Hurry Hurry Physics Question Okay right away, if it s good, add 50.

Question 1.

Question 2. Solution 1 Density of lead =

m= v, and m=54g, then v=<50cm3, so the lead ball is hollow, and the hollow volume v=

Solution 2 The density of lead =,m= v, and v=50cm3, so m=565g>54g

It can be seen that the shot put is hollow.

and hollow volume v=(565-54)=

Solution 3 The mass of the lead ball is 54g and the volume is 50cm, then the density ==<, so the lead ball is hollow.

and hollow volume v=50-(54

It feels weird to do this question, is the data wrong? I think it should be reasonable to have a shot put mass of 540g. But this can also be done. This can basically be copied directly and then uploaded to the homework book.

It's hard to play these symbols, so let's do it.

1) 1kg = 1000g, so.

2) The first method: the density of the ball = m v = less than the density of lead, so it is hollow.

The second method: Assuming that it is solid, then the mass of the ball of 50cm 3 = greater than the mass of the ball, so it is hollow.

In the third method, assuming that it is solid, the volume of the 54g ball is v=m p=less than the volume of the ball, so it is hollow. Hollow part volume =

The team will answer for you.

1) Convert to g. (to write out the steps) ——

2) The mass of the shot put is 54g and the volume is 50cm, ask if this ball is hollow or solid? If it is hollow, how large is its hollow part? (Density of lead = three solutions to be written, detailed steps).

1. V lead = m lead = m lead lead = 54g because it is hollow.

v empty = v ball - v lead =

2. In the same way, the mass is calculated according to the V ball M ball = lead V ball = 565g because 565g 54g is hollow.

The hollow volume will be calculated by repeating the steps in 1.

3. In the same way, according to M ball = 54g and V ball = 50cm to find the ball = M ball V ball =

Because it's hollow.

The hollow volume will be calculated by repeating the steps in 1.

1, 1) Density method.

First of all, the density of this shot put can be calculated from the weight data:

It is much less than the actual density of lead, so the ball is hollow.

2) Volumetric method.

If it is solid.

54 It can only be so large that it is actually 50 cm, indicating that it is hollow.

3) Quality method.

If it is solid. It should have a mass of 565g, which is actually not so heavy, indicating that it is hollow.

Volume of the hollow part: cm

The first article is inserted under the second article, which should be a bad pass 2 sheets**.

1. Conservation of the front compartment and water momentum when separated: after deformation of δv m δm u:

v δm m u (δv is the increased velocity of the forward compartment, u is the velocity of the water ejecting δm relative to the forechamber).

The speed of the obvious front compartment is to be increased.

When not separated, the action between the water and the front and rear compartments is the internal force of the system, so it does not move. To sum up: choose B

The process by which the bullet enters the sandbox and is embedded in it has a conservation of momentum, and the sandbox displacement is considered to be zero due to the extremely short duration of action:

m0×v0=(m0+m)v...1

The bullet and the sandbox are combined with the process of swinging up to the highest point.

Conservation of mechanical energy of the system:

1/2×(m0+m)v^2=(m0+m)gh+1/2×(m0+m+m)v'^2...2

Conservation of momentum in the horizontal direction of the system:

m0×v0=(m0+m+m)v'...3

The maximum height of the 123 formula is h=

2) When the sandbox swings back, its cycloid is perpendicular to the overhead crane, and the sandbox has the largest reverse speed relative to the overhead crane, so the speed of the overhead crane is the largest at this time. This process system is conserved in mechanical energy and horizontal momentum:

m0×v0=(m0+m)v''+mv'''...4

1/2×(m0+m)v^2=1/2×(m0+m)v''^2+1/2×mv'''^2...5

The maximum velocity v of the Sandbox of the Type 245 is obtained''= ；The maximum speed v of the overhead crane'''= .

