As shown in the figure, one end of the light rod with length L = 0.5m is fixed with a small ball with mass m = 0.1kg, and the other end is fixed on the rotation axis O. the small ball rotates uniformly on the horizontal plane around the axis, and the light rod turns 30 degrees every 0.1s. (π ^ 2 = 10) (1) try to find the centripetal acceleration of the small ball. (2) centripetal force of the small ball

As shown in the figure, one end of the light rod with length L = 0.5m is fixed with a small ball with mass m = 0.1kg, and the other end is fixed on the rotation axis O. the small ball rotates uniformly on the horizontal plane around the axis, and the light rod turns 30 degrees every 0.1s. (π ^ 2 = 10) (1) try to find the centripetal acceleration of the small ball. (2) centripetal force of the small ball

1)
R = 30 degree angle = open / 6
W = R / T = (open / 6) / 0.1 = 10 open / 6
Centripetal acceleration a = (W ^ 2) l = 500 / 3 = 166.7rad/s
2)
Centripetal force F = ma = 0.1 * 166.7 = 16.67n

Four and seven fifths equals () tons () kilograms?
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Four and seven fifths equals (4) tons (140) kilograms
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The weight of the solid iron block hanging at the lower end of the spring scale is 7.8n. When all the iron blocks are immersed in water, the buoyancy of the iron block is 1n, then the reading of the spring scale is -- n
When half of the iron block is immersed in kerosene, the reading of the spring scale is - N, and the gravity of the arranged kerosene is - n

(1)
Spring scale reading = gravity buoyancy = 7.8n - 1n = 6.8n
(2)
The specific gravity of kerosene is 0.8,
The weight of half volume kerosene is 0.5 * 0.8 * 1n = 0.4N
The gravity of the separated kerosene is 0.4N
The buoyancy of the iron is 0.4N
Spring scale reading = gravity buoyancy = 7.8n - 0.4N = 7.4n

Write out the following functional relations respectively, and point out the constants, independent variables and functions in them. (1) the relationship between the length of the triangle bottom is 4, the height is h, and the area s and the height H

Write out the following functional relations respectively, and point out the relationship between constant, independent variable and function in them

Two rectangular objects with mass ma and MB are stacked together and slide along the inclined plane with inclination angle of θ. It is known that the dynamic friction coefficients between a and B, B and the inclined plane are μ 1 and μ 2 respectively. The friction force between B and a is calculated

Equation:
mA*g*sin(θ)-μ1*mA*g*cos(θ)=mA*a
mB*g*sin(θ)+μ1*mA*g*cos(θ)-μ2*[mB*g*cos(θ)+mA*g*cos(θ)]=mB*a
Fab = μ 1 * ma * g * cos (θ) diagonally downward

The operation order of the four mixed integers, decimals and fractions

1. If there are brackets, count the one inside the brackets first, and then the one outside the brackets
2. First calculate multiplication and division, then add and subtract
3. If the same level is in the same block, the first one will be counted

As shown in the figure, a disc can rotate around a vertical axis passing through the center of the disc and perpendicular to the surface of the disc. A wooden block is placed on the disc. When the disc rotates at a constant speed, the wooden block moves with the disc, then ()
A. The direction of the friction between the wood block and the disk is opposite to that of the wood block. The direction of the friction between the wood block and the disk is opposite to that of the wood block

The force analysis of wood block shows that the wood block is affected by gravity, supporting force and friction force. Gravity is vertical downward, supporting force is vertical upward, gravity and supporting force are both in the vertical direction. These two forces are balance forces, and only friction force is the centripetal force of circular motion of the object, so the direction of friction force should point to the center of the circle, so B is correct, a, C and D are wrong B

In the arithmetic sequence {an}, the sum of the first six terms is 60, and A6 is the equal proportion middle term of A1 and A21. Find the general term formula an and the first n terms and Sn of the sequence {an}

a1+a2+a3+a4+a5+a6=60
And a1 + A6 = A2 + A5 = A3 + A4
So a1 + A6 = 20
a6²=a1×a21
The equation of A1 and D is
2a1+5d=20
(a1+5d)²=a1(a1+20d)
By solving the equations, A1 = 5 and d = 2 are obtained
So the arithmetic sequence an = 3 + 2n
Sn=n(n+4)

(1 / 2) a car is driven at a constant speed of 10m / s for 10s, and then at a constant acceleration of 1m / S2 for 10s
(1 / 2) a car is driven at a constant speed of 10 m / s for 10 s, and then at a constant acceleration of 1 m / S2 for 10 s, to find out (1) what is the displacement of the car in the 20 s? (2) how large is the displacement of the car

S1=v0t1=10*10=100m
S2=vot+1/2at2^2=10*10+1/2*1*10^2=150m
Displacement within 20s s = S1 + S2 = 100 + 150 = 250m
20s terminal velocity V2 = V0 + at2 = 10 + 1 * 10 = 20m / S
Average velocity in 20s = s / (T1 + T2) = 250 / 20 = 12.5m/s

From home to school, it takes 15 minutes for Xiaohong and 20 minutes for Xiaoli. Xiaohong's speed is () percent faster than Xiaoli's

（20-15）/20=25%