High 1 Physics Urgent 5, High 1 Physical Emergency, etc

K is the thermodynamic temperature scale, let me talk about the difference between the Celsius temperature scale and the thermodynamic temperature scale.

Zero degrees Celsius is the temperature of a mixture of ice and water at a standard atmospheric pressure.

Zero degrees in the thermodynamic temperature scale refers to the temperature at which the molecules stop moving, which is -273 degrees Celsius on the temperature scale. (Of course, it's worth adding: because the molecules are moving irregularly and endlessly, this temperature is not reached.)

So, for every 1k increase in temperature, it is a 1 degree increase. According to "0 k=-273 degrees".

6000k = (6000-273) degrees = 5727 degrees.

6000-273 = 5727 degrees Celsius.

k is the Kelvin temperature scale (also the absolute temperature scale) or the Kelvin problem (absolute temperature), and its conversion to Celsius is that Kelvin temperature = Celsius temperature +. Not too strict, it can be replaced with 273.

6000k = 6000-273 degrees Celsius = 5727 degrees Celsius.

As long as it is the Kelvin temperature, subtract 273 from its value to be the Celsius temperature.

6000 (K) - 273 = 5727 (degrees Celsius).

Formula: Kelvin temperature Celsius.

t=t+

t is the Kelvin temperature.

t is Celsius.

So t=

The ball is subjected to gravity and tension, and the resultant force of these two forces provides centripetal force for the ball, and the centripetal force can be obtained by orthogonal decomposition as the mgtan angle.

2) by f=mv2r

r=lsin angle. Gotta v

The magnitude of the centripetal force is: f

nmgtan, the radius of the circular motion of the ball is: r=lsin, then Newton's second law is obtained: mgtan =

mv2lsinθ

Linear velocity is obtained: v=sin

glcosθ

glsinθtanθ

The larger the ball, the larger the sin and tan, the greater the speed of the ball movement, and the correct option c The ball movement period: t=2 rv

lcosθg

1. The wooden block is in a state of equilibrium, the tensile force and the friction force of the dynamometer are balanced, f=f=4n, and the direction is shouted to the right.

2,f=un=umg,(gravity support force refers to the seepage state only source equilibrium) u=f mg=

The deformation is the same when the AC spring elongation is the longest and the spring compression is the shortest. The elastic potential energy is equal. The mechanical energy of the two wooden blocks and the spring system is conserved. The kinetic energy at the two moments is equal, and when the compression begins, there is only the elastic potential energy, so the elastic potential energy at these two times is less than the beginning.

Because the kinetic energy is equal and the mass is equal, so the velocity is also equal.

This question mainly examines the knowledge of energy conservation, with the help of simple harmonic motion.

Both v and v1 should be 0, otherwise the spring will continue to elongate or shorten, and will not be the longest or shortest;

Then, because the initial spring potential energy is e0 and the kinetic energy of the system is 0, according to the conservation of energy, from the initial kinetic energy = v = v1 = 0, e1 = e2 = e0.

So ab9 hasn't read the question for a long time, and I don't know if it's correct. One thing is not clear, how can the inscription situation appear simple harmonic motion, don't the two wooden blocks slide to the right at equal speed? ）

This question requires a careful analysis of the process of object motion!!

First, when the wooden block 1 is released by rest, the spring is compressed, the wooden block 1 begins to do accelerated motion, the wooden block 2 does not move, and when the wooden block 1 moves until the spring elongates to its original length, the speed of the wooden block 1 reaches the maximum (because it has been subjected to spring thrust).

Second, after the original length, the wooden block 1 continues to move, the spring is elongated, and the wooden block 2 is pulled to move, and the wooden block 1 does a deceleration motion, and the wooden block 2 does an accelerated motion, but the speed of 1 is greater than 2, and the spring is slowly elongated. Thereafter, when the velocity of block 1 and block 2 is the same, the spring elongation is maximum. Block 1 and Block 2 both have a velocity of V.

