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How to make the second grade physics exercise book?
This place is just some difficult communication, not for you to get something for nothing
After all, it's not good for you to ask for an answer if you want to enter high school
/T can be used to express the instantaneous velocity of an object at time T. this definition applies the idea of (cumulative method). (calculus, infinitesimal method)
For example, experiments such as measuring bulb resistance, constant value resistance and bulb power, what variables to control and what role sliding plays, etc,
Electrical test: series parallel circuit, the use of voltmeter and ammeter, the law of current and voltage of series parallel circuit, the factors related to resistance, voltammetry to measure resistance, voltammetry to measure power, Ohm's law
It's a typical control variable method, but it's very simple. It's length, thickness, and material. What's not easy to do is volt ampere method to measure power, Ohm's law, and Joule's law. Volt ampere method to measure power, connect the circuit, adjust the slide of the sliding rheostat to make the voltage representation equal to the rated voltage of the small bulb, read the current representation, Calculate the rated power. Then adjust the slide rheostat vane so that the voltage is slightly less than (slightly larger than) the rated voltage of the small bulb, and calculate the power. Ohm's law connects well with the circuit (same as the voltammetry power) keeps the resistance constant. By adjusting the slide rheostat slide, the number of voltage is increased gradually, and the change of the current is observed, and the resistance is constant. The current is proportional to the voltage. Then adjust the slide of the sliding rheostat to keep the voltage constant. Change different resistors to see the change of the current. It is concluded that when the voltage is constant, the current is inversely proportional to the resistance. Then it is concluded that I = u / R Joule's law is a series circuit. Two different resistors are in two identical bottles with kerosene of the same mass and initial temperature, Through the same time of current, to see which rising temperature is higher, we can get that Q is proportional to R, then increase the current, power on the same time, compare the rising temperature with the same bottle, we can get that Q is proportional to I
Let's give it a score. I've typed so many words
How to do the experiment in the fourth chapter of the first volume of physics?
The exploration experiment of melting and solidification was demonstrated by the teacher first, and then the students were divided into groups to do the experiment in the laboratory. The exploration experiment of melting and solidification is the subject of the key experiment in the senior high school entrance examination, so it is necessary to cultivate the students' practical ability to do the experiment
As for the small experiment in class is very simple, I believe every teacher can!
Experimental activities in the first volume of physics in grade two of junior high school
Place the thermometer in a place where there is no sunshine outside. Record the thermometer every 1 h from 8:00 in the morning until 18:00. Record the measurement results in the table below
Time / h 8:00 9:00 10:00 11:00 12:00 13:00 14:00 15:00 16:00 17:00 18:00
Temperature / ℃
It's very tiring. It's on the Internet
sixteen
eighteen
twenty
twenty-one
twenty-three
twenty-two
twenty-two
twenty-two
twenty-one
twenty
eighteen
Turn on a household electric lamp with the words of "turn on the electric lamp for 20MW / h in a household"?
90 rpm in 20 min, 270 RPM in 1 hour, i.e. P = 3000 / 270 = 100 / 9
220V 100W bulb, resistance R = 220 * 220 / 100 = 22 * 22 ohm
Actual voltage U * u = P * r = 22 * 22 * 100 / 9
U=22*10/3=73.33333
A 2kg flat water bucket with a floor area of 800cm 2 is placed on the horizontal ground. The bucket is filled with 50cm deep water with a volume of 55cubic decimeters. Xiao Ping used a vertical upward force F to lift the bucket, but did not lift it up. At this time, if the pressure of water on the bottom of the bucket is equal to the pressure of the bucket on the ground, he calculated the force F of Xiao Ping to lift the bucket. (g = 10N / kg)
It is known that: M = 2kg, s = 0.08m2, H = 0.5m, v = 0.055 & # 13221; g = 10N / kg, ρ water = 1.0 * 10 & # 179; kg / &# 13221; seeking: the force F of lifting bucket, because from the formula P = ρ water GH, P = f / s, P = ρ water GH = 1.0 * 10 & # 179; kg / &# 13221; * 10N / kg * 0
Two questions and solutions about solid pressure in physics of grade two of junior high school
There should be a detailed process. Thank you
What I need is calculation and process. It's not filling in the blanks or something.
Put 50 N water into a flat bottomed plastic bucket with a bottom area of 0.01 square meters and a weight of 10 n. the water surface is 30 cm away from the bottom of the bucket, and the plastic bucket is placed on a table of 500 square centimeters
Each of them has 5 questions about the calculation of pressure and pressure, solubility and solute mass fraction,
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1. Pressure calculation: an object is a cube, 10cm wide and 1cm thick. It is full of water. In a voyage, it is thrown into 100m deep water to calculate the pressure on each surface of the cube. 2. Pressure calculation: there is a smooth slope with an inclination of 30 degrees, 1m high, and a small object rushes from the bottom of the slope to the slope with a speed of 10m per second
Calculation of atmospheric pressure in the second year of junior high school physics
Under a standard atmospheric pressure, a pressure cooker with an inner surface area of 700cm2 is used. If the pressure in the cooker is 1.2 standard atmospheric pressure, the boiling temperature of water in the cooker is higher than 100 ° C. what is the pressure difference between the inner and outer gases on the lid?
(1.2-1) atmospheric pressure units * 700 * 10.0 ^ (- 4) = 0.2 * 101325 * 0.07n = 1418n
When a bottle is filled with water, the total mass is 1.3KG. When it is filled with kerosene, the total mass is 1.14kg?
Know, seek, solve and answer in full
However, I need more detailed answers, so I chose one that I think is the most detailed and most suitable for our requirements.
It is known that: ρ water = 1.0 × 10 & sup3; kg / M & sup3;, ρ oil = 0.8 × 10 & sup3; kg / M & sup3; m bottle + m water = 1.3KG, m bottle + m oil = 1.14kg. Calculate: m bottle set the volume of bottle as V, from the Title Meaning: m bottle + ρ water v = 1.3KG, that is: m bottle + (1.0 × 10 & sup3; kg / M & sup3;) v = 1.3KG
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