When a certain electric appliance works normally, the electric energy consumed within 15 minutes is 0.3 kW · h, which may be () A. Air conditioner B. refrigerator C. television D. radio

When a certain electric appliance works normally, the electric energy consumed within 15 minutes is 0.3 kW · h, which may be () A. Air conditioner B. refrigerator C. television D. radio

The electric power of electrical appliances is p = wt = 0.3kw · h1560h = 1.2kW. Among the above four kinds of household appliances, the only one with power over 1000W is air conditioner

The factorization of a ^ 2 + 3ab-4b yields

This is , 4b ^ 2
simple form
=(a²+4ab)-(ab+4b²)
=a(a+4b)-b(a+4b)
=(a-b)(a+4b)

How much is a 0.1 ohm resistor that can withstand 1kW power,

I'm afraid you can't buy such a resistor
Do you know how big a 1kW is? It's more than the heating capacity of 40 25W light bulbs (because about 7% of the power of the light bulb is turned into light);
Do you know how much current is required for 0.1 ohm resistance to generate 1kW? 100A!
A resistance of 0.1 ohm is almost equivalent to a short circuit in a circuit. In order for 90% of 1kW power to be borne by a resistance of 0.1 ohm, the total resistance of the circuit must be less than 0.01 ohm

1 2,3,4 5,6,7,8,9 10,11,12,13,14,15,16 …… Seeking the law
It's 234234, it's 56789, it's a triangle. Find the last one in the eighth column and the law
one
two hundred and thirty-four
fifty-six thousand seven hundred and eighty-nine
10，11，12,13,14,15,16，
……

It is not difficult to see that the last number can form a sequence
1,4,9,16,25.
a1=1,a2=4,a3=9,a4=16,.an
a2-a1=3=
a3-a2=5
a4-a3=7
.
.
.
an-a(n-1)=2n-1
Add up to get an
an-a1=3+5+.+(2n-1)
an=n(1+(2n-1))/2
∴an=n^2
a8=64

In series and parallel circuits, what is the relationship between the light and dark of the bulb and the resistance

In a parallel circuit, the higher the resistance, the lower the current
The light and dark of the bulb is proportional to the electric power
The higher the resistance is, the lower the electric power is. P = UI, the darker the lamp is (U is certain)
In a series circuit, the greater the resistance, the greater the voltage at both ends of the resistance
The higher the resistance is, the greater the electric power is, and the brighter the lamp is

Find X in the following expressions: ① X & # 178; = 361, ② X & # 178; + 1 = 1.01, ③ (4x-1) &# 178; = 225, ④ 2 (X & # 178; + 1) = 10

X1 = root 361, X2 = - root 361
x1=0.1,x2=-0.1
x1=4,x2=-3.75
x1=2,x2=-2

An ammeter, internal resistance R = 100 ohm, full bias current I = 3mA, to refit it into a 3V range voltmeter, how much resistance should be connected in series? What is the internal resistance of the voltmeter

900,1000

When k is a value, the solution of the equation 34 + 8x = 7K + 6x about X is 6.5 times larger than that of the equation K (2 + x) = x (K + 2) about X

The solution of the equation 34 + 8x = 7K + 6x is: x = 28K − 38; the solution of the equation K (2 + x) = x (K + 2) is: x = k, according to the meaning of the problem, 28K − 38-k = 6, the solution is k = 5120

When a section of resistance wire with uniform thickness is drawn to 1 / N of the original diameter, what is the resistance?

When r = PL / s, the diameter becomes 1 / N and s becomes 1 / N2
The volume does not change, SL does not change, so l becomes the original N2
So the resistance becomes N4 (the fourth power of n)

F (x) is continuous on [0,1] and differentiable in (0,1). F (0) = f (1) = 0, f (1 / 2) = 1. It is proved that there is at least one point in (0,1) such that f '(x) = 1

Let g (x) = f (x) - x, G (0) = 0, G (1) = - 1, G (1 / 2) = 1 / 2. From the intermediate value theorem (here it can also be the zero point theorem), we can see that there is a point between X = 1 / 2 and 1 that makes g (x) equal to 0. From Rolle's theorem, we can see that there is a point on (0,1) that makes G '(x) = 0, then G' (x) = f '(x) - 1 = 0, that is, f' (x) = 1