The relationship between the displacement of a particle and time is x = 4T + 2t2. If the units of X and T are m and s respectively, the initial velocity and acceleration of the particle are () A. 4m / s and 4m / S2B. 0 and 4m / S2C. 4m / s and 2m / S2D. 4m / s and 0

The relationship between the displacement of a particle and time is x = 4T + 2t2. If the units of X and T are m and s respectively, the initial velocity and acceleration of the particle are () A. 4m / s and 4m / S2B. 0 and 4m / S2C. 4m / s and 2m / S2D. 4m / s and 0

The displacement formula of uniform variable speed linear motion is x = v0t + 12at2, and the relationship between the displacement of particle motion and time is x = 4T + 2t2. By comparison, the initial velocity of the object is v = 4m / s, and the acceleration is a = 4m / S2

The relation of displacement time of a particle is s = (4T + 2T & sup2; + 12) M,

If the relation of displacement time of a particle is s = (4T + 2T & # 178; + 12) m, then its initial velocity VO = 4m / s and acceleration a = 4m / S ^ 2

The relationship between the displacement of a particle and time is: S = 4t-4t2, the unit of S and t is m and s respectively, then the initial velocity and acceleration of the particle are
Process, thank you

The formula is s = VT + 1 / 2at2, compared with your equation, initial velocity 4, acceleration - 8

In the triangular prism abc-a'b'c ', ab ⊥ side bb'c'c, BC = 1, ∠ BCC' = π / 3, BB1 = 2 are proved
1,C1B⊥ABC
2. Try to determine an e (excluding c.c1) on CC1 so that EA ⊥ EB1

1, because ab ⊥ side bb'c'c, so ab ⊥ c'b (because a line is perpendicular to a face, so this line is perpendicular to any straight line on this face), and because, ∠ BCC '= π / 3 = 60 °, BC = 1, BB' = 2, so c'b ⊥ BC (according to the characteristics of right triangle, cos60 ° = 1 / 2 = BC / CC '), because c'b ⊥ BC, a

As shown in the figure, in the triangular prism △ abc-a1b1c1, ab ⊥ side bb1c1c, known BC = 1, BB1 = 2, ∠ bcc1 = π / 3
(I) verification: C1b ⊥ plane ABC;
(II) try to determine the position of a point E on edge CC1 (excluding endpoint C, C1), so that EA ⊥ EB1 (require explanation)
(III) under the condition of (II), if AB = √ 2, find the tangent of plane angle of dihedral angle a-eb1-a1

(I) ∵ abc-a1b1c1 is a triangular prism ∵ CC1 = BB1 = 2 ∵ BC = 1, ∵ bcc1 = π / 3 ∵ BC1 = √ 3 ∵ BC & # 178; + BB & # 178; 1 = CC & # 178; 1 ∵ C1b ⊥ BC ∵ ab ⊥ side bb1c1c, C1b ∩ ab = B ∩ C1b ⊥ plane ABC (II) take the midpoint of CC1 as e, the midpoint of BB1 F

The expanded side view of a regular quadrangular prism is a rectangle with two and four sides. Find its area v
A square cast iron box with cover, the length of each outer edge is 26 cm, and the wall thickness is 0.15 cm. The specific gravity of cast iron is 7.2 g / cm3, and the weight of the box is calculated
In a cuboid, the areas of three faces passing through the same vertex are 3, 9 and 12, respectively
There are three song titles

The expanded side view of a regular quadrangular prism is a rectangle with two and four sides. Find its area v
(1) 2 is the perimeter of the bottom, then the side length of the bottom is 1 / 2, s = 1 / 4, v = sh = 1 / 4 * 4 = 1
(2) 4 is the perimeter of the bottom, then the side length of the bottom is 1, s = 1, v = sh = 1 * 2 = 2
A square cast iron box with cover, the length of each outer edge is 26 cm, and the wall thickness is 0.15 cm. The specific gravity of cast iron is 7.2 g / cm3, and the weight of the box is calculated
Outer ring volume = 26 ^ 3 = 17576
Inner space volume = 25.85 ^ 3 = 17273.55
Wall volume = 242.45
Weight of iron box = 242.45 * 7.2 = 1745.63
In a cuboid, the areas of three faces passing through the same vertex are 3, 9 and 12, respectively
Let three edges passing through the same vertex be a, B and C respectively
be
ab=3
bc=9
ca=12
If you multiply them, you'll get
(abc)^2=3*9*12=324=18^2
The volume of cuboid v = ABC = 18
abc=18 ab=3 c=6
bc=9 a=2
ca=12 b=9/2
Length of diagonal L = √ (a ^ 2 + B ^ 2 + C ^ 2) = 13 / 2

The sides of a triangular prism are all rectangles. Can we judge whether it is a straight triangular prism?

If the bottom surface is abc-a'b'c ', then the three sides abb'a', acc'a ', and bcc'b' are all rectangular. So AA 'is vertical to AB, and AA' / / BB ', BB' is vertical to BC, so AA 'is vertical to BC. Because AA' intersects AB and BC vertically at the same time, AA 'is vertical to ABC

The rectangle of 8cm long and 4cm wide is folded into the side of a triangular prism, making the bottom surface an equilateral triangle, and calculating the bottom area (2 possibilities)

Choose one side as the perimeter of the bottom,
1、 When the length is the bottom, 8 / 3 is the side length of the bottom, so the area is (2 and radical 3) / 3;
2、 When the width is the bottom, 4 / 3 is the side length of the bottom, so the area is (4 and radical 3) / 9;
The area formula of an equilateral triangle s = a square times (root sign 3) / 4 A is the side length of an equilateral triangle

It is known that the side area of a Mitsubishi prism with regular triangle bottom area and rectangular sides is 54cm ^ 2, and its volume is 45 times of root sign 3. The height and side length of the prism bottom are calculated

The answers are as follows:
Let the side length be x and the height be y. 3 * x * y = 54 (radical 3) / 4 * x * x * y = 45 times (radical 3)
The solution is x = 10 / 3, y = 27 / 5;

(2012 Hainan Mathematics) as shown in the figure, in the straight triangular prism abc-a1b1c1, AC = BC = 1 / 2aa1, D is the midpoint of edge Aa1, DC1 ⊥ BD
(1) Proof: DC1 ⊥ BC
(2) Find the size of dihedral angle a1-bd-c1
Vector normal vector method
Vector method
Vector method
Vector method
Vector method

(1) It is proved that: in RT △ DAC, ad = AC, ∩ ADC = 45 ° similarly: ∩ a1dc1 = 45 °, ∩ cdc1 = 90 °; DC1 ⊥ DC, DC1 ⊥ BD ∩ DC = D ∩ DC1 ⊥ face BCD ≁ BC ∩ 8834; face BCD ≁ DC1 ⊥ BC (2) ≁ DC1 ⊥ BC, CC1 ∩ CC1 = C1, ∩ BC ⊥ face acc1a1, ∩ AC ⊂ face