Distress! High school physics questions, urgent!! High school physics questions, ask for detailed an

The gravitational force of an object is divided into f2 along the inclined plane f1 and perpendicular inclined plane, according to s = 10m and h = 6m.

Suppose the object is m and the angle is a

Sin A=6 10 Cos A=8 10F1 mg=Sin A: F1=6m

f2 mg = cos a gives f2 = 8m

The frictional force is f=un=1 8 * f2=m

The resultant force in the inclined direction is f=f1-f=ma, so the acceleration is a=5 m s2

From the displacement formula s=1 2 * at2 to get t=2 s velocity formula vt=at vt=10 m s if you think the steps are detailed, give points!

In addition to the kinematic approach upstairs.

Here's another easy way to do it.

Using the kinetic energy theorem, the work done by the resultant force is equal to the change in kinetic energy mgh-umgcosas=1 2mv squared, and the substituted value v=10m s

Then, according to the momentum theorem, the impulse of the combined external force is equal to the change of momentum, mgsina-mgcosau)t=mv-0, and substitute the value t=2s, and the two methods are used according to their own suitable methods.

It's a bit troublesome, you can do the math yourself, I'll tell you the idea, find the time, decompose the gravity into a component along the inclined plane, f1=mgsina (a is the angle formed by the inclined plane) and f2=mgcosa, then a=(mgsina-mgcosau) m, and then find t according to h=1 2at2, and then find the end velocity according to v end 2-v beginning 2=2ah (initial velocity is 0) If you don't understand it, ask me again, and I will help you answer it carefully.

The law of alternating voltage can be written as u=311sin314t v 220*(root number 2)*sin(100* t) v

It is easy to see that the frequency of this alternating current is f 50 Hz and the period is t seconds.

So in 1 second, there are 50 cycles.

Calculate how much time the tube is glowing in 1 cycle.

In a cycle, there are two equal periods of time when the instantaneous value of the voltage is greater than or equal to 220V to make the tube glow.

The luminescence time in half a cycle is now discussed.

By U220*(root number2)*sin(100* t) V, so U 220 V

Obtain sin(100* t) (root number 2) 2

In half a cycle (i.e., 100* t at 0 to ), 100* t 4, or 100* t 3 4

The corresponding moment is 1 400 seconds for t1 and 3 400 seconds for t2.

That is, in half a cycle, the time that can emit light is t2 t1 (3 400) (1 400) 1 200 seconds.

The luminous time in a cycle is 2*(t2 t1) 1 100 seconds.

It can be seen that in a period of 1 second, the luminescence time has a total of t total 50*(1 100) seconds.

According to the topic, it can be seen that when the voltage is greater than or equal to 220V, that is, 311Sin314TV is greater than or equal to 220, it is satisfied that in a cycle, that is, within 360 degrees, the absolute value of Sin314T is greater than or equal to the root number of two, that is, 314T is greater than 45 degrees less than 135 degrees or greater than 225 degrees less than 315 degrees, and the neon tube emits light in half the cycle time, so it is.

Draw a picture to figure out the condition of the exercise;

1) The velocity of the car is constant v=20m s, which means that the force of the car is always balanced in the tangent direction of the circumference of the nucleus;

At the lowest point A, the power of the motorcycle engine is P=12KW, P=FV, V=20, and the thrust generated by the engine can be solved.

The car sells at a loss, and the force is balanced by the force, and the resistance f = thrust f, n*, which can be solved n;

2) n = centripetal force f + mg, solvable f, mv 2 r=f, solvable r;

3) point B pressure nb + mg = f direction = n - mg, nb = n-2mg;

Resistance fb= = engine thrust.

p=fvunderstand?

1) The car is at the lowest point a, and the positive force n and the resultant force of gravity mg provide the centripetal force. fn=n-mg。It would have been possible to use velocity v, and to find fn is to find the positive pressure, but the answer is that r is unknown.

At point A, there is also a data prompt, p=12kw, and knowing that he is moving in a uniform circular motion, then at point A, the traction force f in the horizontal direction is balanced with the resistance f. According to p=fv, find f=600n. The resistance f and f are equal in magnitude and opposite in the opposite direction.

Then the positive pressure of the erection sail n=10f=10f=6000n.

2) According to the analysis of 1), FN=N-MG=MV2R. Easy to find r.

3) 1) If you can do it, 3) It's a point giveaway.

