On the physics of density and buoyancy The density ratio of ice to water is 9:10, so is the ratio of the volume of ice immersed in water to the total volume 9:10

On the physics of density and buoyancy The density ratio of ice to water is 9:10, so is the ratio of the volume of ice immersed in water to the total volume 9:10

P1v1g = p2v2g, so your statement is correct

Formula of displacement and free falling body motion

The motion of free falling body is a kind of uniform linear motion with zero initial velocity and constant acceleration
Speed formula: final speed = initial speed + speed x time
Instantaneous velocity at the middle moment: (initial velocity + final velocity) / 2, instantaneous velocity at the middle point of displacement = sum of square of initial velocity + square of final velocity divided by the square of 2
2. Average speed formula: average speed = (initial speed + final speed) / 2 (applicable to uniform variable speed linear motion),
Average velocity = displacement (for motion of any nature)
3. Displacement formula: displacement = average velocity x time, displacement = initial velocity x time + (square of acceleration x time) / 2, displacement = (square of final velocity - square of initial velocity) / 2x acceleration,
The velocity of free falling body motion = acceleration of gravity x time, and the displacement of free falling body motion = (the square of acceleration of gravity x time) / 2
If you still have something you don't understand or know, please let me know

In order to shoot a stunt in a TV production center, the stunt actor is required to fall freely from the roof of an 80 m building to a moving car. If the actor starts to fall, the car will move from a standstill to the bottom of the building for 3 s, and then drive to the bottom of the building at a uniform speed. In order to ensure that the actor can fall onto the car safely (regardless of the air resistance, people and cars are regarded as particles, g = 10m / s), the stunt actor is required to fall freely from the roof of an 80 m building, How long does it take for the car to reach the floor? How long does it take for the car to accelerate

Four seconds, three seconds, eight meters per second

On the derivation of x = v0t + 1 / 2at2 and V2 + (V0) 2 + 2aX
The more detailed the better
I don't understand the area yet

On x = V0 T + (1 / 2) a T ^ 2
(sign ^ 2 for square)
The derivation of this formula needs calculus. For those who have studied calculus, it is as easy as 1 + 2 = 3. Since the landlord raised this problem, he should not have studied calculus, right
In the high school physics stage, we should have learned such a basic knowledge:
Then the area of the graph enclosed by x = T0, x = t, X axis and V-T curve is the displacement from t0 to T. If a part of the graph enclosed is below X axis, the area of the part is negative, while the area of the part above X axis is positive
For uniform velocity linear motion,
v = v0 + at
Do V -- t function image, V is Y axis, t is x axis
The V -- t image is a straight line
V0 is the intercept on the y-axis, and let V0 > 0
A is the slope of the line. Suppose a > 0
The figure enclosed by x = 0, x = t, X axis and V = V 0 + at is a right angle trapezoid
The height of the trapezoid is t
The top and bottom of trapezoid is v0
The bottom of trapezoid is V0 + at
The area of trapezoid is (V0 + V0 + at) * t / 2 = v0t + (1 / 2) at ^ 2
So the displacement S = v0t + (1 / 2) at ^ 2
---------------------------------
About: V ^ 2 - V0 ^ 2 = 2As
V = V0 + at
V^2 - V0^2
= (V0 + at)^2 - V0^2
= V0^2 + 2*V0*a*t + a^2 t^2 - V0^2
= 2 V0 a t + a^2 t^2
= 2a[V0 t + (1/2) at^2]
= 2aS

1、 The basic relation v = V0 + at x = v0t + 1 / 2at2 v2-vo2 = 2aX v = x / T = (V0
1、 The basic relation v = V0 + at
x=v0t+1/2at2
v2-vo2=2ax
What do VO2 and 2at2 in v = x / T = (V0 + V) / 2 mean

V = V0 + at; 0 is the subscript
X = v0t + 1 / 2at2; 0 is the subscript, and the following 2 is the square, which should be written as x = v0t + 1 / 2at2;
V2-vo2 = 2aX; the 2 before the equal sign are all squares, which should be written as V & # 178; - V0 & # 178; = 2aX, where 0 is the subscript
V = x / T = (V0 + V) / 2; 0 is the subscript and 2 is the denominator

How to calculate v2-vo2 = 2aX from {v = V0 + at x = v0t + at2 / 2

t=（v-v0）/a
Bring in
x=vo（v-v0）/a+1/2a（v-vo）²/a²
=（vo*（v-vo）/a+1/2（v-v0）²/a
therefore
2ax=2vo（v-v0）+（v-v0）²
=2vo*v-2vo²+vo²+v²-2v*v0
=v²-vo²
We get V & # 178; - V0 & # 178; = 2aX
You see, do you understand? If not,
The most important thing is the method. If we master the method, we can solve similar problems!
Like this question oneself tries under, next time will be able to!

The initial velocity, final velocity, acceleration and displacement of an object moving in a straight line with uniform speed change are V0, V1, a and X respectively. It is proved that V1-V2 = 2aX

Because V1 = V0 + at t = v1-v0 / A
X = v0t = 1 / 2at square x = V0 times v1-v0 / A + 1 / 2A times the square of (v1-v0 / a)
V0 square + V1 square = 2xa

In order to solve 15 calculation problems of physics compulsory one in senior one, 1. Calculate with v = V0 + at, 2. Calculate with x = v0t + at, 2. Calculate with v2-v02 = 2aX
There should be a reward for the quick answers
1. Calculate with v = V0 + at
2. Calculate with x = v0t ＋ & # 189; at2
3. Using v2-v02 = 2aX
5 each

1. Using V = V0 + at to calculate: 1.1. The car runs at a constant speed of 40 km / h, and now accelerates at an acceleration of 0.6 m / S2. How much speed can it reach after 10 s? Analysis: this problem is known as V0, a, t, and VT can be solved by using the speed relationship

How to connect v = V0 + at and x = v0t + 1 / 2at ^ 2 and eliminate t

From v = V0 + at
t=(Vt-Vo)/a
Substituting x = v0t + 1 / 2at ^ 2
x=Vo[(Vt-Vo)/2]+(1/2)a[(Vt-Vo)/a]^2
It is concluded that 2aX = VT ^ 2-VO ^ 2

How to eliminate t in v = V0 + at and x = v0t + 1 / 2at ^ 2
It is concluded that V ^ 2-v0 ^ 2 = 2aX,

t=(v-v0)/a
Substituting x = v0t + 1 / 2at ^ 2
x=v0*(v-v0)/a+1/2*a*((v-v0)/a)^2
The result is v ^ 2-v0 ^ 2 = 2aX