A physics problem in the second year of junior high school, who can help me?

As can be seen from the title, the real temperature is 0° when the thermometer reading is 5°, and 100° at 95°, so the temperature difference of each grid is 100 90°=10 9°, so (1) when the reading is 23°, the actual temperature is.

2) When the scale reading is x, it is the same as normal.

So 0+(x-5) 10 9°=x°

So x=50

The scale reading is the same as normal at 50°.

, so the temperature of 100 degrees is divided into 90 parts by the thermometer, so the temperature of each scale is 10-9 degrees.

When the scale value of the liquid measured by this thermometer is 23°, 23-5=18,18*10 9=20 degrees, so the actual temperature is 20 degrees.

2.(95+5) 2=50 degrees Celsius, the correct temperature value is 50 degrees Celsius which is the same as the original scale value.

Let the measured value y and the actual value x.

y = ax + b

x = 0, y = 5;x = 100, y = 95. Bring in the equation to get:

5 = b； 95 = 100a + b，===》b = 5， a = ， y = + 5。

1) When y = 23, 23 = + 5, so x = 202) y = + 5 = x, so x = 50

Because it is descending at a constant speed, the resistance experienced by the athlete is equal to the gravitational force of the athlete and the umbrella.

f=mg=70×

The time for a listener 20m away is 20 340 = 1 17 seconds.

The time of the listener in front of the radio at 200 km from the auditorium is 200000 3 * 10 8 = 1 1500 seconds < 1 17 seconds.

So, the listener in front of the radio, hear the report first.

t1=20/340

t2=200/3*10^5

The calculation shows that the listener in front of the radio hears the report first.

I guess it's a listener in front of the radio.

1 All (1) Let the density of 60 degree liquor be wine, then wine = m alcohol + m water v alcohol + v water.

60cm3× / 100cm3==

2) If the volume of 60 degrees of wine is V, then the volume of 30 degrees of wine is (1000ml-V), according to the title: 60100 V + 30100 (1000ml-V) = 420ml

The volume of 60 degrees of liquor v = 400ml, then the volume of 30 degrees liquor is 600ml

What's the question! Say. will do some help, they are all like-minded people.

The ignition wire burns at a velocity of v1 and a burning time t=s v1=100 v1.

The safety distance s=v*t=500, i.e. t=s v=500 5=100(s) Substituting the combustion time t into the equation gives v1=1(cm s)!

In fact, in the above calculation, the length of the lead has been ignored in the distance that the person needs to run, because the length of 100cm is too short for the length of 500m, and it is allowed to be ignored in simple calculations of physics.

The time it takes for the ignition man to come back.

t=x/vt=500m/5m/s

t=100s

It's just a point, and then 100 seconds to do it**.

v=x/tv=100cm/100s

v=1cm/s

It cannot exceed 1cm s

Let the burning rate be v, assuming that the person just reaches the safe zone, which can be obtained from the same time, 1 v = 500 5

The solution is v=1cm s, so the combustion rate cannot exceed 1cm s