1.The lowest imaginary silver point: P1 = F1V to get F1 = 600NF1 = to get N1 = 6000N

2.N1-mg=mV2 R=4000Nr=20m3The highest point mg+n2=mv2 r

Get n2=2000n f2=then p2=f2v=4000w

This question is calculated in 2 situations.

Let the distance between AB be x, then the horizontal distance between the aircraft and A when the bomb is dropped for the first time is:

The distance flown by the aircraft after x1 = v0 root (2h1 g) = 300 * 20 = 6000m5s is x0 = 300 * 5 m = 1500m When the aircraft drops the bomb for the second time, the bomb travels in the air for a time of :

t=open root 2(h1-h2) g=16s

The horizontal distance of the aircraft from b at the time of the second bomb drop was;

x2=300*16=4800m

then the total horizontal distance from b when the plane first started dropping bombs:

x1+x=x0+x2

x=x0+x2-x1=1500+4800-6000=300m

For the wooden block, the bullet does positive work on it, and the actual distance of the work is d, so the energy increment of the wooden block is fd, and for the bullet, the wooden block does negative work on it, and the total distance of the bullet is l+d, in this displacement, the wooden block always does work on it, so the energy change of the bullet is -f(d+l).

So the energy change of the system is fd-f(d+l)=-fl to select b c

The bullet does positive work on the wooden block, the work distance is d, and the energy increment of the wooden block is fd.

The wooden block does negative work on the bullet, and the total distance of the bullet is l+d.

During this displacement, the block always does work on it, and the energy of the bullet changes to -f(d+l) and the energy of the system changes to fd-f(d+l)=-fl to select b c

Let the acceleration phase be a and the deceleration phase b be b

vt^2-v0^2=2as

vt^2=2ax1

vt^2=2bx2

2ax1=2bx2

a:b=2:1

Since I don't know, I'll give you a lesson:

vt^2-v0^2=2as

In this problem, v0 = 0 because the initial velocity = 0;

s=(at^2)/2

s=(at)^2/2a "The numerator denominator pain is multiplied by the acceleration a, because at=vt (terminal velocity).

So s=vt 2 2a

i.e. vt 2 = 2as

The object has a stationary start to do a uniform acceleration linear motion, the displacement that occurs is x1, let the acceleration be a, and the final velocity is b

then b 2=2a*x1

Then do a uniform deceleration linear motion and know that the displacement that stops is x2, let the acceleration be c, and the initial velocity is d=b

Then b 2=2c*x2=2c*2x1=4c*x1, so a c=2:1

2 to 1 uses the average velocity as the average velocity, and the same displacement ratio is equal to the time ratio and is equal to the inverse proportion of acceleration.

According to the image of the sock pure excavation, the acceleration: the nuclear a=1m s, the displacement l=8m, after the force analysis, Newton's second law: f- mg=ma, can be used to find the work of f=6n tensile force: w=fl=48j

Instantaneous power p=fv=6 4w=24w

You first calculate the only thing about the object in 0 to 4 seconds, and at first glance, it is a uniform acceleration, Zheng Zi, and multiplying half of the final velocity by time is the displacement, multiplying the displacement by UMG, plus the kinetic energy of the dust mode at the fourth second, is the work done by the pulling force, and you can see clearly, it is horizontal, and the initial velocity is 0, and the velocity at the end of the fourth second is 4 meters per second, multiplied by the magnitude of the pulling force, which is the instantaneous power at the end of 4 seconds, and the pulling force minus UMG is equal to the mass of the object multiplied by the acceleration, and the acceleration is the slope.

First find the acceleration a 1m s 2, then find the combined external force f 2n, and then find the friction force f = Xiangzi slip 4n, so that you can find the tensile force f f 6n, the displacement x of the object, so that the work done by the tensile force is w1 6 8j 48j, and the instantaneous power at the end of 4s can be p = fv 6 4w Qi noisy 24W

V=AT---A=1M Bucket Dismantling S 2 S=1 2AT2---S=8M F- UMG=MA---F=6N WPull Good Pin Chain=FS=48J PInstant=FV---Yousun P=24W

Similar to this kind of problem,I guess you're too lazy to do your homework.,These questions just need to draw an image of speed and time.,It's very simple.。 There are several images of this kind of topic, and I personally think it is acceleration and time, and speed and speed.

75 root number 3 + 25 root number 3 = 100 * root number 3 (about 173n), the force of the former left rope, the latter right, mainly consider the horizontal force when the left rope is flat and horizontal.