Third, after reaching the original length, the wooden block 1 and the wooden block 2 are both stretched, but the wooden block 1 does a deceleration motion, the wooden block 2 does an accelerated motion, and the spring begins to shorten and slowly reaches the original length.

Fourth, at this time, the speed of the wooden block 2 is greater than that of the wooden block 1, the spring begins to compress, the wooden block 1 accelerates, and the wooden block 2 decelerates, and when the two velocities are the same, the spring compression is the shortest, and the velocity of the wooden block 1 and 2 is v1.

From the conservation of momentum in the horizontal direction, v=v1 can be obtained

When the two states can be obtained from the conservation of mechanical energy, the kinetic energy is equal, and the total energy is equal, so e1=e2.

The total energy is e0, e1 + kinetic energy is equal to e0, so e1 e0

So choose AC

c。。This problem looks at the end tense of the last two processes, one is the longest spring, the other is the shortest, then the elastic potential energy of the two is the same, and then the intermediate process is the conversion of elastic potential energy into kinetic energy, so v1 and v2 are not the same, and because of the conservation of energy, the kinetic energy of e1 + wooden block is = e0, so e1 is less than e0 In this question, it is not said that the wooden block is the same, so a cannot be selected.

, I haven't done a physics problem in 8 years. I don't know if it's correct.

Because the vertical upward throwing motion is symmetrical.

So t1 2 is needed from p to the highest point

From o to the highest point, t2 2 is required

According to s=at 2

So (t2) g 8=(t1) g 8+ha

The vertical upward throwing motion is a synthesis of the inverse motion of the free fall motion and the free fall motion, which is a symmetrical motion of two segments, so the time of rise and the time of descent are the same.

The time taken to descend from the highest point to the point p is t1 2, which is obtained by the formula s=gt2: s1=gt1 8

The time taken to descend from the highest point to the point o is t2 2, s2=gt2 8, and the value of g can be found as a option by using h=s2-s1.

The answer to the first two questions m is the mass of the steel ball, r is the length of the line, g is the acceleration due to gravity 10m s 2, v1 is the velocity of the steel ball at the highest point, and v2 is the velocity of the steel ball at the lowest point.

From the title "The steel ball can pass through the highest point", it can be obtained that the centripetal force of the steel ball at the highest point is equal to the gravitational force, that is, (mv1 2) r=mg, and the velocity of the steel ball at the highest point is 1m s

The process of the steel ball from the highest point to the lowest point can be obtained by the kinetic energy theorem (mv2 2-mv1 2) 2=2mgr

The steel ball at its lowest point is subjected to the tensile force f=mg+(mv22) r

Substituting the value yields the tensile force f=15n

If you switch to a light rod with negligible mass, v3 is the speed of the ball at its lowest point.

The velocity of the ball is zero when it passes through the highest point.

The process of the steel ball from the highest point to the lowest point can be obtained by the kinetic energy theorem (mv3 2) 2=2mgr

The tensile force of the steel ball on the rod at the lowest point f1=(mv3 2) r+mg

Substituting the value to obtain the maximum bearing capacity of this rod should be at least f1=

Hope this answer is useful to you o( o haha

The velocity of the steel ball through the highest point is 1m s

The pulling force experienced by the steel ball when it turns to the lowest point is 15n

The maximum bearing capacity of the rod is at least.

The ball does a flat throwing motion, there are:

Vertical: h=(gt 2) 2

Horizontal: s=vt

Touching the plate at B will be:

s=r easy to get:

t=(2h/g)^(1/2)

v=r×(g/2h)^(1/2)

And because of the collision with b, the disc rotates a total of n times (n 360 degrees) during the movement of the ball in the air, and its angular velocity:

w=360 degrees n t=n(g 2h) (1 2)rad sn, take the positive integer.

