A load D of mass m, having received an initial speed v0 at point A, moves in a curved pipe ABC located in a vertical plane; sections of the pipe are either both inclined, or one is horizontal and the other is inclined (Fig. D1.0-D1.9, Table D1). In Figure AB, in addition to gravity, the load is subject to the constant force Q (its direction is shown in the figures) and the resistance force of the medium R, which depends on the speed of the load v (directed against the movement). At point B, the load, without changing the value of its speed, moves to the section BC of the pipe, where, in addition to the force of gravity, it is acted upon by a variable force F, the projection of which Fx on the x axis is given in the table. Considering the load to be a material point and knowing the distance AB = l or the time t 1 of the movement of the load from point A to point B. Find the law of movement of the load on the section BC, i.e. x = f(t), where x = BD. Neglect the friction of the load on the pipe.
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From the points A And B, the distance between which is l, at the same time two bodies began to move towards each other: the first with a speed v 1, second - v 2. Determine how long after they will meet and the distance from the point A to their meeting place. Solve the problem also graphically.
Solution
1st method:
Dependence of body coordinates on time:
At the moment of meeting, the coordinates of the bodies will coincide, i.e. This means that the meeting will occur after a time from the beginning of the movement of the bodies. Find the distance from the point A to the meeting place as .
2nd method:
The velocities of the bodies are equal to the tangent of the angle of inclination of the corresponding graph of the coordinates versus time, i.e., , . The moment of the meeting corresponds to a dot C graph intersections.
After what time and where would the bodies meet (see problem 1) if they were moving in the same direction A→B, and from the point B the body began to move through t 0 seconds after it starts moving from the point A?
Solution
Graphs of the dependence of body coordinates on time are shown in the figure.
Based on the figure, let’s create a system of equations:
Having solved the system for t C we get:
Then the distance from the point A to the meeting point:
.
A motorboat travels the distance between two points A And B along the river in time t 1 = 3 hours, and the raft - in time t= 12 hours. What time t 2 will the motor boat spend on the return trip?
Solution
Let s— distance between points A And B, v is the speed of the boat relative to the water, and u— flow speed. Expressing distance s three times - for a raft, for a boat moving with the current, and for a boat moving against the current, we obtain a system of equations:
Having solved the system, we get:
A metro escalator takes a person walking down it in 1 minute. If a person walks twice as fast, he will descend in 45 seconds. How long does it take for a person standing on an escalator to descend?
Solution
Let us denote by the letter l escalator length; t 1 — descent time of a person walking at speed v; t 2 — descent time of a person walking at speed 2 v; t— time of descent of a person standing on an escalator. Then, having calculated the length of the escalator for three different cases (a person walks at a speed v, at speed 2 v and stands motionless on the escalator), we obtain a system of equations:
Solving this system of equations, we get:
A man runs along an escalator. The first time he counted n 1 = 50 steps, the second time, moving in the same direction at three times the speed, he counted n 2 = 75 steps. How many steps would he count on a stationary escalator?
Solution
Since with increasing speed the person counted more steps, it means that the directions of the speeds of the escalator and the person coincide. Let v— the speed of a person relative to the escalator, u— escalator speed, l- length of the escalator, n— the number of steps on a stationary escalator. The number of steps that fit in a unit length of the escalator is equal to n/l. Then the time a person spends on the escalator when he moves relative to the escalator at a speed v equals l/(v+u), and the distance traveled along the escalator is equal to vl/(v+u). Then the number of steps counted on this path is equal to . Similarly, for the case when the speed of a person relative to the escalator is 3 v, we will receive .
Thus, we can create a system of equations:
Eliminating the attitude u/v, we get:
Between two points located on a river at a distance s= 100 km from one another, there is a boat cruising, which, going with the flow, covers this distance in time t 1 = 4 hours, and against the current - for the time t 2 = 10 hours. Determine the speed of the river. u and speed of the boat v regarding water.
Solution
Expressing distance s twice, for a boat going with the current and for a boat going against the current, we get a system of equations:
Solving this system, we get v= 17.5 km/h, u= 7.5 km/h.
A raft passes by the pier. At this moment in a village located at a distance s 1 = 15 km from the pier, a motor boat departs down the river. She reached the village in time t= 3/4 hours and, turning back, met the raft at a distance s 2 = 9 km from the village. What is the speed of the river current and the speed of the boat relative to the water?
Solution
Let v- speed of the motor boat, u— river flow speed. Since from the moment the motor boat departs from the pier until the moment the motor boat meets the raft, the same time will obviously pass for both the raft and the motor boat, the following equation can be drawn up:
where on the left is the expression of the time elapsed until the moment of meeting for the raft, and on the right - for the motor boat. Let's write the equation for the time it took the motor boat to cover the distance s 1 from the pier to the village: t=s 1 /(v+u). Thus, we obtain a system of equations:
Where do we get it from? v= 16 km/h, u= 4 km/h.
