Focus of parabola y = 4 (x) squared

Focus of parabola y = 4 (x) squared

A:
y=4x^2
x^2=y/4=2py
2p=1/4
p=1/8
p/2=1/16
So the focus is (0,1 / 16)

How many tons per square meter is 120kPa? How to calculate the key?

1pA = 1n / m2
120kPa = 120000n / m2 = 12244.9kg/m2 = 12.2449t/m2

The square of the parabola y is equal to the distance from the focus of 4x to the collimator (given condition)
The square of the parabola y is equal to the distance from the focus to the collimator of 4x. The square of the parabola y is equal to the distance from the focus to the collimator of 4x

y²=4x
Focal length = 4 / 2 = 2
That is, the square of the parabola y is equal to 4x, and the distance from the focus to the collimator is 2

A light bulb printed with "pz220 25" connected to 110V voltage circuit, what is the power?

1, r = u & sup2 / P is obtained from P = UI and I = u / R, and the resistance of the filament is calculated: RL = uo & sup2 / PO = (220 V) & sup2 / 25W = 1936 Ω,
2, P = u & sup2 / R is obtained from P = UI and I = u / R to calculate the actual power: P '= u & sup2 / RL = (110V) & sup2 / 1936 Ω = 6.25w

If the mass of the car is m and the horizontal curve is a circular arc with a radius of 50m, the maximum static friction between the car and the ground is 0 or 2 times of the weight of the car, a
If the mass of the car is m and the horizontal curve is a circular arc with a radius of 50m, the maximum static friction between the car and the ground is 0 or 2 times of the weight of the car. What is the maximum speed of the car when it is bent to make the car not scratch?

When the car bends to the maximum speed, the maximum static friction between the car and the ground provides centripetal force
0.2mg=mv^2/r
v=(0.2gr)^0.5=10m/s

The area of an isosceles right triangle is 10. Find the right side length of the triangle

Let the length of the right angle be a,
A & sup2 / 2 = 10, a = radical 20 = 2, radical 5
Right angle side length 2 root sign 5

Two small balls with different mass are suspended in the car. The ball on the top is heavier than the ball on the bottom. When the car turns to the right
When making uniform acceleration motion, the following figures Why does this question propose that the quality of the upper ball is greater than that of the lower ball? If it is less than that, what will happen?

First of all: the answer is that the three points and two lines are in the same line, leaning to the left
Secondly: why does this question propose that the quality of the upper ball is greater than that of the lower ball? If it is less than that, what will happen?
This is a kind of habitual thinking of the problem maker, which makes you feel that the quality of the two balls should be different. In this way, you have to set up two unknown qualities, otherwise it's meaningless for everyone to solve it, or it's a gimmick. In fact, it's a trap. The quality of the upper ball is greater than that of the lower ball, or the quality of the lower ball is greater than that of the upper ball, or the quality of the two balls is equal. It's the same result
Why?
First of all, the holistic analysis: the two balls are regarded as a whole, that is to say, the two balls are regarded as a ball. The ball is only subject to tension and gravity. The resultant force produces acceleration to the right. According to the composition and decomposition of the force, I think you should understand that the upper rope should tilt to the left. The premise of holistic analysis is that the acceleration of the two balls must be the same, that is, the same state of motion
Then, in the isolation method, let's first set the upper ball a, mass m, and the lower ball B, mass m, assuming that two ropes are in the same straight line
First, analyze B
And then analyze a
Stress analysis diagram, and then decomposition
It can be seen from the above that the acceleration has nothing to do with the mass of the object

Use some of the four symbols to make the following calculation true
1（）2（）3（）5＝10

1（X）2（+）3（+）5＝10
[1（-）2（+）3]（X）5＝10

Under the action of 5N horizontal tensile force, a block of wood advances 2m along the tensile force direction in 10s on the horizontal plane. How many J? 2 and W? Is the work done by the tensile force?

Let Sn be the sum of the first n terms of the arithmetic sequence {an}, if s3s6 = 13, then s6s12 = ()
A. 310B. 13C. 18D. 19

Let the first term of the arithmetic sequence {an} be A1 and the tolerance be d. from the summation formula of the arithmetic sequence, we can get s3s6 = 3A1 + 3d6a1 + 15d = 13, A1 = 2D and D ≠ 0, s6s12 = 6A1 + 15d12a1 + 66d = 27d90d = 310