Why didn't I start to publish it?。。。

1,b After separation, the front compartment moves backward due to the conservation of momentum, then the front compartment must move forward. And it is constantly stressed, that is, it accelerates the movement. When there is no separation, it is also due to the conservation of momentum, the front compartment has a forward trend, and the rear compartment has a backward trend, and there is no movement overall.

2. First, m0*v0=(m0+m)v is used to find the initial velocity.

1) The maximum height is the kinetic energy lost in a completely inelastic collision, and this part of the kinetic energy is all converted into gravitational potential energy.

First, (m0+m)v=(m0+m+m)v

Then there is (m+m0)gh=1 2*(m0+m)*v 2-1 2*(m0+m+m)v 2

2) The maximum speed of the car must be that the sand is in the vertical direction of the car, and the direction of movement of the sand relative to the car is backward. Because at this time, the rope will not give resistance to the car, and it has just swung from the front to the rear, it must have just finished accelerating. And at this time, there is no energy loss, elastic collision.

m0+m)v=(m0+m)v'+mv'

1/2* (m0+m)v^2=1/2* (m0+m)v'^2+1/2*mv'^2

Just solve it.

3) What is a wooden block?

Solution: 1, here we must first understand the title, the title says that the front cabin and the rear cabin are not separated, it means that there is no partition, not really separated, otherwise the ship will sink!!

If the front cabin is separated from the rear cabin, with the passage of time, the center of gravity of the ship will move at a uniform speed from the front cabin to the rear cabin, which is equivalent to a small boat, a person walks from the left to the right at a constant speed, and the whole hull moves forward at a constant speed;

If it is not separated, then the position of the center of gravity remains the same, which is equivalent to a person standing at the midpoint and not moving, and the ship is also not moving.

But I don't think there's this option, it's not a problem with the question.

2, (1) Momentum is conserved during the bullet entering the sandbox m0v0 (m0 m)v1

During the swing, the momentum of the whole system is conserved in the horizontal direction, and the mechanical energy is also conserved.

When the sandbox reaches its maximum altitude, the system has the same speed, (because as long as the sandbox and the crane have relative motion, the altitude will change) is set to v2.

m0＋m）v1＝（m0＋m＋m）v2 ②

1/2（m0＋m）v1²＝1/2（m0＋m＋m）v2²＋（m0＋m）gh ③

The maximum height at which the sandbox can be assigned.

h=m0²mv0²/[2(m0+m)²(m0+m+m)g]

2) When the bullet and the sandbox swing back to the lowest point, the speed of the crane is the maximum, (because in the process of swinging the sandbox from the highest place back to the lowest place, the rope has always had a component of the crane in the direction of speed) Let the speed of the crane be v3 at this time, and the speed of the sandbox will be v4, which is obtained by the conservation of momentum.

m0＋m）v1＝mv3＋（m＋m）v4 ④

It is obtained by the conservation of mechanical energy of the system.

m0＋m）v1²＝mv3²＋（m＋m）v4² ⑤

Lianli to find the maximum speed of the crane.

v3=2(m+m0)v1/(m+m0+m)=2m0v0/(m+m0+m)

3) This one is a bit not very good, wooden blocks, it's a sandbox!

Of course, when the bullet is first fired, the speed of the wooden block is the maximum, and the mechanical energy is running out of them, that is, the maximum velocity of the wooden block v1=m0v0 (m0 m).

o(∩_o~

Question 1: Choose B, because when the water is ejected, there will be a reaction force (e.g., rockets, the magnitude of the force is related to the size and velocity of the emitted material), but if the front and back are not separated, the front and back two compartments are a whole, just like if you lie on the ground and step on the other foot with one foot, it will not move.

Question 2 (1) No picture, no good boy.

Landlord, the calculation is not the hardest.

The most rare thing is to call you the process of calculation.

It took me no less than half an hour for this process, and finally gave up (helplessly......I don't want to do the other questions anymore......

Let's just announce the first question.

Answer: The power supply voltage is 15V; r0 is 10

Idea: If R0 remains unchanged, you can get R1:R2=1:3 before and after the current in the sliding rheostat circuit: I1:I2=2:1

Then there is columnar calculation.

