Senior 1 Physics Questions with changes Please ask for Senior 1 Physics questions

The force of the ox lifts the weight, according to the vector synthesis law, it should be 50n is the hypotenuse of the weight of 80n, the resultant force is the right-angled side, then the left and right ropes each contribute 50, gravity takes half of them, forming a triangle, at this time the hypotenuse is 50n, the gravity is 40n (half of 80), so that the rope is 1m long, and each side of the left and right hands is only calculated from the lowest point The height difference in the direction of gravity is the horizontal distance. So the distance between the hands.

2.Similarly, similar to the triangle rule, take half of the gravity force as the right-angled side, and the tensile force is the hypotenuse, which can be calculated as one, 1, 2, the triangle of the root 3, half the length of the rope corresponds to 2, and the horizontal distance corresponds to 1.

3.In addition, you can draw a vector analysis diagram by yourself. This kind of question should be the homework or exam question around the third monthly exam of the first year of high school. Hope that helps.

In the force analysis, only the horizontal direction is considered, and the three objects A, B, and C are only subjected to centripetal force and friction. The condition for an object to be at rest is that the centripetal force is equal to the static frictional force. The centripetal force increases with the increase of rotational speed, and when the centripetal force is greater than the maximum static friction force, the object produces centrifugal motion.

According to the title, if the coefficient of dynamic friction is the same, then the maximum static friction (set to f) is proportional to the mass. So: fa = 2fb = 2fc

According to the law of circular motion, the formula for centripetal force and centripetal angular velocity is:

f = mv r = m r (m is the mass, r is the radius).

a = v²/r = ω²r

1) All points on the table have the same angular velocity, then the centripetal angular velocity is only related to the radius, because 2ra = 2rb = rc, so option a is correct;

2) When at rest, the frictional force is equal to the centripetal force. fa = (2m)ω²r；fb = mω²r；fc = m (2r), so option b is correct.

3) The sliding condition is that the centripetal force is greater than the maximum static friction force. From Fa = 2FB and Fa = 2Fb, it can be seen that the maximum static friction force and centripetal force of A are both twice that of B, so A and B will slide at the same time. Option C is incorrect.

4) From fb = fc and fb = fc, it can be seen that b and c have the same maximum static friction force, but the centripetal force of c is 2 times that of b, so c slides first. Option d is correct.

1min=60s, 1s=30 times, 1min=1800 times. Because there is no feeling of rotation, the blade may rotate 120N(N Z) at a time, when N = 1, 1800 120 = 216000°, 216000° 360° = 600R, so a to the cavity early limb, other options are counted by themselves.

1.One carriage is L in length and has a total of n cars.

l=a2²/2

nl=a6²/2

The solution is n=92The first 4s have a total of n sections passed.

nl=a4²/2 n=4

So the carriages that passed by him in the last 2 seconds had 9-4=5 cars.

3.The time required for the first 8 sections is t, i.e., there is 8l=at 2 and the first equation is solved in conjunction with t=4 2s

The time it takes for the last carriage to pass through the observer is 6-4 2s

Let the total time t, then:

s1=1/2a*3^2=

s2=1/2at^2-1/2a(t-3)^2=1/2a(2t-9)=s2-s1=6

s1:s2=3:7

So: :(7

Solution: a=1m s2, t=15s

Slope length l=1 2at 2=1 2*1*15 2=

Convert to find the area of a right triangle, the base is the time, the height is the velocity, and the total length is the area of the triangle.