Zhao Yang had an electric heater in his home. He wanted to know how much power the electric heater had when it was working. He turned off all the other appliances in his home and only let the electricity run Zhao Yang has an electric heater at home. He wants to know how much power the electric heater has when it works. He turned off all other electrical appliances in his home and only let the electric heater work. After 3 minutes, he observed that the electric energy meter turned 12 turns, and the words "220V 5A 300r / kW. H" were on the electric energy meter of his home. Please help him figure out the power of the electric heater? Find the formula

Zhao Yang had an electric heater in his home. He wanted to know how much power the electric heater had when it was working. He turned off all the other appliances in his home and only let the electricity run Zhao Yang has an electric heater at home. He wants to know how much power the electric heater has when it works. He turned off all other electrical appliances in his home and only let the electric heater work. After 3 minutes, he observed that the electric energy meter turned 12 turns, and the words "220V 5A 300r / kW. H" were on the electric energy meter of his home. Please help him figure out the power of the electric heater? Find the formula

Record the value of the electric energy meter. After a period of time, observe the value of the electric energy meter. Divide the increased work by the time (seconds) to get the electric power

As shown in the figure, a uniform thin rod with mass m and length L rotates on the axis at o-point, from rest to the vertical position at the angle of θ
The angular velocity of the uniform rod m and the velocity of the object m after collision are calculated

First, the gravitational potential energy of the uniform rod is transformed into kinetic energy, and the velocity of the rod in the vertical position is calculated
The following is a collision problem. Because it is an elastic collision, we can use the invariant kinetic energy and the conservation of angular momentum. Pay attention to j of a uniform thin rod and its angular kinetic energy. Two equations, two unknowns, can be solved

How many kilos is equal to one ton

1000

An iron ball with a volume of 100cm3 is hung on the spring scale. If the iron ball is immersed in water, the indication of the spring scale is 5N. Find out: (1) buoyancy of the iron ball; (2) gravity of the iron ball; (3) is the iron ball hollow or solid? If it is hollow, what is the volume of the hollow? (known as ρ Fe = 7.9 × 103kg / m3, g = 10N / kg)

(1) Because the metal ball with volume of 100cm3 is completely immersed in water, at this time v matter = V row = 100cm3 = 10-4m3, f float = ρ g, V row = 1.0 × 103kg / m3 × 10N / kg × 10-4m3 = 1n. (2) from F float = G-F pull, g = f float + F pull = 1n + 5N = 6N. (3) volume of iron in iron ball: V iron = m ρ = g ρ g = 6n7.9 × 103K

If the perimeter of a rectangle is 12cm, find its area SCM & # 178; find the functional relationship between it and one side length xcm, and find the area of the rectangle when one side length is 2cm

Since the length of one side is x, the length of the other side = (12-2x) / 2 = 6-x
Area s = x (6-x)
When the length of one side is 2,
S=2×(6-2)=8cm²

The dynamic friction coefficient between the object and the horizontal plane is 0.2, and the maximum static friction and dynamic friction are determined by
When the box with a mass of 20kg is still on the horizontal plane, the dynamic friction coefficient between the object and the horizontal plane is 0.2, and the maximum static friction and dynamic friction are regarded as equal. The force F acts on the box. When f = 20n, f = 48.5n and along the horizontal direction, what is the friction between the ground and the object?

Maximum friction = MGU = 20kg * 10N / kg * 0.2 = 40n
When the force is 20n, it is less than the maximum friction, so the friction is equal to the horizontal force, which is 20n
When the force is 48.5n, it is greater than the maximum friction, so the friction is the maximum friction, which is 40n

-How much is 1.25 times 0.4
What is 27-19 + 4
What is 4 + 32 + 76-52

The result of multiplication or division between a negative number and another negative number is a positive number, and the result of multiplication or division between a negative number and a positive number is a negative number

The ball with mass m moves in a circle in a vertical circular orbit with radius R. when the ball passes through the lowest point of the orbit at a certain time, the pressure on the orbit is 7Mg. When it reaches the highest point after half a cycle, it can just pass through the highest point. In this process, the work of the ball overcoming the orbital resistance is ()
A.mgR/8
B.mgR/4
C.mgR/2
D.mgR

Just passing the highest point means that at the highest point, the orbit has no support for it. Just remember that the centripetal force is completely provided by gravity. Let's set the ball speed at v2
Mg = MV2 square / R
At the lowest point of gravity, the resultant force of the downward supporting force and the upward supporting force is equal to the centripetal force
F support - Mg = MV1 square / R
The work done by the resistance is equal to the kinetic energy difference between the two states
W = 1 / 2mv2-1 / mv1-2mgr (gravity work)
The solution is w = C

If a1 + a3 + A8 = A4 ^ & # 178;, then the maximum value of a5.s4 is
If a1 + a3 + A8 = A42, then
The maximum value of a5.s4 is______________ .
The answer is 36, but I get 75 / 2,

a1+a3+a8=3a1+9d=3﹙a1+3d﹚=﹙a1+3d﹚²
a4=a1+3d=3
s4=﹙3+a1﹚*4/2=2﹙3+a1﹚=6+2a1=12-6d
a5=a4+d=3+d
a5*s4
=﹙3+d﹚﹙12-6d﹚
=36+12d-18d-6d²
=36-6d²-6d
=-6﹙d+0.5﹚²+75/2
an＞0
a1=3-3d＞0
d＜1
a1+﹙n-1﹚d＞0
-a1＜﹙n-1﹚d
d＞a1/﹙1-n﹚

When the car runs at the speed of 18m / s, after the emergency brake, it makes a uniform deceleration linear motion, and its acceleration is 6m / S2 (square)
When the car is running at the speed of 18m / s, after the emergency brake, it makes a uniform deceleration linear motion, and its acceleration is 6m / S2 (square). 1. Find out how much time the car stops after. 2. Find out the car's speed at the end of 2S. 3. Find out the car's displacement after braking for 6S?
Urgent, urgent,

1. V = V0 + at, a = - 6m / S2, V0 = 18m / s, v = 0, so t = 3S
2. V = V0 + at, a = - 6m / S2, V0 = 18m / s, t = 2, so v = 6m / S2
3. From the first question, we can see that the car stops in 3 seconds, so s = v0t + 2 / 1At ^ 2, t = 3, so s = 27m