6 200 10-4m2=

Pressure on the road surface: p card = f s = =

P card P accept.

will damage the road surface

Therefore, when the car is driving on the road, it will cause harm to the road

Is this called difficult? I'll show you my answer for the third year of junior high school!

Question 1 6 2 The speed of sound is equal to the answer because the car hears an echo after 6 seconds. Explanation: The sound is sent to the wall plus the time it takes to return to the car equals 6 seconds.

Question 2 Because the sound travels twice in the air and in the steel pipe.

It's the speed of sound and the speed of sound in the iron substance.

1 solution: the distance traveled by the car: s1 = vt = 15 * 4 = 60m sound propagation distance: s2 = 60m + 340 * 4 = 1420m The distance between the car and the cliff when the horn is sounded: s = 1420 2 = 710m2 solution:

Tap on one section of the longer steel pipe and hear (2) sounds at the other end. The first time it was propagated through (steel pipes). The second propagation through (air) because the speed of propagation of sound in a solid is greater than the propagation speed in a gas.

1 (speed of sound*6+15*6) 2

2 The long pipe transmits the sound once, and the air transmits the sound once.

The next question should be how long is the long pipe?

Generally speaking, people can distinguish between the above intervals, and we can consider this interval time to be a, and the speed of sound propagation in the long pipe is b

Then the shortest length of the long tube is (B-380)*a

1. The speed of sound is 340m s

Drawing, speed of sound x time + speed x time = 2 times the distance.

The distance is (340+15)*6 2=1065m2, and (1) the two sounds are because the speed of sound transmission in the steel pipe is different from the speed of transmission in the air, resulting in a time difference.

2) Length of the steel pipe The speed of sound in the air - the length of the steel pipe The speed of sound in the steel pipe = the duration of the sound.

2s-6*15=340*6 s=1065

The primary sound is transmitted through the steel pipe, and the primary sound is transmitted through the air.

seconds so that the ears can tell.

Don't forget to give extra points!

Facing **, the right mouse button "**Save As" is very clear after saving and enlarged.

Trust me, yes.

Analysis: When the branch switch is disconnected, only R1 is connected to the circuit, so at this time, the voltage at both ends of R1 is equal to the power supply voltage 18V, and 3A is the current of R1 i1, when the branch switch is closed, R1 and R2 are connected in parallel to the circuit, the branch voltage is equal to the power supply voltage, which is 18V, and the ammeter measures the current of the dry circuit, and the current of R2 is more than the current I2, so the current of R2 is - 3A=.

According to Ohm's law, you can calculate what the value of each resistance is.

Solution: r1 = u1 i1 = 18v 3a = 6 i2 = i-i1 = =

r2 = u2/i2 = 18v/ = 12ω

1. When the switch is disconnected, R1 and the ammeter are connected in series, and the indication of the ammeter is 3A, indicating that the current through R1 is 3A, and the power supply voltage is 18V, and R2 is not connected to the circuit at this time, so the voltage added to both ends of R1 is 18V. Then, the resistance of R1 is (18V 3A) = 6 ohms.

When the switch is closed, R1 and R2 are connected in parallel in the circuit, and the voltage added to both ends of R1 and R2 is 18V, then the current through R1 is (18V 6 ohms) = 3A; At this time, the indication of the ammeter is, then the current through R2 is, then the resistance value of R2 is (18V ohm.

2. The maximum allowable current in the circuit needs to be considered. The ammeter is on the main road, and its range is , so the maximum current allowed on the main road is . The current through R1 is (9V 18 ohms) = ; Then the maximum current that can pass through the sliding rheostat is, that is to say, the resistance value of the sliding rheostat reaches the minimum at this time, and the resistance value of the sliding rheostat connected to the circuit at this time is (9V ohm; Then the value range of r2 is 90-100 ohms.

Don't forget the range of values of the sliding rheostat given by the known conditions in the question! ）

When the current is 3A, the switch on the road is open, and the switch on the dry road is closed, so that the fixed resistance (R1) = 18 3=6 ohms can be found, and when the current is the switch is all closed, the total resistance in the circuit R=18 ohms and because R=R1XR2 R1+R2

So r2=12 ohms.

1. Because r=u i, r1=18v 3a=6 ohms r total = ohms 1 r2+1 r1=1 4 r2=12 ohms.

When the first question is disconnected; i=u/r1=3a

When closed; i=u/（r1+r2)/r1r2；

According to what is known, it is found.

Choose A, because if A is a voltmeter, the light on the right will be short-circuited and not on, so A is an ammeter; If B is also an ammeter, the current will flow directly from the positive pole of the power supply through the two ammeters to the negative pole, causing a short circuit, so B is a voltmeter (plus points!). ）

40 equals;

whether it is equal to 40;

Whether it is equal to 158

Hollow volume = 40-158

1 is the ammeter.

2 is the voltmeter.

Answer: a

The ammeter is equivalent to a wire, and the voltmeter is quite disconnected.

a. Two lamps in parallel;

b. The power supply is short-circuited;

cOnly the left light is on;

d The right light is short-circuited, and only the left light is on.