Let the value of the sliding rheostat be r1 for the first time and r2 for the second time.

Then, in the first time, the current in the circuit i1 = 10 ro, and in the second time, the current in the circuit i2 = 5 ro, and the two comparisons show that the current in the second time is 1 2 of the first current

The following is an analysis of the sliding rheostat: because the value of the sliding rheostat changes in the two processes, when calculating its power, the formula with r cannot be used, only p=ui, and the voltage at both ends is U1 and the current is I1 when the sliding rheostat is used for the first time;

The voltage at both ends of the second time is U1+5, and the current passing through it is columnar: the power at the first time is obtained: U1*I1=10......From the power at the second time, we get: (u1+5)*(

Solving this system of equations, we get: u1=10, i1=1 from the first situation: power supply voltage=u1+10=20v;

And from the previous equation i1 = 10 ro, we get: ro = 10 i1 = 10 1 = 10

Final summary: the power supply voltage U = 20V, the fixed resistance R0 = 10 Rmax Variable Resistance=3*6=18 I=U R=U unchanged, that is, R lamp*1+6=R lamp*

Get p1 = p = i square r lamp = 6w

p2 = i square r lamp =

9. According to the voltage unchanged, the front and back are equal.

The formula ur01+ur1=ur02+ur210+10 (10 r0)=5+

R0 is 10 and the voltage is 20V

10. For convenience, a is the resistance of the bulb, b is the maximum resistance of the sliding rheostat, and b is 18, u=a+6

u=(9/18)（18+a)

Get u=12, a=6

p1=6w p2=3w

12. Since the electrical power consumed by the resistor r1 is large, the connection method is to solve the parallel equation in series first. Let the resistors r1 and r2 be ab(4b) = 16a, respectively

4b/(a+b))²a=9

Get b = 3 a = 9

Voltage 12V13 ammeter A2 is 6 20 = R160 The maximum number of ammeter A1 is , so the minimum resistance of the sliding rheostat connected to the circuit is 6 (0, 6

The maximum power that the circuit is allowed to consume is:

9.According to u r=i. 10V R0=I1 5V R0=I2 So I1=2I2 And because P=UI U power is unchanged, so P1=2P2

10v*i1+10w=( Calculate the rest of i.

Light can be transmitted in optical fibers, mainly due to the joint action of the core and cladding. According to the above principle of light refraction, we will understand that the core of the optical fiber and the cladding outside it must be two substances with different densities, and the density of the fiber core should be greater than the cladding. In this way, as long as one ray of light enters at the right angle, the light will be fully reflected inside the fiber and transmitted to the other end.

So how does it transmit all kinds of words, images, and sounds? It turns out that with the use of electronic technology, people can convert text, images, sounds and other information into electronic signals, so that they all become digital strings composed of "1" and "0", which is what we often call "digital technology" now. In digital technology, 1 and 0 represent the opening and closing of circuits, and when applied to optoelectronic technology, they can achieve both light and no light.

Therefore, people send a series of light and dark light signals to the optical fiber through the optical transceiver (the device that inputs the optical signal into the optical fiber), and after the other end of the optical fiber receives these optical signals, it is restored into a digital signal through a special equipment, and finally the digital signal is restored into text, image, sound, etc. by television, radio, computer, etc.

What happens in the fiber optic cable is total reflection, and it and the information it carries can be transmitted from one end of the cable to the other without loss.

It is easy to know the cube side length for this, I believe needless to say).

So, the minimum pressure p=79n (this is easy).

The maximum pressure is to place the iron on the edge of the table, and at most half of it should be suspended in the air, otherwise it will fall (because the center of gravity cannot exceed the edge of the table). That is, the area is half of the minimum pressure area, that is, the maximum pressure is twice the minimum pressure. (I use a ...... on my phone.)It's painful to play ......I'm sorry ).