High School Physics Questions, Physics Emperor Advance

The first question is a right-angled triangle with an angle of 60 degrees, where the 60 angle is you. When the plane flies, it takes the same time as the sound reaches your ears, s1 t=v1 s2 t=v2

s1 is the side corresponding to the 60-degree angle, which is the distance the plane has flew. S2 is the side corresponding to 30 degrees, which is the distance from the plane to the ground, that is, the distance at which the sound flies.

According to the geometry, we can know that s1:s2 = root number 3

So the speed of the aircraft should be 3 times the root number of the speed of sound.

The second question uses water as a frame of reference, and such an equation can be listed.

v water * [(v boat - v water) + x] (v boat + v water) = x (the formula behind v water is equivalent to finding the time it takes from the beginning of the fall to the time it catches the object. The time of the v-water* should be the distance traveled by the lifebuoy during this time).

However, you didn't give the relationship between the V boat and the V water in your title, so I suspect you are left behind. If you know, you can find out the V water.

v fly v tone = tan30o

V fly 3 3v tone.

v water = x 2 t

Let the time of the back-chase be t1, the back-up time be s, the speed of the ship in still water is v, and the distance traveled by the ship in time t is s1.

The speed of the counter-current vessel is: v-v water; The speed of the boat is V+V water.

Then: the distance of the lifebuoy drift: x=v water*(t+t1) 1 The distance of the ship's pursuit: s=(v+v water) * t1

The distance traveled by the ship at time t: s1 = (v-v water) * t

So: x=s-s1

v+v water)*t1-(v+v water)*t 2

Joint 1 formula, solution:

vt1=vt

So: t1=t

Then: v water = x 2t

Draw a triangle in the first question, answer 3 3. Question 2 v water = x 2t

The workpiece first does a uniform acceleration linear motion with zero initial velocity on A, and then does a uniform motion after the speed reaches 2m s.

It is not possible to accelerate all the time on A, because if you keep accelerating, its average speed cannot be greater than 1 meter second. And its actual average speed reached 10 meters 6 seconds = 5 3 (meters and seconds).

Let the acceleration time be t, v 2*t+v(6-t)=10m

t=2sa=v/t=1m/s＾2

a g = the frictional force experienced by the object on the inclined plane is ff = mgcos 30° and the component of gravity is mgsin30° ff

a′=（mgsin30°-ff）/m

The object moves in a uniform accelerated linear motion with an initial velocity of 2 m s on an inclined plane, and the time from b to c is t

10m=2t′+1/2*a′*t′＾2

It can be calculated that t is not good to enter, so calculate it yourself)

The time from A to C is 6 seconds + t

First of all, it can be judged that the object must be uniformly accelerated in the AB section and then move at a constant speed of v=2m s (which can be proved by the arctic hair).

In the ab section, let the uniform acceleration time be t and the acceleration is a, then there is:

a*t=2m/s

The solution is a=1m s2, so the dynamic friction factor between the object and the conveyor belt is in the BC section, and the force analysis knows that the acceleration of the object a = GSIN30° so the initial velocity of the object is 2m s, and the acceleration is a uniform acceleration linear motion, there is the following equation:

10m=2m/s*t′+

Solution t = time for the workpiece to arrive at c from A = 6s + t

Solution: First of all, the water flowing out of 1s is taken as the research object, the flow rate of 1s is 10 cubic meters, the mass is 10 4kg, and the moment when these waters fall below 20m, before touching the turbine, the gravitational potential energy is all converted into kinetic energy, and there is mgh = 1 2mv

m about to go, 10 20 = 1 2 v, v = 10 m sWhen it hits a turbine, the turbine works a negative effort on the water, and the flow velocity becomes half of the original.

v=5m s, the kinetic energy reduction is the amount of electricity. 1 2 10 4 10 -1 2 10 4 5 =, because it is a flow rate of 1s, so the power is.

Gravitational potential energy is converted into kinetic energy, and the amount of kinetic energy reduced is the amount of electricity generated!

A little hint, mgh, no stranger!

Is there a mistake that "the flow velocity becomes 5m to the third power s"?

This problem can be analogized to the alternator model, where the maximum electromotive force e nbs , because the rotation starts from the "neutral plane" and therefore produces a "sinusoidal" alternating current, so the RMS is equal to the maximum value divided by the root number two. The effective value for heat and the average value for electric charge. The specific solution is shown in the figure below.

Analysis: If the velocity of the object at the first point of e is ve, then ve has a minimum limit. Let this minimum be v0

That is, when the velocity of the object at e is v0, the orbit just has no elastic force on the object, and gravity completely provides the centripetal force.

Mg m* v0 2 r is obtained

v0 root number (g r).

