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Seventh grade math problems (may be easy for you) 1. There are 10 bags of rice, the standard mass of each bag is 50kg, the number exceeding the standard mass is positive, and the number less than the standard mass is negative. The records of a spot check (unit: kg) are as follows: + 0.5, + 0.3, - 0.2, - 0.3, + 1.1, - 0.7, 0, - 0.2, + 0.6, + 0.7, the average mass of these 10 bags of rice is_____ Kg, the total mass is______ kg. 2. The operation steps of adding two rational numbers are generally divided into two steps______ First, determine the sum________ And then determine and________ . 3. The positions of rational numbers a and B on the number axis are shown in the figure (a on the left side of 0, the distance is about "" 0, B on the right side of 0, the distance is about "") （1）a+b_____ 0； （2）a-b_____ 0； （3）-a+b______ 0； （4）（-a）+（-b）______ zero 4. If the sum of two numbers is positive, then () A. These two numbers are both positive numbers B. these two numbers are both negative numbers C. two numbers have different signs, and the absolute value of positive number is smaller than D. at least one of the two numbers is positive 5. The following statement is correct () A. For any rational number, if a + B = 0, then a = B. for any rational number, if a = B, then a + B = 0 C. for any rational number, if a ≠ 0, B ≠ 0, then a + B ≠ 0 D. if the sum of two rational numbers is positive, then both rational numbers must be positive 6. If we know that a = 5 and B = 2, then the value of a + B is () A. 3 or - 3 B.7 or - 7 C.3 or 7 d. ± 3 or ± 7 All the above are in the addition of rational numbers
1.50.18 501.8
2.2 sign absolute value
3.＜ ＜ ＞ ＞
4.D
5.A
6.D
Seventh grade is very simple mathematics problem, I am in a hurry to help!
Fill in the blank: 37 ° 20'42 ″=______ 53.32°=___ °___ ′___ 〃
Calculation (main process): 1. 42 ° 17 ′ - 20 ° 3 ′ + 35 ° 25 ′
2： (98 ° 12 ′ 25 ″ 5 + 20 ° 5 ′ 16 ″) × 2
Thank you. I'm in a hurry!
37 ° 20 ′ 42 ″ = 37.345 o 53.32 ° = 53 ° 19 ′ 12 ″: 42 ° 17 ′ - 20 ° 3 ′ + 35 ° 25 ′ = (42-20 + 32) ° (17-3 + 25) ′ = 54 ° 39 ′ 2: (98 ° 12 ′ 25 ″ 5 + 20 ° 5 ′ 16 ″) × 2 = [95 ° (3 * 60 + 10) ′ (2 * 60 + 25) ″ 5 + 20 ° 5 ′ 16 ″] * 2 = [19 ° 38
A small ball is thrown from a high building 15m above the ground at a speed of 10m / s. The air resistance is ignored and the ground is taken as the zero potential energy surface. The gravitational potential energy and kinetic energy of the ball are equal at different heights. G = 10m / s
Initial kinetic energy 1 / 2mV * V = 50m potential energy MGH = 150m
According to the conservation of energy, the potential energy of gravity is equal to kinetic energy, MGH = 100m, H = 10m
A physics problem, need formula, process
When an object with a mass of 0.5 kg is thrown up vertically, the resistance in the process of motion is opposite to the direction of motion. If the resultant force of the object in the process of rising is 8 N, the magnitude and direction of the resistance in the process of rising are calculated?
Gravity g = mg = 0.5 * 10 = 5N
F = G + F
F resistance = f combination - g = 8-5 = 3N
The resistance is in the same direction as G, downward
Write the first question
(12) a crane will lift the cargo of 3600n for 4m. If the mechanical efficiency of the crane is 60%, the useful work of the crane is_________ J. The total contribution is_________ J. What's the bonus_________ J.
(11) in the process of spraying water to the flowers and trees on both sides of the road, the green sprinkler moving at a constant speed on a section of straight road has its kinetic energy_________ This is because___________________________ .
W = GH = 3600n * 4m = 14400j
Wtotal = w useful / η = 14400j / 60% = 24000j
W extra = w total - W useful = 24000j-1440j = 9600j
reduce
Although the speed does not change, the total mass of the car and water decreases (the kinetic energy is related to the mass of the object and the speed of movement)
What if you have to use it and there is no formula to apply? Let's take an example to see what the formula is?
1 M & sup3; water contains 3.35 × 10 & sup2; & sup2; water molecules, and the volume of a drop of dew is 2 × 10 - & sup2; CM & sup3;. How many water molecules does it contain?
Just a formula. And
We must use formulas to solve physics. (I am a physics major, at least in my undergraduate course, but I think it's the same with my master's and doctor's degree.)
2 × 10 - & sup2; * 0.000000 1 * 3.35 × 10 & sup2; & sup2;, that's it
Change the volume of dew into cubic meters, and then multiply it by the number of molecules per cubic meter
Junior three physics problems, speed, requires a detailed solution process and formula, thank you
The mass of the hollow copper ball with the volume of 30cm & sup3; is 89g, and the total mass of the ball is 105g after filling the hollow core with some liquid
The density of copper is 8.9g/cm & sup3;, so the volume of 89g copper is 10cm & sup3;
The volume of the added liquid is 30-10 = 20cm & sup3; and the mass is 105-89 = 16g
So density = mass / volume = 16 / 20 = 0.8g/cm & sup3;
The "economic fuel consumption (90km / h) 8L / 100km" of a car means that the car runs 100km at 90km / h and consumes 8L of gasoline
Q: the efficiency of the car is 31.25% when the car runs 25km in a straight line with economic fuel consumption. How many liters of gasoline is consumed and the power of the car (the calorific value of gasoline is 3.2 * 107 J / L)
According to the fact that the car runs 100km at the speed of 90km / h, the consumption of gasoline is 8L
It can be seen that the actual fuel consumption v = (8 / 100) * 25 = 2L when the vehicle runs 25km with economic fuel consumption
Heat release of 2l gasoline after complete combustion q = QV = 3.2 * 10 ^ 7 * 2 = 6.4 * 10 ^ 7J
Work done by car w = η q = 31.25% * 6.4 * 10 ^ 7 = 2 * 10 ^ 7J
∵V=90km/h=25m/s
The driving time t = s / v = 25000 / 25 = 1000s
∴P=W/t=2*10^7/1000=2*10^4W
A: the car consumes 2L gasoline; the power of the car is 2 * 10 ^ 4W
The heat calculation formula of junior high school physics should be solved in detail
① After a certain process, the temperature change is △ T, which absorbs (or releases) the heat,
Q=c×m×△t．
Q suction = C × m × (t-t0)
Q amplifier = C × m × (t0-t)
(t0 is the initial temperature; t is the final temperature)
Where C is the specific heat (capacity) associated with this process
formula
The cylinder piston area of a single cylinder four stroke gasoline engine is 30cm2 power, the distance of a stroke piston moving in the cylinder is 50mm, the average pressure of the gas in the full load working stroke is 9.0 * 10 power PA, and the flywheel rotates 1800 cycles in 1 min. when the gasoline engine works at full load (without friction), what is the power of the gasoline engine?
(1) average pressure of gas to piston:
F=Ps=9.O×105Pa×30×10-4m2=2.7×103N
(2) Work done by gas to piston: w = FS = 2.7 × 103n × 50 × 10-3 M = 135j
(3) The number of times the gas works on the piston in one minute: n = 1800 / 2 = 900 (Times)
Power of gasoline engine: P = wtotal / T = 135 × 900 J / 60 s = 2025w
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