A column of troops during a march moves at speed v 1 = 5 km/h, stretching along the road for a distance l= 400 m. The commander, located at the tail of the column, sends a cyclist with an order to the lead detachment. The cyclist sets off and rides at speed v 2 = 25 km/h and, having completed the assignment on the move, immediately returns back at the same speed. After how long t did he return after receiving the order?
Solution
In the reference frame associated with the column, the speed of the cyclist when moving towards the lead column is equal to v 2 -v 1, and when moving back v 2 +v 1 . That's why:
Simplifying and substituting numerical values, we get:
.
Car width d= 2.4 m, moving at speed v= 15 m/s, was pierced by a bullet flying perpendicular to the movement of the car. The displacement of the holes in the walls of the car relative to each other is equal to l= 6 cm. What is the speed of the bullet?
Solution
Let us denote by the letter u bullet speed. The flight time of a bullet from wall to wall of the car is equal to the time it takes the car to travel the distance l. Thus, we can create an equation:
From here we find u:
.
What is the speed of the drops v 2 vertically falling rain, if the driver of a car notices that raindrops do not leave a mark on the rear window, tilted forward at an angle α = 60° to the horizon when the vehicle speed v 1 more than 30 km/h?
Solution
As can be seen from the figure,
To ensure that raindrops do not leave a mark on the rear window, it is necessary that the time it takes for the drop to travel the distance h was equal to the time it took the car to cover the distance l:
Or, expressed from here v 2:
It is raining outside. In which case will a bucket in the back of a truck fill with water faster: when the car is moving or when it is stationary?
Answer
Same.
At what speed v and at what course should the plane fly so that in time t= 2 hours fly exactly to the North way s= 300 km if during the flight the wind blows from the north-west at an angle α = 30° to the meridian with speed u= 27 km/h?
Solution
Let's write the system of equations according to the figure.
Since the plane must fly due north, the projection of its speed onto the axis Oy v y is equal to y- wind speed component u y.
Having solved this system, we find that the plane should head northwest at an angle of 4°27" to the meridian, and its speed should be 174 km/h.
Moves along a smooth horizontal table at speed v Black board. What shape of mark will be left on this board by chalk thrown horizontally at speed u perpendicular to the direction of movement of the board, if: a) the friction between the chalk and the board is negligible; b) is friction high?
Solution
The chalk will leave a mark on the board, which is a straight line making an angle arctan( u/v) with the direction of movement of the board, i.e. it coincides with the direction of the sum of the speed vectors of the board and chalk. This is true for both case a) and case b), since the friction force does not affect the direction of movement of the chalk, since it lies on the same straight line with the velocity vector, it only reduces the speed of the chalk, so the trajectory in case b) may not reach the edge of the board.
The ship leaves the point A and goes at speed v, making an angle α with line AB.
At what angle β to the line AB should have been released from the point B a torpedo to hit a ship? The torpedo must be released at the moment when the ship was at the point A. The speed of the torpedo is u.
Solution
Dot C in the picture this is the meeting point between the ship and the torpedo.
A.C. = vt, B.C. = ut, Where t— time from the start to the moment of the meeting. According to the sine theorem
From here we find β :
.
To the slider, which can move along the guide rail,
attached cord threaded through the ring. The cord is selected at a speed v. At what speed u the slider moves at the moment when the cord makes an angle with the guide α ?
Answer and solution
u = v/cos α.
In a very short period of time Δt the slider moves a distance AB = Δl.
During the same period of time, the cord is selected to a length A.C. = Δl cos α (angle ∠ ACB can be considered right, since the angle Δα very small). Therefore we can write: Δl/u = Δl cos α /v, where u = v/cos α , which means that the speed of rope retraction is equal to the projection of the speed of the slider onto the direction of the rope.
Workers lifting loads
pull ropes at the same speed v. What speed u has a load at the moment when the angle between the ropes to which it is attached is 2 α ?
Answer and solution
u = v/cos α.
Projection of load speed u the direction of the rope is equal to the speed of the rope v(see problem 15), i.e.
u cos α = v,
u = v/cos α.
Rod length l= 1 m articulated with couplings A And B, which move along two mutually perpendicular slats.
coupling A moves at constant speed v A = 30 cm/s. Find speed v B couplings B at the moment when the angle OAB= 60°. Taking the moment when the clutch A was at the point O, determine the distance O.B. and clutch speed B as a function of time.
Answer and solution
v B= v Actg α
= 17.3 cm/s; , 
.
At any point in time, velocity projections v A and v B ends of the rod
on the axis of the rod are equal to each other, since otherwise the rod would have to be shortened or lengthened. So we can write: vA cos α = v B sin α . Where v B = vA ctg α .
