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Given that the product of polynomials (x-a) and (x to the second power + 2x-b) does not contain the first and second terms of X, find a and B
(x-a) (the second power of X + 2x-b)
=x^3+2x^2-bx-ax^2-2ax+ab
=x^3+(2-a)x^2-(b+2a)x+ab
So 2-A = 0
b+2a=0
So a = 2
b=-4
The sum of 3x's second power + 5x-2 and another polynomial is x's second power - 2x + 4. Find the difference between the two polynomials
This polynomial is:
(x²-2x+4)-(3x²+5x-2)
=x²-2x+4-3x²-5x+2
=-2x²-7x+6
So the difference between them is:
(3x²+5x-2)-(-2x²-7x+6)
=3x²+5x-2+2x²+7x-6
=5x²+12x-8
Note: the difference between two numbers means that the former minus the latter, so there is only one result
Why is the light bulb with low power in series circuit brighter and the light bulb with high power darker?
The comparison of bulb brightness is based on the "actual power", not the "rated power". In series, two bulbs with the same rated voltage are used. According to P = UI, when the rated voltage is the same, the bulb with higher rated power has lower resistance, and because the current in the series circuit is equal everywhere, the greater resistance is obtained according to P = I square R
If point P (3m-2, m-1) is in the fourth quadrant, then the value range of M is in the fourth quadrant
Point (x, y) is in the fourth quadrant,
Then: x > 0, Y0
m-12/3
m
In a DC circuit, there is an ideal constant voltage power supply, which is connected with a constant resistance R, and the R power is p. if a constant resistance R 'is connected in series, what is the relationship between the total power P and the power p' of R and the resistance of R '?
The voltage of point source is constant, u = PR under root sign. After series connection, the voltage of R 'is (PR under root sign multiplied by R') divided by (R + R '), and the electric power can be calculated
As shown in the figure, the stiffness coefficient of a light spring is 500N / m. one end of the light spring is fixed on an object a with a mass of 10kg. Through the light spring, a horizontal force F is applied
As shown in the figure, the stiffness coefficient of a light spring is 500N / m, one end of which is fixed on the object a with a mass of 10kg. Through the light spring, a horizontal force F is used to pull the object. 1 (when f = 1n, the object does not move. Then how much friction force does the object bear? 2 (when the spring elongation is 4cm), the object just moves in a uniform straight line on the horizontal plane, Find the dynamic friction factor 3 between the object and the horizontal. When the elongation of the spring is 6cm, what is the friction force on the object
In horizontal direction:
1: At rest, tension and friction are balanced, so friction equals tension equals 1n
2: Uniform speed: the tension and friction are balanced, so the friction is equal to the tension, KX = 500N / m * 0.04m = 20n
3: The pulling force is greater than the friction force, and the friction force is constant at this time. The friction force is 20n when sliding
In AC circuit, resistor, inductor and capacitor are connected in parallel to seek current and voltage?
I = u / Z1 / z = 1 / R + 1 / ZL-1 / ZC
On a smooth horizontal plane, there is a stationary car a with mass m, and on the car there is an object B with mass M. pull the car with horizontal force F to the right, and ab remains relatively stationary
B what is the force of right acceleration? What is the direction? How is it provided?
First, find out the acceleration of the same acceleration,
a=F/（m+M）
According to the law of two cattle
f=am
Friction provided by a to B
The direction is the same as f
What is the area of an isosceles right triangle if its base length is 5
s=5*5/2/2=25/4
A light spring is fixed at the upper end and a heavy object is hung at the lower end
When balancing, the spring is extended by 5cm, then the weight is pulled down by 1cm, and then let go. Then the acceleration of the object at the moment of release is (G is 10)
A.2
B.7.5
C.10
D.12.5
How to thank
Equilibrium is determined by equilibrium conditions
mg=kx
among
M -- mass of heavy object
K -- spring stiffness coefficient
x=5cm
So K / M = g / X
Then pull the weight down for 1cm
Combined external force F = K (x + △ x) - Mg
among
△x=1cm
From Newton's second law
ma=k（x+△x）-mg
So acceleration a = [K (x + △ x) - Mg] / M = K (x + △ x) / m-g
=[（x+△x）/x-1]*g
=2
So choose a
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