Solve a high school physics problem. The masters are coming soon, and they are grateful

The speed of the conveyor belt v1 = wr =

From B to the relative stationary process between the object and the conveyor belt, it takes time t1

The acceleration of the object doing a uniform deceleration motion a = - ug = -4m s Let the velocity of the object at point b vb

v1^2 - vb^2 = 2al1 --1)v1 = vb +at1 --2)

The object moves with the conveyor belt at a uniform speed t2 = (l-l1) v1 --3)t1+t2 = t --4).

Joint (1) (4).

Solve vb = process from ramp to b.

mgh +w = mvb^2

Solve w but there seems to be a problem with the data you gave, should h be 15m?

There is a problem with the data you wrote, assuming there is no friction, the velocity of a-b v1 = root number 2gh = root number 30

The speed of the conveyor belt v=wr=

Start from point B to do a uniform deceleration motion, before reaching point C, and the conveyor belt is relatively stationary, indicating V1>V, and the above assumption can see V1, so say see if there is a problem with the data of the question, and then do it for you after the data is changed.

The speed of the horizontal conveyor belt is v0=r =3m s

From Newton's second law, mg=ma

again v0=vb-at1

l-l1=v0t2

t1+t2=t

From the kinetic energy theorem, we get the simultaneous solution wf=

First of all, when an object slides down a slope, you can ignore its velocity by using the conveyor belt as a reference.

The work done by gravity plus the work done by drag is equal to the amount of change in kinetic energy.

So. mgh+wf+umg=0

wf=-15-20=35

Conveyor belt velocity v= from a---b there is conservation of mechanical energy: mgh = 1 2mv2 + work done by friction Then you can solve the equation!

Let's start with the first scenario, "when his speed is 5 km h, he feels the wind blowing from due north".

At this time, it feels like it is blowing from due north, indicating that the wind and the speed of the car are the same in the direction of due east. The speed at which the wind blows due east is 5 km h

Looking at the acceleration process, "when the speed of the bicycle increased to 15 km h, he felt the wind blowing from 45 degrees east-north".

In the first case, the component of the wind due east is 5 km h. At this time, the speed of the car is 15 km h, so the speed of the car in the eastward direction is 10 km h relative to the wind at this time (that is, people feel the wind blowing from the east) However, the title says that the wind is blowing from 45 degrees east of north, so the wind also has a weight of blowing from the north. Since the angle is 45, the wind speed blowing from the north (due south) is also 10, so the speed of the wind is under the root number (5*5+10*10) and the direction is south-west a degree tana=

The wind direction is a combination of 10 km h north wind and 5 km h west wind.

The wind direction is vector, the wind direction is 5km h to the east and 10km h to the south.

Solution: 1) The dispersion slope is 4m long and pushes to the top in 10 seconds, so the speed is 4 10=

2) The gravitational potential energy added by the useful work for the wooden box.

3) The force of gravity is 400 4* and the thrust is 75n

Thus the equilibrium equation of the force in the direction of Li You along the inclined plane is.

f+50=75

So f=25n

So the friction is 25 Newtons.

4) The size of the useless work is 25*4=100j

The size of the useful work is 200j

So the efficiency is 200 (100+200)=

According to the conditions, let the elliptic equation be (x a) 2+(y b) 2=1 (a b). The focus of the general order is the right focal point ( ( ( a 2-b 2),0 ) and the slope of the straight line is k, so the equation for the straight line is y=k[x-( a 2-b 2)].

Simultaneous two equations are simplified by subtracting y:

ka) 2+b 2]x 2-2(ka) 2 (a 2-b 2)x+k 2a 4-k 2a 2b 2-a 2b 2=0, which is a quadratic equation about x. The two solutions x1 and x2 are the abscissa of p and q, respectively. It is calculated that δ=(k 2+1)a 2b 4 0 is constant.

From Vieta's theorem: x1+x2=2(ka) 2 (a 2-b 2) [(ka) 2+b 2],x1x2=(k 2a 4-k 2a 2b 2-a 2b 2) [(ka) 2+b 2].