At any time for a triangle OAB The Pythagorean theorem is true: l 2 = O.A. 2 (t) + O.B. 2 (t). Let's find it from here O.B.(t): . Because the O.A.(t) = v A t, then we will finally write down the expression for O.B.(t) So: .
Since ctg α
at any moment is equal to O.A.(t)/OB(t), then we can write an expression for the dependence v B from time: .
The tank moves at a speed of 72 km/h. At what speed do they move relative to the Earth: a) the upper part of the caterpillar; b) the lower part of the caterpillar; c) the point of the track that is currently moving vertically relative to the tank?
Answer and solution
a) 40 m/s; b) 0 m/s; c) ≈28.2 m/s.
Let v- speed is the speed of the tank relative to the Earth. Then the speed of any point on the track relative to the tank is also equal to v. The speed of any point on the track relative to the Earth is the sum of the vectors of the speed of the tank relative to the Earth and the speed of the point on the track relative to the tank. Then for case a) the speed will be equal to 2 v, for b) 0, and for c) v.
1. The car drove the first half of the journey at a speed v 1 = 40 km/h, second - at speed v 2 = 60 km/h. Find the average speed over the entire distance traveled.
2. The car drove half the distance at speed v 1 = 60 km/h, the rest of the way he walked at speed half the time v 2 = 15 km/h, and the last section at speed v 3 = 45 km/h. Find the average speed of the car along the entire journey.
Answer and solution
1. v av =48 km/h; 2. v avg =40 km/h.
1. Let s- all the way, t- time spent to cover the entire path. Then the average speed along the entire path is s/t. Time t consists of the sum of the time intervals spent covering the 1st and 2nd halves of the journey:
.
Substituting this time into the expression for the average speed, we get:
.(1)
2. The solution to this problem can be reduced to solution (1.), if you first determine the average speed in the second half of the path. Let's denote this speed vср2 , then we can write:
Where t 2 - time spent to overcome the 2nd half of the journey. The path traveled during this time consists of the path traveled at speed v 2, and the distance covered at speed v 3:
Substituting this into the expression for vср2 , we get:
.
.
The train traveled for the first half of the journey at a speed of n=1.5 times greater than the second half of the path. Average speed of the train along the entire journey v cp = 43.2 km/h. What are the speeds of the train at first ( v 1) and second ( v 2) half way?
Answer and solution
v 1 =54 km/h, v 2 =36 km/h.
Let t 1 and t 2 - time for the train to travel through the first and second half of the journey, respectively, s- the entire distance covered by the train.
Let's create a system of equations - the first equation is an expression for the first half of the path, the second - for the second half of the path, and the third - for the entire path traveled by the train:
By making a substitution v 1 =nv 2 and solving the resulting system of equations, we obtain v 2 .
Two balls began to move simultaneously and at the same speed along surfaces having the shape shown in the figure.
How will the speeds and times of movement of the balls differ by the time they arrive at the point? B? Ignore friction.
Answer and solution
The speeds will be the same. The first ball will take longer to move.
The figure shows approximate graphs of the movement of the balls.
Because the paths traveled by the balls are equal, then the areas of the shaded figures are also equal (the area of the shaded figure is numerically equal to the distance traveled), therefore, as can be seen from the figure, t 1 >t 2 .
The plane flies from the point A to point B and returns back to point A. The speed of an airplane in calm weather is v. Find the ratio of the average speeds of the entire flight for two cases when the wind blows during the flight: a) along the line AB; b) perpendicular to the line AB. The wind speed is u.
Answer and solution
Airplane flight time from point A to point B and back when the wind blows along the line AB:
.
Then the average speed in this case is:
.
If the wind blows perpendicular to the line AB, the aircraft's velocity vector must be directed at an angle to the line AB so as to compensate for the influence of wind:
The round-trip flight time in this case will be:
Aircraft flight speed to point B and vice versa are the same and equal:
.
Now we can find the ratio of the average speeds obtained for the cases considered:
.
Distance between two stations s= 3 km a metro train travels at average speed v avg = 54 km/h. At the same time, it takes time to accelerate t 1 = 20 s, then goes evenly for some time t 2 and it takes time to slow down to a complete stop t 3 = 10 s. Graph the speed of the train and determine the highest speed of the train v Max.
Answer and solution
The figure shows a graph of the speed of a train.
The distance traveled by the train is numerically equal to the area of the figure limited by the graph and the time axis t, so we can write the system of equations:
From the first equation we express t 2:
,
then from the second equation of the system we find v Max:
.
The last car is unhooked from a moving train. The train continues to move at the same speed v 0 . How will the distances traversed by the train and the car be related to the moment the car stops? Assume that the car was moving at equal speed. Solve the problem also graphically.
Answer
At the moment when the train started, the person accompanying him began to run evenly along the train at a speed v 0 =3.5 m/s. Assuming the motion of the train to be uniformly accelerated, determine the speed of the train v at the moment when the person seeing off reaches the person seeing him off.
