It is known that the velocity of the object passing through point B is 3 times that of the velocity passing through point C 4. The distance between BC is 7m ① What's the velocity of the object passing through C? ② The distance between a and C? （g=10m/s2）

It is known that the velocity of the object passing through point B is 3 times that of the velocity passing through point C 4. The distance between BC is 7m ① What's the velocity of the object passing through C? ② The distance between a and C? （g=10m/s2）

Let the velocity of point C be v. according to the formula of velocity displacement relation, for AB process, there are: (3
4v)2＝2g(x−7)
For AC process, V2 = 2gx
The simultaneous solution is: v = 8
5m/s
x=16m
Answer: ① the velocity of the object passing through point C is 8
5m/s；
② The distance between a and C is 16m

About time and time, the following statement is correct () A. When the object is in 5S, it refers to the time when the object is at the end of 5S B. When the object is in 5S, it means that the object is at the beginning of 5S, which is the time C. The object within 5S refers to the 1s time from the end of 4S to the end of 5S D. The object in the 5S is the time from the end of 4S to the beginning of 5S

A. When the object is in 5S, it means the time when the object is at the end of 5S;
C. The object within 5S means that the object is in the 5S time from zero time to the end of 5S, so C error;
D. The object in 5S refers to the 1s time from the end of 4S to the end of 5S, so D is wrong;
Therefore, a

A problem about friction force in high school physics When a piece of wood is placed on a horizontal surface, the external force F1 = 10N (right) and F2 = 2n (left) is applied in the horizontal direction, and the wood block is in a static state. If the external force F1 is removed, the friction force on the wood block is () n, and the direction is () Why not 2n? The static friction force changes from 8N to 2n, Why the answer is 10N

Believe in yourself, the answer is wrong

A problem of power in senior one physics In order to speed up the railway, many technical problems have to be solved. Usually, the resistance of the train is proportional to the square of the speed, that is, F1 = kV ^ 2. If the train wants to run fast, it must be pulled by a high-power locomotive. Try to calculate the wallpaper of locomotive power when the train runs at a constant speed of 120km / h and 40km / h respectively

P = F1 * V F1 = kV ^ 2 deduces that P = k * V 3 (that is, the power ratio equals to the third power of speed)! While the speed ratio is 3:1, the power ratio is 27:1

If the aircraft flies at a constant speed V and the engine power is p, then when the aircraft flies at a constant speed of NV, the engine power is______ .

From the problem, the expression of resistance f = Kv2, K is the proportional coefficient
When the aircraft flies at a constant speed V, the power of the engine is p, then p = FV, while the traction force F = f = Kv2, then p = Kv3
When the aircraft flies at a constant speed of NV, the engine power is: P '= f ′ NV = f ′ NV = K (NV) 2nv = n3kv3 = n3p
So the answer is: n3p

A body is in equilibrium under the action of several forces. If a constant force is applied to physics, the object may do () a static or uniform linear motion B uniform speed linear motion C curve motion D and variable speed curve motion the answer is BCD why is not a constant force not a equilibrium force? Should the motion state not be changed after application

A balance force is relative to more than two forces. A constant force is a constant force, that is, it does not change in size or direction
When a body is in equilibrium, it may be at rest or moving in a straight line at a constant speed. If I add a constant force to a body which is still subjected to gravity and support, it will still be in a static state, It is impossible to keep a uniform linear motion. It can either accelerate or decelerate uniformly, or move in a curve

When a vehicle with a mass of 4T moves in a straight line from standstill with rated power P0 = 100kW on a horizontal road, and the maximum speed of the vehicle can reach 50M / s, what is the resistance of the vehicle in the process of motion? What is the acceleration when the vehicle speed is 20 m / s? If the vehicle is started at a constant acceleration of 0.5 m / S ^ 2, how long can the acceleration be maintained?

1. Because the rated power P = FV, in order to make V maximum, f is the minimum. In order to enable the advance, Fmin must be equal to the resistance, so f = f = 100000 ﹤ 50 = 2000n2. F1 = 100000 ﹤ 20 = 5000n, f  f  f = 5000-2000 = 3000na = f  closing ﹣ M = 3000 ﹤ 4000 = 0. 75m / S23

Block a is stacked on Block B, and block B is stacked on a horizontal table top. When a horizontal force F is used to push block a, a and B still remain static. Suppose that the friction force of B against a is F1, and that of table top against B is F2 A．f1=F,f2=0 B．f1=O,f2=F C,f1=F/2,f2=F/2 D．f1=F,f2=F/2

The simplest analysis method: because a and B are relatively static, and AB and the table top are relatively static, so when you study a, you take B and the table top as a whole. At this time, because a and B are relatively static, so the horizontal thrust of a and the friction force F1 of F and B against a are equal, and the direction is opposite

As shown in the figure, the four identical springs are in horizontal position, and their right ends are subject to the pulling force of F, while the left ends are different ① The left end of the middle spring is fixed on the wall; ② The left end of the middle spring is also affected by the tensile force of F; ③ The left end of the middle spring is tied with a small block, which slides on a smooth table top; ④ The left end of the middle spring is tied with a small block, which slides on the friction table If the mass of the spring is considered to be zero and the elongation of the four springs is represented by L1, L2, L3 and L4, then () A. l2＞l1 B. l4＞l3 C. l1＞l3 D. l2=l4

The tension of the spring is directly proportional to its elongation (the difference between the actual length of the spring and the original length); and the indication of the dynamometer represents the force acting on the hook of the dynamometer; since the magnitude of the force on the right end is f, it has nothing to do with the force on the left end; therefore, the elongation of the four springs is the same, that is, L1 = L2 = L3 = L4; therefore, ABC is wrong and D is correct;
Therefore, D

Physics of senior one -- balance of objects under the action of common point force If a patient's ESR is measured, it is helpful for the doctor to make a judgment on the patient's condition. If the blood is a suspension composed of red blood cells and plasma, and the suspension is placed in a vertical tube, the red blood cells will sink uniformly in the tube, and its sinking rate is called ESR. The value of erythrocyte sedimentation rate (ESR) of a person is about 10 mm / h. If the red blood cell is approximately regarded as a small cell with radius of R, The results show that the viscosity resistance of the red blood cell is f = 6 π η RV. At room temperature, η = 0.0018 PA · s. The density of plasma ρ. = 0.001 kg / m3 and the density of red blood cell ρ = 0.0013 kg / m3. The radius of red blood cell was estimated by the above data To have a detailed calculation process or calculation ideas!

Because the red blood cells sink at a constant speed, the gravity of red blood cells is balanced with buoyancy and resistance
4πR^3(ρ-ρ')g/3=6πηRv
R ^ 2 (ρ - ρ ') g = 4.5 η V is obtained
R={4.5ηv/[(ρ-ρ')g]}^(1/2)
v = 10mm/h =2.778*10^(-6) m/s g=9.8 m/s^2
Substituting the known data such as ρ - ρ '= 0.0013-0.001 = 0.0003 kg / m ^ 3, η = 0.0018 PA · s and other known data into the above formula, r = {4.5 η V / [(ρ - ρ') g]} ^ (1 / 2)
={4.5*0.0018*2.778*10^(-6)/[0.0003*9.8]}^(1/2) = 0.00277 m = 3 mm ?
I'm afraid the data is wrong
"ρ. = 0.001 kg / m3, red blood cell density ρ = 0.0013 kg / m3", will the density be so small?