Who s going to do these physics problems, give 50 points first, and if it s good, then chase the poi

Solution: When the tension of 1 N pulls the spring dynamometer hook, the pointer of the spring scale points to the second position of the 5 black lines below the scale mark; 2 N tension pull the hook of the spring dynamometer, the elongation of the spring, the pointer of the spring dynamometer points to the position of the scale mark" which indicates the tension spring dynamometer 1n7 cm cm = cm.

3n Rally Race Pull Spring Dynamometer Meter The pointer of the spring dynamometer of the hook should be 7 cm, and there is some distance below the next spring dynamometer 8 cm cm cm cm cm = cm.

Tick marks, spring pointers can be stretched to the centimeter position for further growth out of range.

The spring dynamometer can measure the maximum and minimum ranges, and the difference between the next spring dynamometer engraving line is irrelevant.

So, the answer is: 0?3n;No change.

Invariant solution: "When pulling the hook of the spring dynamometer with the pull of 1 N, the pointer of the spring scale points to the second black line below the scale 5; When the hook of the spring dynamometer is pulled with a tension of 2N, the needle of the spring dynamometer is pointed exactly at the position of scale mark 7" This shows that the elongation of the spring when the spring dynamometer is subjected to a tensile force of 1N.

When pulling the hook of the spring dynamometer with a tension of 3N, the pointer of the spring dynamometer should point to the scale mark next to the 7cm+ spring dynamometer, although it is up to 8cm, there is still some distance below, and the pointer of the spring is the position that can be stretched to, and if it is longer, it will exceed the range

The difference between the maximum and minimum values that can be measured by the spring dynamometer is the range, and it has nothing to do with the engraved line next to the spring dynamometer

So the answer is: 0 3n; No change.

Known: t1 = 100, t2 = 80, m = 4kg, q amp = 336000j

Find: c=?Solution: q amp = cm (t1-t1).

c=q put m(t2-t1).

336000j/4kg×(100℃-80℃)=4200j/(kg.℃)

Known: m=3t=3000kg, t1=20, t2=100, c=4200j (kg.).℃)

Seeking: Q Suck = ?

Solution: q suction = cm (t2-t1).

q suction = 4200 J (kg.)℃)3000kg×(100℃-20℃)=1008000000j

to the 5th power (4kg 20) = 4200j (kg·).

3000kg to the power of 80 = 9 J

At maximum velocity, f=f=p=fv=p=vp=50000v, kinetic energy theorem wf= mv2 pt

So v = 30m s p = 15000000w at this time the traction force f = p v = 1500000n f = f f = ma a = 1m s2

It's not a place to do your homework, this question is relatively simple, turn the book yourself.

1.Force analysis, the component of gravity is balanced by the supporting force and the elastic force.

gsina=kx i.e. x=mgsina k

2.Because the elevator has a vertical acceleration, that is, the net force of the small object must provide this upward acceleration in order to rise with the elevator. We might as well think of small blocks as having a downward ma'(better explained if you have studied inertial forces).

The greater this downward force, the greater the downward trend of the small block, so there is a maximum value for this acceleration.

At this time, the critical state force is analyzed.

mg+ma')sina=u*(mg+ma'cosa+k*2x to find a'That's the maximum.

If you have questions, you can ask them.

1.The force provided by the inclined plane in the oblique direction above is equal to the net force of the spring tension force and the gravitational force, which is reversed.

2.Because the slider has a tendency to move, in the direction of the spring, the friction force provides the resistance, and the spring elastic force is added to the resultant force of the oblique upward force provided by the inclined plane and the opposite force of gravity, and the combined force of the two provides the acceleration of the slider vertically upward.

The heat of vaporization of liquid oxygen is 215kj kg, the boiling point is -183, the density is 1140kg m3, the specific heat capacity of liquid oxygen is Celsius, the specific heat capacity of oxygen is Celsius, the specific heat capacity of water is Celsius, the mass of liquid oxygen is not m1 = 224kg = 8kmol, the mass of water m2 = 10000kg, the heat released by the complete gasification of liquid oxygen is, and then calculated by the heat balance, theoretically 10 cubic 30 °C water can be reduced by 5 °C

When a block is first placed on the conveyor belt, Newton's second law:

Mg ma gets a 4 M S2

After the block lasts t1, it will speed with the conveyor belt, then.

at1＝v0＋a0t1

Get t1 1 s

The displacement of the block during this process is:

s1＝ at12/2

S1 2 m so the object has not yet reached point B, and in the process thereafter, due to A0< G, the block will move with the same acceleration as the conveyor belt, and then it will take T2 to reach point B, then.

s－s1＝at1t2＋ a0t22/2

Get t2 1 s

Object duration from A to B: T t1 t2 2 s

1 Because it floats.

So f buoyancy = mg of the wooden block

And because of the buoyancy of the land chamber f = mg of the water being discharged

So p wood v wood g = p water v water g

Because three-quarters of the wood is in the water.

So v water = three-quarters of v wood.

So P wood = P water of three quarters of the early macro.

P-wood = 750kg per m

The symbol of that density could not be typed, so P was used instead.

The second course. The idea is the same as the first course.

From the question, it can be obtained that the water discharged by v: v wood = 72:96 = 3:4 is exactly the same as the first one.

The process is the same.

The answer is the same.

Buoyancy = f object gravity.

f buoyancy = g * v water * density of water = m object * g So m object = v water * density of water = 1 4v object * density of water.

Object density = object mass Object digging god body volume = 1 4 Density of water = 3rd power.

So the mass of the object = the density of the object = divided by 96*10 to the minus 6 power = 3 power.

Absolutely, yes!

Hear the first echo = closer to the mountains.

l1=340*

Second echo = farther mountain.

l=340*(

The distance between the two mountains = l1 + l2

Let the two mountains be A, B, Xiao Ming is X meters away from A Mountain, and B Mountain Y is Y, then X=

y=(Note: It's over again.)

Therefore, the distance between the two mountains is x+y=68+51=119m

Let him be distance from mountain A xm then t=2x 340= x=51m set distance from mountain b ym t=2y 340= y=68 s=x+y=119

The location of Xiao Ming station and the location of the nearest mountain is a round-trip journey of the second of sound, so the time multiplied by the speed of 340 to get 102 is a round-trip journey, as long as half a round-trip journey, the distance between Xiao Ming and the nearest mountain is 51 meters, and then the echo of the farthest mountain is heard in seconds, and what is heard in the second after the second is heard after the second, and the distance can be calculated to be 69 meters with the method just now. The distance between Xiao Ming and the near mountain is 51, and the distance from the far mountain is 69, and the distance between the two mountains is 17 meters.