Answer
v=7 m/s.
A graph of the velocity of a certain body versus time is shown in the figure.
Draw graphs of the acceleration and coordinates of the body, as well as the distance traveled by it, versus time.
Answer
Graphs of the acceleration, coordinates of the body, as well as the distance traveled by it versus time are shown in the figure.
The graph of the acceleration of a body versus time has the form shown in the figure.
Draw graphs of speed, displacement and distance traveled by the body versus time. The initial velocity of the body is zero (at the discontinuity site, the acceleration is zero).
The body begins to move from a point A with speed v 0 and after some time gets to the point B.
What distance did the body travel if it moved uniformly with an acceleration numerically equal to a? Distance between points A And B equals l. Find the average speed of the body.
The figure shows a graph of the body coordinates versus time.
After the moment t=t 1 curve of the graph is a parabola. What kind of movement is shown in this graph? Draw a graph of the body's speed versus time.
Solution
In the area from 0 to t 1: uniform movement with speed v 1 = tg α ;
in the area from t 1 to t 2: uniform slow motion;
in the area from t 2 to t 3: uniformly accelerated movement in the opposite direction.
The figure shows a graph of the velocity of a body versus time.
The figure shows velocity graphs for two points moving along the same straight line from the same initial position.
Known moments in time t 1 and t 2. At what point in time t Will 3 points meet? Construct motion graphs.
During what second from the beginning of movement is the path traveled by a body in uniformly accelerated motion three times greater than the path traveled in the previous second, if the movement occurs without an initial speed?
Answer and solution
In a second.
The easiest way to solve this problem is graphically. Because the path traveled by the body is numerically equal to the area of the figure under the line of the speed graph, then from the figure it is obvious that the path traveled in the second second (the area under the corresponding section of the graph is equal to the area of three triangles) is 3 times greater than the path traveled in the first second (the area is equal to the area one triangle).
The trolley must transport the cargo in the shortest possible time from one place to another located at a distance L. It can accelerate or decelerate its movement only with the same magnitude and constant acceleration a, then moving into uniform motion or stopping. What is the highest speed v must the trolley reach to fulfill the above requirement?
Answer and solution
It is obvious that the trolley will transport the cargo in the minimum time if it moves with acceleration for the first half of the journey + a, and the remaining half with acceleration - a.
Then we can write the following expressions: L = ½· vt 1 ; v = ½· at 1 ,
where we find the maximum speed:
Jet plane flies at speed v 0 =720 km/h. From a certain moment the plane moves with acceleration for t=10 s and at the last second the path passes s=295 m. Determine acceleration a and final speed v airplane.
Answer and solution
a=10 m/s 2, v=300 m/s.
Let's draw a graph of the speed of the airplane in the figure.
Airplane speed at time t 1 is equal v 1 = v 0 + a(t 1 - t 0). Then the distance traveled by the plane in the time from t 1 to t 2 is equal s = v 1 (t 2 - t 1) + a(t 2 - t 1)/2. From here we can express the desired acceleration value a and, substituting the values from the problem conditions ( t 1 - t 0 = 9 s; t 2 - t 1 = 1 s; v 0 = 200 m/s; s= 295 m), we get the acceleration a= 10 m/s 2. Airplane terminal speed v = v 2 = v 0 + a(t 2 - t 0) = 300 m/s.
The first carriage of the train passed the observer standing on the platform behind t 1 = 1 s, and the second - for t 2 =1.5 s. Car length l=12 m. Find acceleration a trains and their speed v 0 at the beginning of observation. The movement of the train is considered to be uniformly variable.
Answer and solution
a=3.2 m/s 2, v 0 ≈13.6 m/s.
The distance traveled by the train to the moment in time t 1 is equal to:
and the path to the moment in time t 1 + t 2:
.
From the first equation we find v 0:
.
Substituting the resulting expression into the second equation, we get the acceleration a:
.
A ball launched up an inclined plane passes successively two equal segments of length l everyone continues to move on. The ball passed the first segment in t seconds, the second - in 3 t seconds Find speed v ball at the end of the first segment of the path.
Answer and solution
Since the motion of the ball under consideration is reversible, it is advisable to choose the common point of the two segments as the starting point. In this case, the acceleration when moving in the first segment will be positive, and when moving in the second segment - negative. The initial speed in both cases is equal v. Now let’s write down the system of equations of motion for the paths traversed by the ball:
Eliminating acceleration a, we get the required speed v:
A board divided into five equal segments begins to slide down an inclined plane. The first segment passed the mark made on the inclined plane in the place where the front edge of the board was at the beginning of the movement, beyond τ =2 s. How long will it take the last section of the board to pass this mark? The movement of the board is considered to be uniformly accelerated.
