Conversion of calorific value of coal between kcal and Joule

Conversion of calorific value of coal between kcal and Joule

Kcal = kcal = 4184 joules
In the international system of units, it is stipulated that the unit of heat and work is Joule, so the thermal work equivalent is meaningless. However, large calorie is still widely used in some industries

Calculate x-3 / X & # 178; + X divided by (X & # 178; - 6x + 9)

X-3 / X & # 178; + X divided by (X & # 178; - 6x + 9)
=[x-3/x(x+1）]*[1/(x-3)²]
=1/[x(x²-2x-3)]
1/(x³-2x²-3x)

If the two fixed value resistors R1 and R2 are connected with the power supply in some form, the electric power of R1 is 9 watts;
If the two resistors are connected to the power supply in another way, the power of R1 is 16 watts, and the current through R2 is 4 A. then, what are the power supply voltage and the resistance values of R1 and R2? (set the power supply voltage unchanged)

For resistance R1:
Because the second power is higher than the first power, the second power is parallel and the first power is series
For R1, the second power is 16:9 times of the first power, then the voltage obtained by the resistor in the second time is 4:3 times of the first time, in other words, the voltage obtained in the first time is 3 / 4 times of the second time, that is, in series, R1 gets 3 / 4 of the total voltage, of course, R2 gets 1 / 4 of the total voltage, so it is concluded that R1: R2 = 3:1;
Look at the parallel circuit again: the current on R2 is 4a, and the current on R1 is (4 / 3) a of course
From the second R1 power is 16 watts, the equation can be listed as follows:
(4 / 3) ^ 2 * R1 = 16 solution: R1 = 9 Ω, R2 = 3 Ω
Power supply voltage U = R2 * 4 = 12V

When an object moves in a straight line with uniform acceleration, the displacement in the 5th second is 10m, and the displacement in the 7th second is 20m
Using the inverse method, explain why it is not 10m / (1s) & sup2; but 10m / 2 × (1s) & sup2;

The error lies in △ t = at & sup2; the applicable case is the displacement passing through in continuous and equal time. The 5th second and 7th second of this question are not continuous and equal time. So the correct answer is 5m / S & sup2;

The greater the resistance of the appliance, the greater the power, right?

Not necessarily, in some places, P = u * U / R is suitable, and the electric power will decrease with the increase of resistance; in some places, P = I * I * r is suitable, and the electric power will increase with the increase of resistance

A uniform thin rod, m in mass and l in length, with one end fixed and falling freely from the horizontal position,
Then in the horizontal position, the acceleration of the center of mass C is ()
A.g
B.0
C.3/4g
D.g/2

The moment of inertia of the thin rod passing through the rotation axis perpendicular to the rod at the end point J = m (L ^ 2) / 3
From the law of rotation
Moment of gravity M = moment of inertia J * angular acceleration w
And M = MGL / 2
So (1 / 2) MGL = (1 / 3) m (L ^ 2) w
W=(3/2)g/L
The acceleration of the center of mass C is
a=W*L/2=(3/4)g

How many kilos is a ton?

Ton is a commonly used unit of mass, expressed by the English letter (T), 1 ton = 1000 kg. Kilogram is the international unit of mass, expressed by (kg), there are also commonly used units of mass, such as gram (g), milligram (mg), their rate of progress is 1000

The volume is 100cm3 and the weight is 7.6n. What is its density? How buoyant is it when it is completely submerged in water? At this time, if the iron block is hung on the spring scale, what is the reading of the spring scale? G is 10N / kg

(1) M = GG = 7.6n10n / kg = 0.76kg, v = 100cm3 = 1.0 × 10-4m2, ρ = MV = 0.76kg, 1.0 × 10 − 4m3 = 7.6 × 103kg / m3, (2) f floating = ρ water GV drainage = 1.0 × 103kg / m3 × 10N / kg × 1.0 × 10-4 & nbsp; M3 = 1n (3) f reading = G-F floating = 7.6 & nbsp; n-1n = 6.6n

If the perimeter of the rectangle is 16, then the functional relationship between the area y of the rectangle and the length x of one side of the rectangle is

y=x(16÷2-x)=8x-x²(0＜x＜8)
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The mass of a and B are ma = 6kg, MB = 2kg respectively. The dynamic friction coefficient between AB is μ = 0.2,
At the beginning, f = 10N, and then gradually increases. In the process of increasing to 45N, then ()
A. When the tension f
Object a is on B, and f acts on a

Because it is smooth, the friction between B and the horizontal plane is 0
If two objects remain relatively static, they can be regarded as a whole
view in its entirety
2 × Ma = FB to a
F-f=（ma+mb）a
F=（ma+mb）a+0.2×ma
From B to see f '= MB × a
From a to see f '' = f-f '= ma × a = (MA + MB) a
When two objects start to slide
F=6×8=48>45
So AB is wrong
Because there is relative motion only when it is powerful, so C is wrong