Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 70
You are missing trading opportunities:
- Free trading apps
- Over 8,000 signals for copying
- Economic news for exploring financial markets
Registration
Log in
You agree to website policy and terms of use
If you do not have an account, please register
(4) There are two electric kettles with thermostats. They work like this: when the temperature drops to 70°, the heating coil turns on and a light bulb comes on; when the temperature reaches 90°, both the heating device and the light bulb turn off. On one of the kettles the light bulb comes on, on the other it does not. Which one is most likely to have hotter water and why?
If the light goes on => temperature is less than 90, i.e. it can be 80, it can be 70, it can be 40, it can be 10.....
If not lit => temperature higher than 70
Answer: where the bulb is not lit the water is"most likely" hotter ?
О ! I have an option )))
The one with the light bulb.If light illuminates => temperature less than 90, i.e. 80 can be 80, 70 can be 70, 40 can be 40, 10.....
If not lit => temperature higher than 70
Answer: where the bulb is not lit"most likely" the water is hotter ?
They have the same temperature with or without bulbs
Besides, it can't be lower than 70
If light illuminates => temperature less than 90, i.e. 80 can be 80, 70 can be 70, 40 can be 40, 10.....
If not lit => temperature higher than 70
Answer: where the bulb is not lit"most likely" the water is hotter ?
But that's too easy))
In general, the water cools down unevenly depending on the difference with the temperature in the room. Therefore, when the bulb is switched off, the possible temperature will be longer at lower values and the average temperature will be below 80. And the heating time is linear with the temperature difference. That is, the probability distribution of temperature from time to time is uniform when the light bulb is on. And the average is 80. So the hotter one is probably the one with the light bulb on. I.e. we have to compare the heating and cooling time curves here
Then the problem would be too simple.
It is very likely that the thermopot is already in the "fired up" state, i.e. the light has already gone out at least once. The temperature cannot be lower than 70. This is a default, which is not written in the conditions, but which makes the task more difficult. Here's my solution (I don't claim to be correct, it hasn't been tested yet):
RATIONALE:
Water heats from 70 to 90 much faster than it cools: this is a logical requirement of practice. During heating, the energy flow to the water is much higher than the heat exchange flux of the water with the air through the walls of the thermostat, so the heat exchange with the surroundings can be neglected. At the same time, the rate at which the water is heated by the coil is determined by the heat capacity of the water, which is weakly dependent on its temperature. Therefore, we can assume that the heating of water is practically linear in time. Therefore, the average temperature of the water when the bulb is switched on is 80 degrees with an acceptable accuracy.
Cooling down of the water after switching off the bulb occurs differently, and in this process the main role is played by the heat exchange of the water with its surroundings through the walls of the thermostat. This heat transfer is proportional to the temperature difference between the water and the surroundings, so at first, at 90 degrees, the water cools down faster. Later on, the cooling rate decreases as the temperature difference between the water and the surround decreases. Therefore, the curve of water temperature versus time during cooling is concave.
This is the main factor that we will take into account. Consider the curve of temperature change from 90 to 70 degrees. If we connect the points of the curve corresponding to 90 and 70 degrees with a segment, we obtain a function according to which the water would cool down if it cooled down linearly. The average temperature of the water would be the same 80 degrees.
The real curve is lower because it is concave. The real cooling function can be represented as the difference of the "linear" and some non-negative function.
Consequently, the average temperature of the water over its entire cooling interval will be lower than the average for "linear" cooling, since it is equal to the integral of the temperature function divided by the cooling time. The difference is unlikely to be large, but it is significant.
The first graph is the cooling from 90 to 70, the second is a linear interpolation of this cooling, the third is the difference between the interpolation and reality.
It's a bit messy, but it's the only option I've got. The assumption is that the hot water consumer is not aware of the broken bulb.
Then, he is more likely to take water from a kettle with a working bulb if the water intake coincides with the moment of heating from 70 to 90
Then the problem would be too simple.
It is very likely that the thermopot is already in the "fired up" state, i.e. the light has already gone out at least once. The temperature cannot be lower than 70. This is a default, which is not written in the conditions, but which makes the task more difficult. Here is my solution (I don't claim to be correct, it hasn't been tested yet):
Here's my solution (I don't claim to be right, it hasn't been tested yet):
(5) The cake is shaped like an arbitrary triangle. Two megabrains divide it in the following way: the first points a point on the cake, the second makes a rectilinear cut through this point and takes the largest part. How much of the cake can the first megabrain get for himself? The cake is thought to have the same thickness everywhere.
Mischek: Ну и при чем тут лампочка ?
It just shows that the kettle is heating up, not cooling down. The curves are different.
That's it, I fucked up at the beginning.
Mathemat:
(4) Есть два электрочайника с термостатами. Они работают так: когда температура опускается до 70°, включается нагревательная спираль и загорается лампочка; когда температура достигает 90°, выключается и прибор подогрева, и лампочка. На одном из чайников лампочка горит, на другом — нет. В каком из них вода скорее всего горячее и почему?