Answer and solution
τ n =0.48 s.
Let's find the length of the first segment:
Now let's write down the equations of motion for the starting points (time t 1) and end (time point t 2) fifth segment:
By substituting the length of the first segment found above instead of l and finding the difference ( t 2 - t 1), we get the answer.
A bullet flying at a speed of 400 m/s hits an earthen shaft and penetrates it to a depth of 36 cm. How long did it move inside the shaft? At what acceleration? What was its speed at a depth of 18 cm? At what depth did the bullet's speed decrease by a factor of three? The movement is considered uniformly variable. What will be the speed of the bullet by the time the bullet has traveled 99% of its path?
Answer and solution
t= 1.8·10 -3 s; a≈ 2.21·10 5 m/s 2 ; v≈ 282 m/s; s= 32 cm; v 1 = 40 m/s.
We find the time of movement of the bullet inside the shaft from the formula h = vt/2, where h- full depth of immersion of the bullet, from where t = 2h/v. Acceleration a = v/t.
A ball was allowed to roll from bottom to top on an inclined board. On distance l= 30 cm from the beginning of the path the ball has visited twice: through t 1 = 1 s and after t 2 = 2 s after the start of movement. Determine the initial speed v 0 and acceleration a movement of the ball, considering it constant.
Answer and solution
v 0 = 0.45 m/s; a= 0.3 m/s 2.
The dependence of the ball speed on time is expressed by the formula v = v 0 - at. At a moment in time t = t 1 and t = t 2 the ball had the same velocities and opposite in direction: v 1 = - v 2. But v 1 =v 0 - at 1 and v 2 = v 0 - at 2, therefore
v 0 - at 1 = - v 0 + at 2, or 2 v 0 = a(t 1 + t 2).
Because the ball moves uniformly accelerated, then the distance l can be expressed as follows:
Now you can create a system of two equations:
,
solving which, we get:
A body falls from a height of 100 m without initial velocity. How long does it take for a body to travel the first and last meters of its path? How far does the body travel during the first and last second of its movement?
Answer
t 1 ≈ 0.45 s; t 2 ≈ 0.023 s; s 1 ≈ 4.9 m; s 2 ≈ 40 m.
Determine the open time of the photographic shutter τ , if, when photographing a ball falling along a vertical centimeter scale from the zero mark without an initial speed, a strip was obtained on the negative extending from n 1 to n 2 scale divisions?
Answer
.
A freely falling body traveled the last 30 m in a time of 0.5 s. Find the height of the fall.
Answer
A freely falling body has covered 1/3 of its path in the last second of its fall. Find the time of fall and the height from which the body fell.
Answer
t≈ 5.45 s; h≈ 145 m.
At what initial speed v 0 you need to throw the ball down from a height h so that he jumps to a height of 2 h? Neglect air friction and other losses of mechanical energy.
Answer
With what time interval did two drops break away from the roof eaves, if two seconds after the second drop began to fall, the distance between the drops was 25 m? Neglect air friction.
Answer
τ ≈ 1 s.
The body is thrown vertically upward. An observer notices a period of time t 0 between two moments when the body passes the point B, located at a height h. Find the initial throwing speed v 0 and the time of the entire body movement t.
Answer
; 
.
From points A And B, located vertically (point A above) at a distance l= 100 m from each other, two bodies are thrown simultaneously with the same speed of 10 m/s: from A- vertically down, from B- vertically up. After how long and in what place will they meet?
Answer
t= 5 s; 75 m below point B.
A body is thrown vertically upward with an initial velocity v 0 . When it reached the highest point of the journey, from the same starting point at the same speed v 0 the second body is thrown. At what altitude h from the starting point will they meet?
Answer
Two bodies are thrown vertically upward from the same point with the same initial speed v 0 = 19.6 m/s with time interval τ = 0.5 s. After what time t after throwing the second body and at what height h will bodies meet?
Answer
t= 1.75 s; h≈ 19.3 m.
The balloon rises vertically upward from the Earth with acceleration a= 2 m/s 2. Through τ = 5 s from the beginning of its movement an object fell out of it. After how long t will this object fall to earth?
Answer
t≈ 3.4 s.
From a balloon descending at a speed u, throw the body up at speed v 0 relative to the Earth. What will be the distance l between the balloon and the body at the moment of the highest rise of the body relative to the Earth? What is the greatest distance l max between body and balloon? After what time τ from the moment of throwing the body will be level with the balloon?
Answer
l = v 0 2 + 2uv 0 /(2g);
l max = ( u + v 0) 2 /(2g);
τ = 2(v 0 + u)/g.
A body located at a point B on high H= 45 m from the Earth, begins to fall freely. Simultaneously from the point A, located at a distance h= 21 m below point B, throw another body vertically upward. Determine initial speed v 0 of the second body, if it is known that both bodies will fall to the Earth at the same time. Neglect air resistance. Accept g= 10 m/s 2.
