An object slides from the top of the slope at a constant speed, and the sliding acceleration is 2m / S2. If the object slides to the bottom, the sliding distance is 3% of the length of the slope 4. What is the slope length?

Let the sliding time of the object be t and the length of the slope be L
L＝1
2at2… (1)
One
4L＝1
2a(t−2)2… II.
By
②=4＝t2
(t−2)2
So the time for the object to slide is: T = 4S
Slope length: l = 1
2at2＝1
2×2×42m＝16m
Answer: the slope is 16m long

On the calculation of physics in senior one 1. Some aircraft carriers are equipped with ejection systems to help take off. It is known that the maximum acceleration that a certain fighter may generate when accelerating on the runway is 5m / S? And the take-off speed is 50M / s. if the aircraft is required to taxi for 100m before taking off, what initial speed must the aircraft have with the ejection system? Assuming that there is no ejection system, in order to make the aircraft take off from the aircraft components, the maximum acceleration that can be generated is 5m / S? And the takeoff speed is 50M / s, How long is the taxiway at least? 2. Car a and car B move in a straight line in the same direction through an intersection at the same time. A moves at a uniform speed of 4m / s, while B starts from stationary at the intersection and accelerates uniformly in a straight line with an acceleration of 2m / S? 2 (1) How long does it take for them to meet for the first time? How far is it from the intersection when they meet? (2) When is the maximum distance between two objects before meeting? What is the maximum distance?

a=5m/s2,vt＝50m/s,s=160m．
Vt2-v02 = 2As
v0=vt2－2as
2 according to the formula VT 2 = 2As'
s'=vt2/2a
3. Since the aircraft carrier sails at a constant speed along the take-off direction of the aircraft, if the speed is V1 and the displacement in the process of aircraft starting is x1, then: X1 = V1 · t
In the starting process, the aircraft does the uniform acceleration movement with the initial speed of V1, and the displacement is set as x2
According to the formula, X2 = v1t + 1 / 2at2
The time of motion is t = (v2-v1) / A
According to the displacement relation, l = x2-x1 = 160m, i.e,
Substituting the data, V1 = 10m / s
It can be seen that the speed of the aircraft carrier along the take-off direction is at least 10m / s: (1) for the first time, the displacement of the aircraft carrier is the same
B's displacement: according to the formula x = v0t + 1 / 2A (T) ^ 2, the displacement is T ^ 2, and a formula is obtained: 4T = T ^ 2
T = 4
T = 4S v = 4m / s find x
x=vt=16m.
(3) Before the speed of B catches up with that of a, the distance increases. Therefore, when the speed of B is 4m / s, the distance is the largest
According to v = V0 + at, 4 = 0 + at, at = 4, and because a = 2m / S ^ 2, t = 2S
(4) T = 2S, v = 4m / s, so x = 8m

In order to measure the height of clouds, an explosion is carried out at a distance of 3.0KM from the observer on the horizontal ground. The difference between the explosion sound directly transmitted from the air and the explosion sound reflected by the cloud layer is △ t = 6.0s. Try to estimate the height of the lower surface of the cloud layer. The velocity of sound in the air is known as v = 1 3km/s．

As shown in the figure, a is the explosion site, O is the reflection point, s is the observer's location, and H is the height of the cloud's lower surface (1)
T2 is the time taken for the explosion sound to arrive at s through the cloud reflection. Since the incident angle is equal to the reflection angle, there is
Two
(D
2)2+h2=vt2 … II.
It is known that t2-t1 = △ t 3.
We can get the formula ①, ② and ③
H=1
Two
(v△t)2+2dv△t
Substituting the value, H = 1
2 x
(1
3×103×6)2+2×3×103×6=2.0×103m
Therefore, the height of the lower surface of the cloud layer is 2.0 × 103m

A car drives at a speed of 90km / h for 2min on a straight expressway, and then it takes another 1min to advance 2km. The speed is increased to 120km / h, and the average speed of the car in the three minutes is calculated

Total distance of car s = 90
6 × 2 × 60 + 2000 = 5000m, average velocity v = s
T＝5000
3×60=27.8m/s
A: the average speed of the car in these three minutes is 27.8m/s

A body is stationary at the bottom of the smooth slope. From a certain moment on, a constant force along the slope acts on the object, making the object slide upward along the slope. After a period of time, the constant force is suddenly removed, and after the same time, the object returns to the bottom of the slope with a kinetic energy of 120J (1) What work does this constant force do to the object? (2) What is the kinetic energy of an object when the constant force is suddenly removed?