Considering that B is in equilibrium when it is about to leave the inclined plane, the spring force it is subjected to is equal to its sliding force: mbgsin = kx

So the displacement of block A is x=d+d (back to equilibrium position); Therefore, there is (MB+MA)gsin = kd

So d=mbgsin k

And the resultant force it is subjected to must be along the inclined plane, because it has acceleration along the inclined plane, so the net force is equal to f'=f-(mbgsinθ+magsinθ)

That is, the tension along the inclined plane minus the spring tension minus the sliding force generated by itself.

In the original state, the spring is compressed, and the spring force is f1 k*x1, and f1 ma*g*sin

Get k*x1 ma*g*sin ,x1 ma*g*sin k

Later state (i.e., when B is about to leave C), the spring is stretched, and the spring force is f2 k*x2, and f2 mb*g*sin, resulting in x2 mb*g*sin k

Later state for a object, the resultant external force of a is f f ma *g*sin f2

i.e. f f ma*g*sin mb*g*sin f (ma mb)*g*sin

The displacement of the sought block a is d x1 x2 (ma*g*sin k) (mb*g*sin k) (ma mb)g*sin k

First calculate the force on the spring at some time according to the initial resting state, take a as the breakthrough point, a is subjected to the spring elastic force, support force, and gravity at this time, and the elastic force f=ma*g*sin, and the elastic force f=k*x, so in the initial case, the spring is compressed x1=ma*g*sin k.

When B just leaves the baffle C, B is in the state of force equilibrium at this time, first carry out force analysis on B, B is supported by force, gravity, spring tension, at this time the tensile force of the spring can be obtained F=MB*G*Sin, at this time you can find the distance of the spring being stretched X2=MB*G*SIN K, and the displacement of A at that time D=X1+X2=(MA+MB)*G*Sin K.

At this time, the force analysis of A can be obtained, and the resultant external force F = f-f bomb-ma*g*sin = f-(ma+mb)*g*sinThe direction is upwards along the inclined plane.

By the way, how did you type it?

This problem is trapped on displacement d. d should be equal to the amount of the previous a compression spring d1 plus the amount of the last elongation of the spring d2

At first, the system is stationary, and it can be obtained.

kd1=mag*sin hence d1=mag*sin k

In the end, when B just leaves C, that is, B is not supported by C, then who balances the force of gravity along the inclined plane, the answer is given by the spring.

kd2=mbg*sin hence d2=mbg*sin k

Q2: D=(MA+MB)G*SIN K

The first question is to analyze the force of A, the component force of F that A is subjected to upward and downward gravity along the inclined plane mag*sin and the downward tensile force of the spring mbg*sin.

A is supported by an inclined plane in the direction of the perpendicular to the inclined plane, and the force n is equal to the component of gravity in this direction mag*cos

I haven't forgotten the concept of external forces, you can add it yourself.

If the spring tension received by block B is f1=mbgsin, then the resultant force experienced by block A is f=f-(f1+magsin)=f-(ma+mb)gsin direction along the inclined plane.

The distance at which the spring is compressed is d1=magsin k The displacement of the spring that is stretched is d2=f1 k=mbgsin k

The displacement of the block d=d1+d2=magsin k +mbgsin k=(ma+mb)gsin k

When you are just about to leave, you think that a is at rest and the resultant external force is k

Block B is about to leave C when it has 0 support. Etiquette Analyze the elasticity, and then analyze a on o.

Displacement with hook top green.

Hello, I am a high school student.

Your question is of the object motion type.

1 First find the acceleration a=f m=30 10=3m s 2, so v=v0-at=18-3*2=12m s2 if you add another force a to the left'=(f+f)/m=6m/s^2v'=v-a't=12-6*1=6m/s^2s=(v^2-v'^2)/2a'=9m

And that's the process, and when you write it, you can add some narrative language and the reason is that it's the formulas, and if you want to ask how the formulas come from, you can read the book, and there's a derivation process in the book.

Glad to help you.