Answer
v 0 = 7 m/s.
A body falls freely from a height h. At the same moment another body is thrown from a height H (H > h) vertically down. Both bodies fell to the ground at the same time. Determine initial speed v 0 second body. Check the correctness of the solution using a numerical example: h= 10 m, H= 20 m. Accept g= 10 m/s 2.
Answer
v 0 ≈ 7 m/s.
A stone is thrown horizontally from the top of a mountain with a slope α. At what speed v 0 a stone must be thrown so that it falls on a mountain in the distance L from the top?
Answer
Two people play with a ball, throwing it to each other. What is the greatest height the ball reaches during a game if it flies from one player to another for 2 s?
Answer
h= 4.9 m.
The plane flies at a constant altitude h in a straight line at speed v. The pilot must drop the bomb on a target in front of the aircraft. At what angle to the vertical should he see the target at the moment the bomb is dropped? What is the distance from the target to the point over which the plane is located at this moment? Do not take into account air resistance to the movement of the bomb.
Answer
Two bodies fall from the same height. On the path of one body there is a platform located at an angle of 45° to the horizon, from which this body is elastically reflected. How do the times and speeds at which these bodies fall differ?
Answer
The falling time of the body on the path of which the platform was located is longer, since the vector of the speed accumulated at the moment of impact changed its direction to horizontal (with an elastic collision, the direction of the speed changes, but not its magnitude), which means the vertical component of the speed vector became equal to zero, while like another body, the velocity vector did not change.
The falling velocities of the bodies are equal until the moment of collision of one of the bodies with the platform.
The elevator rises with an acceleration of 2 m/s 2 . At the moment when its speed became equal to 2.4 m/s, a bolt began to fall from the ceiling of the elevator. The height of the elevator is 2.47 m. Calculate the time the bolt falls and the distance traveled by the bolt relative to the shaft.
Answer
0.64 s; 0.52 m.
At a certain height, two bodies are simultaneously thrown from one point at an angle of 45° to the vertical with a speed of 20 m/s: one down, the other up. Determine height difference Δh, on which there will be bodies in 2 s. How do these bodies move relative to each other?
Answer
Δ h≈ 56.4 m; bodies move away from each other at a constant speed.
Prove that when bodies move freely near the Earth's surface, their relative speed is constant.
From point A the body falls freely. Simultaneously from the point B at an angle α another body is thrown towards the horizon so that both bodies collide in the air.
Show that the angle α does not depend on the initial speed v 0 body thrown from a point B, and determine this angle if . Neglect air resistance.
Answer
α = 60°.
Body thrown at an angle α towards the horizon at speed v 0 . Determine speed v this body is on top h above the horizon. Does this speed depend on the throwing angle? Ignore air resistance.
At an angle α =60° a body is thrown towards the horizon with an initial speed v=20 m/s. After how long t it will move at an angle β =45° to the horizon? There is no friction.
From three pipes located on the ground, jets of water shoot out at the same speed: at an angle of 60, 45 and 30° to the horizon. Find the ratio of the greatest heights h the rise of the jets of water flowing from each pipe, and the fall distances l water to the ground. Do not take into account air resistance to the movement of water jets.
From a point lying at the upper end of the vertical diameter d of a certain circle, along gutters installed along various chords of this circle, loads simultaneously begin to slide without friction.
Determine after what period of time t the loads will reach the circle. How does this time depend on the angle of inclination of the chord to the vertical?
Initial speed of a thrown stone v 0 =10 m/s, and after t=0.5 s stone speed v=7 m/s. To what maximum height above the initial level will the stone rise?
Answer
H max ≈ 2.8 m.
At a certain height, balls are simultaneously thrown from one point with equal speeds in all possible directions. What will be the geometric location of the points where the balls are located at any given time? Neglect air resistance.
Answer
The geometric location of the points where the balls are located at any moment in time will be a sphere whose radius v 0 t, and its center is located below the starting point by an amount GT 2 /2.
The target located on the hill is visible from the gun location at an angle α to the horizon. The distance (horizontal distance from the gun to the target) is equal to L. Shooting at the target is carried out at an elevation angle β .
Determine initial speed v 0 projectile hitting the target. Ignore air resistance. At what elevation angle β 0 will the firing range along the slope be maximum?
Answer and solution
, .
Let's choose a coordinate system xOy so that the reference point coincides with the tool. Now let’s write down the kinematic equations of projectile motion:
Replacing x And y to the target coordinates ( x = L, y = L tgα) and excluding t, we get:
Range l projectile flight along a slope l = L/cos α . Therefore, the formula we received can be rewritten as follows:
this expression is maximum at the maximum value of the product
That's why l maximum with maximum value = 1 or
At α = 0 we get the answer β 0 = π /4 = 45°.
