The boy measures the time t of falling small pebbles
Task B11 (#27955) After rain, the water level in the well may rise. The boy measures the time t of falling small pebbles into the well and calculates the distance to the water using the formula h=5t
2
, where h is the distance in meters, t is the fall time in seconds. Before the rain, the fall time of the pebbles was 0.6 s. How much must the water level rise after rain in order for the measured time to change by 0.2 s? Express your answer in meters.
Solution.
Find the distance to the water in the well before the rain. Because before rain, the fall time of the pebbles was 0.6 s, we substitute this value into the formula by which the distance to the water is calculated:
h=5(0.6) 2 =1.8 m.
Obviously, after the rain, the water level rises, which means that the time for the fall of the pebble decreases. That is, it becomes equal to 0.6-0.2=0.4 s.
Calculate the distance to water after rain:
h=5(0.4) 2 = 0,8
The water level rose by 1.8-0.8=1 m.
Answer: 1 m .
27955. After rain, the water level in the well may rise. boy measuring timet small pebbles falling into the well and calculates the distance to the water using the formula h=5t 2 , whereh - distance in meters,t - fall time in seconds. Before the rain, the fall time of the pebbles was 0.6 s. How much must the water level rise after rain in order for the measured time to change by 0.2 s? Express your answer in meters.
We determine the distance to the water before and after the rain, and calculate how much the level has changed.
Before rain: h=5t 2 =5∙0,6 2 \u003d 1.8 meters.
After: h=5t 2 =5∙(0,6–0,2) 2 \u003d 0.8 meters.
The water level should rise by 1.8 - 0.8 = 1 meter.
Answer: 1
263802. The distance from an observer located at a low altitude h kilometers above the ground to the horizon line he observes is calculated by the formula:
From what height is the horizon visible at a distance of 4 kilometers? Express your answer in kilometers.
The task is reduced to solving the equation:
The horizon at a distance of 4 kilometers is visible from a height of 0.00125 kilometers.
Answer: 0.00125
28013. A mass of 0.08 kg oscillates on a spring with a speed that varies according to the law
The kinetic energy of the load is calculated by the formula:
Determine what fraction of the time from the first second after the start of movement the kinetic energy of the load will be at least 5∙10 –3 J. Express your answer as a decimal fraction, if necessary, round to hundredths.
Let's pay attention to the fact that the process is considered during the first second, that is, 0< t < 1, следовательно 0 < Пt < П (умножаем все части неравенства на Пи). Отметим, что на этом интервале имеет как положительное, так и отрицательное значение. Далее определяем, какой промежуток времени в первой секунде кинетическая энергия груза будет не менее 5∙10 –3 J, that is:
Substitute v, we get:
We get two inequalities:
We represent the solutions of inequalities graphically:
The periodicity of the cosine is not taken into account, since we consider the angle in the interval from 0 to Pi.
We divide the parts of the inequalities by Pi:
Thus, the kinetic energy of the load will be at least 5∙10 –3 J from the very beginning of the movement to 0.25 seconds, and from 0.75 to the end of the first second. Total time 0.25 + 0.25 = 0.5 seconds.
Answer: 0.5
28011. A skateboarder jumps onto a platform standing on rails with a speed v=3m/s at an acute angle α to the rails. From the push, the platform starts to move at a speed
m = 80 kg is the mass of a skateboarder with a skateboard, and M = 400 kg is the mass of the platform. At what maximum angle α (in degrees) must one jump to accelerate the platform to at least 0.25 m/s?
It is necessary to find the maximum angle α at which the platform accelerates to 0.25 m/s or more, i.e. u ≥ 25. The problem is reduced to solving the inequality:
We represent the solution of the inequality graphically:
The periodicity of the cosine is not taken into account when solving the inequality, since, by the condition, the angle α is acute. In this way:
Thus, the maximum angle at which you need to jump in order to fulfill the set condition is 60 degrees.
Answer: 60
Task: No. 395
After rain, the water level in the well may rise. The boy determines it by measuring the time t of small stones falling into the well and calculating the distance to the water using the formula h=5t2. Before the rain, the fall time of the stones was 0.8 s. What is the minimum height to which the water level must rise after rain in order for the measured time to change by more than 0.2 s? (Express your answer in meters).
Formula h=5t2. Before the rain, the fall time of the stones was 0.8 s. On theWhat is the minimum height the water level must rise after rainmeasured time has changed by more than 0.2 s? (Express your answer in meters).
1) find h1
h1=5*t^2=5*0.64=3.2 m
2) if the level rises, the time will decrease
t2=0.8-0.2=0.6 s
h2=5*t2^2=5*0.36=1.8 m
h1-h2=3.2-1.8=1.4 m
Answer : level should rise by more than1.4 m
Given:
After rain, the water level in the well may rise. The boy measures the time t of small pebbles falling into the well and calculates the distance from the Earth's surface to the water level using the formula h = -5t 2 . Before the rain, the fall time of the pebbles was 0.8 s.
Question:
What is the minimum height to which the water level must rise after rain for the measured time to change by more than 0.1 s? Express your answer in meters.
Solution
By condition, the fall time t can take 2 values:
t 1 = 0.8 - initial, given in the condition of the problem;
t 2 \u003d 0.8 - 0.1 \u003d 0.7 is the new value. Since, according to the condition, the water level rises, which means that the distance from the water to the upper edge of the well becomes smaller. Consequently, the flight time of the stone also decreases.
Now let's substitute these values into the formula h(t) = -5t 2 . So we find the distance from the top of the well to the surface of the water before and after the rain. We have:
h(t 1) = -5 (0.8) 2 = -5 0.64 = -3.2
h(t 2) = -5 (0.7) 2 = -5 0.49 = -2.45
So there are two values: -3.2 meters and -2.45 meters. If we subtract the smaller one from the greater height, we get the desired minimum height ∆ h, to which the water level should rise:
∆h = -2.45 - (-3.2) = 3.2 - 2.45 = 0.75
Summary
determined the distance from the upper edge of the well to the surface of the water before and after rain. We got the following values: -3.2 meters and -2.45 meters;
determined the minimum height to which the water level should rise. This height is 0.75 meters.