From the inscription, the vector op=(x1,y1), the vector oq=(x2,y2), and x1x2+y1y2=0, i.e., (k 2+1)x1x2-k 2 (a 2-b 2)(x1+x2)+k 2(a 2-b 2)=0. Bring the vieta theorem x1x2 and x1+x2 into the above equation and simplify it:

a 4-a 2b 2-b 4)k 2-a 2b 2 = 0, i.e. k 2 = (ab) 2 (a 4-b 4-a 2b 2).

Because k 2 0, there must be a 4-b 4-a 2b 2 0, i.e. 1-(b a) 2-(b a) 4 0. Solution: 0 (b a) 2 ( 5-1) 2.

Eccentricity e=c a= (a2-b 2) a= [1-(b a) 2], so the range of eccentricity is ((5-1) 2,1).

You're talking about math.

2 root number 2 times v 2 g

Brother, you have too many topics.

(1) If the angle is 45 degrees, then at the end of one second, horizontal velocity = vertical velocity = gt = 10m s2 vertical velocity at the end of the second = g2 = 20m s; The horizontal velocity is unchanged = 10 m s, so the velocity size at the end of two seconds is 10v5 m s

2) The initial velocity of the fall in the vertical direction is 0, which can be understood as the opening of the square = 15s from the height of 1125m free fall time t = 2 * 1125 10

Horizontal distance = 50v3 * 15 = 750 v3 m5 vertical velocity at the end of seconds = g * 5 = 50m s; Horizontal velocity = 50v3 m s, the velocity magnitude is 100m s, the direction is tana = 50, 50v3 = 1 v3, that is, it is 30 degrees with the horizontal.

4) 45 degrees, 20m

5） s ;45 degrees horizontally.

Anyway- I remember that there is a formula, what is tangent than what can be set of formulas - - formula I forgot - -

With a mass of two kilograms, the original length of the spring was 5 cm to hang it vertically, and when the object was stationary, the spring length was 15 cm

The elastic coefficient k=mg (15-5)=2n cm is obtained for uniform linear motion.

The elastic force is found to be umg=

Spring deformation = 4 2 = 2 cm

Because of pulling the object, the spring length is 5+2=7cm

According to the title: 2 10 = k(15-5) 100

u*2*10=k(x-5)/100

When an object moves in a straight line at a uniform speed. The length of the spring is x=

Force analysis, column equations. 10k=mg in the vertical direction, k(x-5) = friction in the horizontal direction, the friction force is the sliding friction force, the dynamic friction factor is multiplied by the gravity force = 10k, and the equation is solved.

The gravitational potential energy is conserved, and the water surface is 0 potential energy surface, then (2+4)*10*(-20)=0+4*10*h is h=-30 meters, so the depth of the iron block is 30 meters at this time.

Because it is a horizontal span, the height of the person's center of gravity should be taken as the starting point (the person's center of gravity is equal to half of the height, so the height of the center of gravity is meters).

The height of the jump is h=

Symbol: "Cheng sign: *".

Vertical upward throwing motion solution:

Solution: Known conditions: --g = 10 m per square second --- vt = 0 m s---

From vt square - vo square = 2as gets: vo square = vt square - 2as (I can't type the "root number", so I didn't square the "VO squared", you have to open your own square when you answer.) ）

A with -G generation|S is replaced by H) obtains: VO squared = vt squared 2gh = 02 2 10m s2*

Remember to open the square, vo square I didn't open the square. Answer = root number 18

Rounding and the like answer you took the calculator and pressed it, which is about the same.

Kinetic energy definite comprehension:

Solution: Let the ground be 0 potential energy point.

1 2m "VO squared" = mgh Both sides are eliminated at the same time to obtain: 1 2 "VO squared" = gh

From 1 2 "Vo squared" = gh to get Vo square = gh divided by 1 2 = gh*2 = 10 meters per square second*

Remember to prescribe a prescription too!