An elastic body falls from a height h on an inclined plane. Determine how long t after reflection, the body will fall onto an inclined plane. How does time depend on the angle of the inclined plane?
Answer
Does not depend on the angle of the inclined plane.
From high H on an inclined plane forming an angle with the horizon α =45°, the ball falls freely and is elastically reflected at the same speed. Find the distance from the place of the first impact to the second, then from the second to the third, etc. Solve the problem in general form (for any angle α ).
Answer
; s 1 = 8H sin α ; s 1:s 2:s 3 = 1:2:3.
The distance to the mountain is determined by the time between the shot and its echo. What can be the error τ in determining the moments of a shot and the arrival of an echo, if the distance to the mountain is at least 1 km, and it needs to be determined with an accuracy of 3%? Speed of sound in air c=330 m/s.
Answer
τ ≤ 0.09 s.
They want to measure the depth of the well with an accuracy of 5% by throwing a stone and noting the time τ , through which the splash will be heard. Starting from what values τ Is it necessary to take into account the sound travel time? Speed of sound in air c=330 m/s.
Answer
Option No. 523758
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Three material points begin to move without initial velocity from a point with coordinate x= 0 along the horizontal axis OX. The figures show graphs of the dependences of the kinematic characteristics (velocity projection, acceleration projection and co-ordination) of these bodies on time. Establish a correspondence between the graphs and the dependences of the coordinates of bodies on time: for each element of the first column, select the corresponding element from the second and enter the selected numbers under the corresponding letters in the answer line.
| GRA-FI-KI | ZA-VI-SI-MO-STI | |
Write down the numbers in response, arranging them in a row corresponding to the letter:
| A | B | IN |
Answer:
The body moves along the axis OX. The figure shows a graph of the dependence of the coordinates x of this body from time to time t. The movement with the highest modulus speed corresponds to the section of the graph
Answer:
In which of the listed cases does the conversion of ten-tsi-al energy into ki-ne-ti-che-che-e occur?
1) The car accelerates after a traffic light on a horizontal road
2) The soccer ball flies upward after being hit
3) A stone falls from the roof of the house to the ground
4) The satellite rotates in a constant orbit around the Earth
Answer:
The ball begins to fall to the ground from a height of 20 m with an initial speed equal to zero. At what height above the Earth's surface will the ball be 1 s after the start of the fall? Do not neglect air resistance.
Answer:
A boat floats in a pool of water, and at the bottom of the pool lies a heavy stone. A stone is taken from the bottom of the pool and placed in the boat. How does the water level in the basin change as a result?
1) no matter what
2) you-sha-et-sya
3) not from me
4) there is no definite answer, since the answer depends on the size of the stone
Answer:
The figure shows a graph of coordinates versus time for a body moving along an axis Ox.
Using the graph data, select two correct statements from the list provided. Indicate their number.
1) The section of the sun corresponds to the equally accelerated movement of the body.
2) At the moment of time t 3 the speed of the body is zero.
3) In the period of time from t 1 to t 2 the body changed the direction of movement to pro-false.
4) At the moment t 2 the speed of the body is zero.
5) The path corresponding to section OA is equal to the path corresponding to section BC.
Answer:
A spring was attached to a cart weighing 1 kg and they began to pull on it, applying a horizontally directed constant force, so that in a time of 2 s the cart traveled a distance of 1.6 m. Moreover, during the movement of the cart, the spring was extended by 1 cm. What is the stiffness of the spring? -living? Don't neglect friction.
Answer:
The figure shows the graphs of the heating and melting of two solids - “1” and “2” - one at a time -howl mass taken at the same initial temperature. Samples are heated on the same heaters. Compare the specific heat capacities of these two substances and their melting temperatures.
1) Substance “1” has a higher specific heat capacity and melting temperature than substance “2”.
2) Substance “1” has a lower specific heat capacity, but a higher melting temperature than substance “2”.
3) Substance “1” has a higher specific heat capacity, but a lower melting temperature than substance “2”.
4) Substance “1” has the same specific heat as substance “2”, but higher than the melting temperature.
Answer:
The figure shows graphs of coordinates versus time for two bodies: A and B, moving in a straight line along which the axis is directed Oh. Choose two true statements about the movement of bodies.
1) Body A moves equally accelerated.
2) The time interval between meetings of bodies A and B is 6 s.
3) During the first five seconds, the bodies moved in the same direction.
4) In the first 5 s, body A traveled 15 m.
5) Body B moves with constant acceleration.
Answer:
On the diagram for two substances, when there are values of the amount of heat, there is no need for heating 1 kg of substance at 10 ° C and for melting 100 g of substance, heating to melting temperature le-nia. Compare the specific heat of melting ( λ 1 and λ 2) two substances.
