In △ ABC, the edges of angles a, B and C are a, B and C respectively, if B & # 178; + C & # 178; = A & # 178; + BC, and vector AB, vector AC = 4, Then the area of △ ABC is equal to

cosA=(b2+c2-a2)/2bc=1/2
AB.AC=bccosA=4
bc=8
Area 1 / 2 * bcsina = 1 / 2 * 4 * √ 3 / 2 = √ 3

What are the functions of sliding rheostat in measuring resistance by voltammetry and power by voltammetry?
It's mainly about their differences

A: the function of sliding rheostat in voltammetric resistance measurement is to change the resistance in the circuit, change the current in the circuit, so as to change the voltage at both ends of the resistance to be measured. Several groups of current values and voltage values can be measured, and several resistance values can be calculated, so as to achieve the purpose of multiple measurement and average value to reduce the experimental error

If in the triangle ABC, the point D is on the edge of BC, and the vector CD = 2, the vector dB, the vector CD = S &; the vector AB + T &; the vector AC,
s. If t belongs to R, then S + t=

Drawing first.. easy to get: vector CD = (vector AB vector AC) * (2 / 3)
Then r = 2 / 3 and S = - 2 / 3
So R + S = 0

Measure the power of the small bulb

Principle: P = UI
Method: voltammetry

If point D is on edge BC of triangle ABC, and vector CD = 3, vector DB = x, vector AB + y, vector AC, then x + y =?

Vector DB = 1 / 4, vector CB = 1 / 4 (vector Ba + vector AC) = - 1 / 4, vector AB + 1 / 4, vector AC, so x = - 1 / 4, y = 1 / 4, so x + y = 0

About measuring the electric power of small bulb
1. Why is the light bulb with low power brighter when connected in series?
2. Why is the light bulb with high power brighter in parallel connection?

The brightness of the bulb is determined by the actual power
Set the bulb resistance constant
When the rated voltage is the same, r = u & sup 2 / P is known from P = u & sup 2 / R, that is, the resistance with higher rated power is smaller, and the resistance with lower rated power is larger
When connected in series, the current is equal everywhere. According to P = I & sup2; R, the actual power with large resistance is larger, while that with small resistance is smaller. That is to say, the bulb with large rated power is dark, while the bulb with small rated power is on
In parallel, the voltage at both ends of each bulb is equal. According to P = u & sup2 / / R, the actual power with large resistance is small, and that with small resistance is large, that is, the bulb with high rated power is on, and that with low rated power is dark

If point D is on the BC side of triangle ABC, vector CD = 3, vector DB = m, vector AB + N, vector AC, find m + n

Vector CD = 3 / 4, vector CB = 3 / 4 (vector AB vector AC)
It can be seen that:
m=3/4 n=-3/4
m+n=0

Measure the electric power of "small bulb". The rated current of small bulb is 0.3A, and the filament resistance is between 10 Ω and 15 Ω. The equipment provided are: ammeter, voltmeter, power supply (constant voltage is 6V), switch, sliding rheostat & nbsp; ("5 Ω & nbsp; 2A", "20 Ω & nbsp;); (1) as shown in figure a, it is a physical circuit that has not been completely connected. Please use stroke line instead of wire to complete the physical diagram. (2) the specification of the selected sliding rheostat is (3) close the switch and move the slide. When the indication of the ammeter is 0.3A, the indication of the voltmeter is as shown in Figure B, and the reading is 5. The rated power of small bulb is W. (4) after the experiment, Xiao Ming found that the AP part of the sliding rheostat was hotter than the PD part when he was tidying up the experimental equipment (5) Xiaohong doesn't use a voltmeter, but she uses a switch. Other devices are the same as Xiaoming's, and she also measures the rated power of the small bulb. Please draw the circuit diagram she designed in the dotted line box C, and write down the experimental steps and the expression for measuring the rated power=__ ．

(1) If the rated current of the bulb is 0.3A and the filament resistance is between 10 Ω ~ 15 Ω, the rated voltage of the bulb u = IR is between 3V ~ 4.5V, the range of the voltmeter should be 0 ~ 15V, and the voltmeter should be connected in parallel with the bulb. The circuit diagram is shown in Figure 1: (2) when the circuit current is 0.3A, the voltage at both ends of the bulb can be seen from u = IR

In triangle ABC, if D is a point on the edge of BC and satisfies vector CD = 2, vector dB, vector CD = x, vector AB + y, vector AC, then x + y =?

Are you sure it's a CD?
CD=2/3 CB
CB=AB-AC
x+y=0

Problems encountered in the experiment of measuring the electric power of small bulbs
When I was doing the experiment of measuring the electric power of a small bulb, I encountered such a problem: the power supply was two dry batteries, and the circuit connection was correct. When the ammeter used a small (0-0.6a) range, the ammeter did not show the number, the bulb did not light up, and the voltmeter did not show the number. When I used a large range (0-0.3a), the finger turned too far, the light was on, and the voltmeter showed the number. Then I replaced the sliding rheostat, I can't figure out why. Before the experiment, the students connected the positive and negative poles of the two dry batteries in reverse

Generally, the measuring range of magnetoelectric ammeter is determined by its parallel resistance. The smaller the parallel resistance is, the larger the measuring range is. Therefore, when the measuring range is increased, the internal resistance of ammeter in series in the circuit is reduced, so the bulb is on, and the voltmeter also shows the number

In △ ABC, the edges of angles a, B and C are a, B and C respectively, if B & # 178; + C & # 178; = A & # 178; + BC, and vector AB, vector AC = 4, Then the area of △ ABC is equal to