Let the velocity of the object be V1 when the constant force is removed and V2 (1) when the body returns to the bottom of the slope, we use the kinetic energy theorem WF = 12mv22 = 120J (2) to take the process of sliding from the bottom to withdrawing the constant force F along the upward direction of the slope as follows: S = 0 + v12t the process from removing f to returning to the bottom is: − s =

A driver drives at a constant speed of 30 m / s and finds that the vehicle stops at 70 m ahead. If the driver sees that the vehicle stops in front of him and the reaction time is 0.5 s, will the vehicle have safety problems? It is known that the maximum acceleration of the brake is 7.5m/s

During the reaction time, the vehicle moves in a straight line at a constant speed, and the displacement is 0
s1=vt1=30×0.5m=15m
The car brake makes even deceleration movement until the displacement of the stop is S2 = 0 − v2
2a＝0−302
−2×7.5m＝60m
So the total displacement of the vehicle is s = S1 + S2 = 75m > 70m
Therefore, the driver must encounter the car in front, there is a safety problem
A: the driver will encounter the car in front of him, and there is a safety problem

At the initial velocity of V0 = 4m / s, the ball glides upward from the bottom of the inclined plane with an angle of 30 degrees. The maximum distance of upward sliding is 1 m and the mass of the ball is 2 kg. What is the kinetic energy of the ball when it slides back to the starting point? My algorithm is to calculate the dynamic friction coefficient first, and then calculate the velocity at the bottom of the slide to find the kinetic energy. But after calculating the dynamic friction coefficient, I found that the maximum static friction force is larger than the component force of gravity downward along the slope, and it seems that it can't slide down by itself

Let the friction force be f, H = 1 * sin30 = 1 / 2
At first, the kinetic energy of the ball: 1 / 2 * MV ^ 2 = 16J
MGH at standstill after sliding up
According to the conservation of energy:
Wf+1/2*mv^2=mgh
When the ball glides down, the friction force still does work WF
So the kinetic energy of the ball back to the starting point is:
W = MGH WF
Wf=f*L1
The final solution is as follows:
Moving 4J = 4W
You're not right. The ball is sliding on the slope, not sliding friction, but rolling friction, so it can't be calculated by the friction coefficient, so the ball will surely slide down, get it!

1. The vehicle with initial speed of 12m / s starts to brake at 32m away from the stop plate, and the acceleration of 2m / S ^ 2 is obtained (1) Find the speed at the end of 2S and the displacement within 2S after braking (2) How long does the car pass the stop sign after braking? How fast is the car at this time?

(1) 2S final velocity: v = V0 + at = 12m / S + (- 2) m / S ^ 2 * 2 = 8m / S
Displacement in 2S 2aX = V ^ 2-v0 ^ 2
2*(-2)m/s^2*x=(64-144)m^2/s^2
x=20
(2) x=v0t+1/2*a*^2
32m=12m/st-m/s^2*t^2
T = 4 or 8 (12-16 less than 0, round off t = 8)
v=v0+at=12-8=4m/s

The elevator starts from a standstill and rises at a constant speed of A1 for 2 seconds, and then reaches a speed of 3m / s. then it moves at a constant speed for a period of time. Finally, it slows down and rises at an acceleration of 1 m / S? 2 to stop. Calculate the time T2? Of the elevator?

According to the problem, the initial velocity of uniform deceleration motion is 3m / s, the acceleration is - 1m / S? And the final velocity is 0
So t = (0-3) / - 1s = 3S

If the mass of the two planets is M1 and M2, and the radius ratio of their orbits around the sun is R1: R2, then the ratio of their periods around the sun T1: T2 is () A. M1R22 M2R12 B. M1R12 M2R22 C. R13 R23 D. R23 R13

According to the formula of centripetal force and cycle of circular motion provided by universal gravitation, it is obtained that:
GMm
r2=m•4π2r
T2
T=2π
R3
GM, where R is the orbital radius, M is the centroid mass, and the period is independent of the mass of the planet
The ratio of the radii of the orbits of the two planets around the sun is r1:r2,
The ratio of the periods of their orbits around the sun, T1: T2, is
R13
R23．
Therefore, C

An object slides from the top of the slope at a constant speed, and the sliding acceleration is 2m / S2. If the object slides to the bottom, the sliding distance is 3% of the length of the slope 4. What is the slope length?