Answer:
The student placed a metal line on a switched-off electric lamp, brought it to its end, without hesitation, from-the-re-tsa-tel-but for-the-wives' stick and began to carefully move the stick in an arc around -no-sti. At the same time, Li-na-ka ran after the stick. This is about-is-ho-di-lo in a way that
1) between the stick and the line there is a gravitational force
2) at the end of the line closest to the stick, a precise positive charge is generated and it is applied cha-gi-va-et-sya to pa-loch-ke
3) at the end of the line closest to the stick, a precise charge is generated and it is applied cha-gi-va-et-sya to pa-loch-ke
4) the whole li-ney-ka-re-ta-et-is-from-precise-po-lo-living-charge and is-attracted-to-pa-loch -ke
Answer:
In an electrical circuit (see ri-su-nok) there is a voltmeter V 1 shows 2 V voltage, volt meter V 2 - voltage 0.5 V. The voltage on the lamp is
Answer:
To the north of the place there is a small magnetic arrow. Point out the ri-su-nok, on which the right-vil-but-for-the-installation-of-the-sight-of-the-mag-nit -th arrow.
Answer:
In the process of rubbing against the silk, the glass ruler acquired a positive charge. How did the number of charged particles on the ruler and silk change, if we assume that there was no exchange of atoms between the ruler and silk during friction?
For each physical quantity, determine the corresponding nature of change:
1) increased
2) decreased
3) didn’t change my mind
Write down the selected numbers for each physical quantity in the table. The numbers in the answer may be repeated.
Answer:
The bicycle is equipped with a generator that generates electrical energy for two series-connected lamps. In each lamp the current is 0.3 A and the voltage on each lamp is 6 V. What is the work done by the generator current in 2 hours?
Answer:
The nucleus of the potassium atom contains
1) 20 pro-to-nov, 39 neu-tro-nov
2) 20 pro-to-nov, 19 neu-tro-nov
3) 19 pro-to-nov, 20 neu-tro-nov
4) 19 pro-to-nov, 39 neu-tro-nov
Answer:
In the process of rubbing against the wool, the ebony stick acquired a negative charge. How did the number of charged particles on the stick and wool change under the condition that there was no exchange of atoms during friction? Establish a correspondence between physical ve-li-chi-na-mi and their possible consequences. Write down the selected numbers under the corresponding letters in the table. The numbers in the answer may be repeated.
| A | B | B |
Answer:
The teacher in the lesson, using two identical sticks and a piece of fabric, subsequently conducted experiments on electricity. The description of the teacher’s actions is presented in the table.
What statements correspond to the results of the ex-peri-mental-on-the-blues? From the given list of statements, you take two correct ones. Indicate their numbers.
1) Both the pa-loch and the fabric are electric when rubbed.
2) When rubbing, the pad and the fabric re-form in equal amounts.
3) When rubbing, the stick and the fabric re-generate charges of different signs.
4) Pa-loch-ka-re-ta-et-re-charges.
5) Electricity is connected with the transfer of electrons from one body to another.
Answer:
On-observation, a source of light is approaching, fixing
1) increasing the speed of light and decreasing the wavelength of light
2) increasing the speed of light and increasing the wavelength of light
3) decreasing the light wavelength
4) increasing the wavelength of light
ABOUT λ 0 . Observers at points A And B λ v A B
The change in light wavelength depends on the speed of the source relative to the observer (along the line of sight) and is determined by the Doppler formula: .
The Doppler effect has found wide application, in particular in astronomy, to determine the velocities of radiation sources.
Answer:
About 100 years ago, the American astronomer Vesto Slifer lived that the wavelengths in the spectra were due to the large number of The ga-lak-tik state is shifted to the red hundred. This fact may be due to the fact that
1) ga-lak-ti-ki raz-be-ga-yut-sya (All-flax-ra-shi-rya-et-sya)
2) ga-lak-ti-ki bring-together (The whole flax is compressed)
3) The universe is endless in space
4) Everything is not one-of-a-kind
Doppler effect for light waves
The speed of light is not affected by the speed of the light source or the speed of the observer. The constancy of the speed of light in a vacuum is of great importance for physics and astronomy. However, the frequency and wavelength of light change with the speed of the source or observer. This fact is known as the Doppler effect.
Let us assume that the source located at the point ABOUT, emits light with a wavelength λ 0 . Observers at points A And B, for which the light source is at rest, will record radiation with a wavelength λ 0 (Fig. 1). If the light source begins to move at a speed v, then the wavelength changes. For the observer A, to which the light source approaches, the wavelength of the light decreases. For the observer B, from which the light source moves away, the light wavelength increases (Fig. 2). Since in the visible part of electromagnetic radiation the shortest wavelengths correspond to violet light, and the longest to red, it is said that for an approaching light source there is a shift in wavelength to the violet side of the spectrum, and for a receding light source - to the red side of the spectrum.