How to get to the east gate of the Olympic Center in the new cit

1. The South extension line on the first floor is to Jiangning, take Line 1 to the end, Olympic west gate
line 2 can reach the east gate of Olympic sports
both stations are on the way to the Olympic Sports Center.

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4. When the object moves at a constant speed on the inclined plane, the inclined plane is not subject to friction. When the object accelerates on the inclined plane, if the mass of the object is m, the acceleration is a, and the inclination angle of the inclined plane is B, then the horizontal acceleration is acosb, and the friction force is macosb. The direction is along acosb. If the object decelerates along the inclined plane, the direction is opposite, and the size is the same as above

5. When an object is stationary on an inclined plane, the upward static friction is balanced with the component force of gravity along the inclined plane. If the static friction is equal to mgsin30 = 25N

sliding at a constant speed, the upward sliding friction along the inclined plane is balanced with the component force of gravity along the inclined plane. That is to say, the sliding friction is equal to mgsin30 = 25N. From the vertical slope direction, the supporting force is balanced with the component force of gravity in the vertical slope direction. The supporting force is equal to mgcos30. The pressure n of the object on the inclined plane is equal to the supporting force
then μ N= μ mgcos30=mgsin30， μ= Tan30 = sqrt (3) / 3
[Junlang lieying] team answers for you

6. Analysis: it is assumed that there are two inclined planes with inclination angles of 0 α、β, The friction coefficient is 0 μ. If the object with gravity g is placed on the two inclined planes, it can slide at a constant speed and the friction force is equal μ Gcos α＝μ Gcos β. So there is: cos α＝ cos β.
because α、β Should be less than 90 °. It must be: α＝β
that is to say, if the slope satisfies the condition, the inclination angle is unique

7. 1. The object is supported by gravity and vertical slope, and the resultant force is downward along the slope. According to Newton's second law, f = mgsin θ= Ma, so a = GSIN θ
2. The object is subject to gravity, the supporting force of the vertical inclined plane and the upward friction along the inclined plane. According to Newton's second law, f = mgsin θ- umgcos θ= Ma, so a = GSIN θ- ugcos θ< According to Newton's second law, f = mgsin θ+ umgcos θ= Ma, so a = GSIN θ+ ugcos θ

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