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Student C: it is wrong to dip the pH test paper directly into the sodium chloride solution and compare the color displayed on the test paper with the standard colorimetric card. The pH = 7 is measured But why can't Oh test paper be directly immersed in sodium chloride solution Can pH test paper be reused
It will pollute the reagent. In the analytical chemistry experiment, this bottle of sodium chloride will be thrown away
The standard method is to use glass rod or drip irrigation to take out a little, drop it on the pH test paper on the glass plate, stand for 3-5min, and then compare with the colorimetric card
Scientific calculation in grade three of junior high school (utilization of electric energy)
An electric kettle with rated voltage of 220 V, rated power of 840 W, frequency of 50 Hz and capacity of 4 L. how much current does Xiao Ming use this kettle? If all electric energy is converted into internal energy, how much electric energy does a kettle of water need to consume when it is boiled from 20 ℃? The heat absorbed by the kettle itself is not included
I=P/U=840/220=3.82A
W=c*m*（t2-t1)=4.2*10^3*4*(100-20)=1344000J
I is the current
P is the rated power
U is the rated voltage
C is the specific heat capacity of water
M is the mass of water
T is the temperature of water
T2-t1 is the temperature difference
W is the total electrical energy consumed
Well, that's it,
The use of electric energy
1. If an electric appliance consumes 200000 joules of electric energy and the current does () work, then () of electric energy will be converted into other forms of energy
2. The power of electric lamp a is 60W, which means that the electric energy of () is converted into heat energy and light energy within 1 second () through the dead current. If the power of electric lamp B is 40W, compared with lamp A and lamp B, the one that converts the electric energy into heat energy and light energy within 1 second is () lamp, and the brighter one is () lamp. Therefore, the brightness of the bulb is determined by the size of () lamp
200000 joules
200000 joules
Work 60 J
60J
nail
nail
Actual power
As shown in the figure, AC ', BD', Ca ', DB' are bisectors of the four interior angles of the parallelogram ABCD. Please find out all parallelograms except ABCD (no longer shown)
The quadrilateral ac'ca ', bd'db' and four bisectors are parallelograms. There are three parallelograms
In quadrilateral ABCD, AB / / CD, AE bisection ∠ bad intersects BC at point E, and ab = EB
Certification:
∵AB=EB
∴∠BAE=∠BEA
∵ AE bisection ∠ bad
∴∠DAD=∠BAE
∴∠DAE=∠BEA
∴AD//BC
∵AB//CD
The quadrilateral ABCD is a parallelogram
As shown in the figure, in the parallelogram ABCD, e is the point on AB, de and AC intersect at point F, AF: FC = 3:7, then AE: EB=______ ．
∵ quadrilateral ABCD is a parallelogram, ∥ ab ∥ CD, ∥ AEF ∥ CDF, ∥ AFFC = AEDC = 37, and ∵ AB = DC, ∥ aEAb = 37, ∥ aeeb = 34
In △ ABC, ab = 8, BC = 10, CA = 12, and the shortest side of △ DEF is de = 2. When EF and FD are equal to each other, the two triangles are similar
2:8=1:4
EF:10=1:4
EF=2.5
FD:12=1:4
FD=3
When EF and FD are 2.53 or 32.5 respectively
It is known that in ABC and def, ab = 2cm BC equals 3cm CA = 4cm de = 7.5cm FD = 5cm. These two triangles are similar
Because AB / FD = 2 / 5
BC/DE=3/7.5=2/5
CA/EF=4/10=2/5
So AB / FD = BC / de = Ca / ef
So triangle ABC is similar to triangle FDE
It is known that in △ ABC and △ def, ab = 2 cm, BC = 3 cm, CA = 4 cm, de = 7.5 cm, EF = 10 cm, FD = 5 cm. Are these two triangles similar?
Why?
AB = 2 cm, BC = 3 cm, CA = 4 cm, de = 7.5 cm, EF = 10 cm, FD = 5 cm
∴AB/DF=BC/DE=AC/EF
∴△ABC∽△FDE
As shown in the figure, ∠ C is a right angle in △ ABC, ab = 12cm, ∠ ABC = 60 °, rotate △ ABC clockwise with point B as the center, and make point C rotate to point D on the extension line of AB, then what is the area of the figure (shadow part) swept by AC edge? （π=3.14159… , the final result retains three significant numbers)
In ∵ ABC, ≌ C is a right angle, ab = 12cm, ≌ ABC = 60 °, AC = 63cm, BC = 6cm ∵ rotate ∵ ABC clockwise with point B as the center, so that point C rotates to point D on the extension line of ab ≌ ABC ≌ EBD. According to the title image, s shadow = s sector Abe + s △ bde-s △ abc-s sector BCD = 120 π· 122360 + 12 × 6 × 63 − 12 × 6 × 63 − 120 π· 62360 = 120 π (122 − 62) 360 ≈ 113 A: AC side swept The area of the figure (shadow part) is about 113cm2
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