In order for an object to be thrown flat from point E and fall on the BC orbit, then VE has a maximum value limit, let this maximum value be vm

It is obtained by the law of flat throwing.

bc vm * t, t is the time of motion of the object in the air.

2r＝g * t^2 / 2

From the above two formulas, vm bc * root number [ g 4r ) 4r * root number [ g 4r ) 2 * root number (g r).

Visible, the root number (g r) ve 2 * the root number (g r).

After the object is released from a certain place in AB, the process of going up to point E for the first time is obtained by the kinetic energy theorem.

mgh－μ mg * bc＝m * ve^2 / 2

i.e. H (VE2 2 G * BC) (2G).

Substitute bc 4r , and root number (g r) ve 2 * root number (g r) into the above equation.

r ≦ h ≦ r

Just past point E.

Mg(H-2R)- Mg4R=1 2mV2 acts as a centripetal force in gravity at point E.

mg=mv2r to solve the minimum h

The maximum value is that the small slider flies out from E to B

2r=1/2gt^2

4r=vt solves v

Then replace the top equation to solve the maximum h

The corner mark is not easy to play, so they all use the same h,v, please Haihan. Beg.

Falls on the BC segment.

So the minimum velocity of the small slider needs to be exactly to point E, where gravity provides the centripetal force, and the gravitational potential energy, kinetic energy, and the energy consumed by friction on BC are listed to calculate a height of a according to the conservation of energy.

The maximum velocity is exactly from the flat toss of e to point a, and the velocity of point e is calculated from the flat toss, and a is calculated according to the conservation of energy

I think it should be ...

The first one asked.

As the title suggests, particles are only affected by the force of the electric field in the electric field.

Electric field strength e=u d. Therefore, the electric field force is f=e*q=uq d

From the kinetic energy theorem, 1 2mv = f*d

So, v=(2uq m) (1 2) ps: i.e. open squared.

The second asked. It is derived from the impulse theorem.

f*t=mv

Therefore, t=mv f=d*(2uqm) (1 2) (uq) results are more difficult to type, students can write their own results. That's the rough process.

The strength of the electric field between the plates e=u d, the force on the charged particles f=e q,1) the electric field does work on the particles w=f d=(u d) q d=u q, and all the work done is converted into kinetic energy (m (v 2)) 2=u q, then the velocity v=( ( (2 u q)) m

2) The force in the uniform electric field is the same, all are E q, so the particles do a uniform acceleration motion, the average velocity is half of the final velocity, that is, half of the velocity when leaving plate B, so the time t = d (v 2) = (2 d m) ( ( (2 u q) ).

1.Formula: uq = (1 2) mv square The energy gained by the object in the electric field is uq and all is converted into kinetic energy 1 2) mv square, and v = under the root number (2uq m) is obtained

2.Formula: s=(v0+vt) t 2 (uniform acceleration linear motion, average velocity multiplied by time equals displacement).

That is, d=v t 2, and t=root2*(dm) is obtained under the root number (uqm).

The electric field strength is e=u d, the force of the particle is f=q*e, then the particle acceleration is a=f m=q*e m, according to 2as=v 2, the velocity of the charged particle from the b plate can be obtained, t=v a, the time of the charged particle moving in the electric field can be obtained.

s is the distance traveled by the particles, which in the case is d

electric field e=u d;

Force f=e*q;

acceleration a=f m;

d=, t can be found;

v=at

Before the yard buried B leaves the ground, A is only subjected to 3 forces, f g and the spring tension f1 acceleration is unchanged, indicating that the empty mode is transported to the net force without edge, that is, f-f1-g is unchanged, and g is unchanged in the process of A rising, ignore.

F1=kx, F1 becomes larger with x uniform bucket beam, so f also becomes larger with x. Choose A

Assuming that the masses of A and B are both m, and the original length of the spring is L, then at the beginning f=ma, the elastic force of the spring is the gravitational force mg of a; After that, a should maintain the acceleration calendar next to a, let its rise distance x (the spring is still in compression), then there is f=ma+kx; When the limb mountain spring returns to its original length, then f=ma+kx, so a is selected

Acceleration: a=f (m+m);

Because the quality of the shell does not affect the result of weighing (look at the spring scale, the shell is set from top to bottom, the weight of the shell is directly borne by the hand, and the mass added to the hook will be displayed), so the reading of the spring scale is only the gravity of the weight).

Then: g'=ma=fm/(m+m)

g'This is the reading of the spring scale.

f-(m+m)g to find the net external force fo of the system, fo (m+m) to find the acceleration of the system a, a*m is the resultant external force fa of the hanging weight, and then use fa+mg is the indication of the spring scale, which is fm (m+m).

First find the acceleration a=f m=f (m+m) and then the reading =f+mf (m m).