Kinetic energy theorem 1. The mass m of the vehicle starts from standstill, and reaches the maximum speed VM after moving x along the horizontal road. Assuming that the traction power is constant and the resistance is k times of the vehicle weight, the traction power of the vehicle and the time required for the vehicle to move from standstill to uniform speed can be calculated 2. The object slides from the top a of the high H inclined plane to the top B of the horizontal plane. The horizontal distance of AB is s, and the dynamic friction coefficient between the object and the contact surface is calculated

Kinetic energy theorem 1. The mass m of the vehicle starts from standstill, and reaches the maximum speed VM after moving x along the horizontal road. Assuming that the traction power is constant and the resistance is k times of the vehicle weight, the traction power of the vehicle and the time required for the vehicle to move from standstill to uniform speed can be calculated 2. The object slides from the top a of the high H inclined plane to the top B of the horizontal plane. The horizontal distance of AB is s, and the dynamic friction coefficient between the object and the contact surface is calculated

When the traction and resistance are equal, the speed is the maximum
P = FV = FVM, f = KMG, so p = kmgvm
Pt-fx=1/2*m*Vm^2 t=[(1/2*m*Vm^2)+fx]/P
2. Use the kinetic energy theorem for the whole process, and let the inclination angle of the inclined plane be θ (draw your own picture)
mgh-umgcosθ*h/sinθ-umg(s-h/tanθ)=0
So u = H / s

Calculating normal acceleration and tangential acceleration
The initial velocity of a particle is V, the angle of projection is 60 degrees, and the air resistance is not considered. The normal acceleration and tangential acceleration of the throwing point O, the highest point a and the landing point B are calculated

For the throw out point, the angle between the velocity direction and the horizontal direction is 60 degrees, and the acceleration is vertical downward, so the tangential acceleration is - √ 3G / 2, and the normal acceleration is g / 2; the same is true for the landing point, but the tangential acceleration direction is the same as the velocity direction, which is √ 3G / 2
The tangential acceleration at the highest point is 0, and the normal acceleration is g

How to deduce the formula of normal acceleration of curvilinear motion in College Physics

Derivation of normal acceleration formula of curvilinear motion in College Physics

It's on any college physics textbook. Turn over the books, OK

A = DV / DT, a is the tangential acceleration, V is the velocity
If v = 14-2t, t ∈ [0,7], then a is calculated to be - 2, but a should not have positive and negative. What's the matter?
I'm writing scalar. That is to say, why is the module of loss a positive or negative? In one dimension, the expressions of velocity and velocity are the same when they are greater than 0. Is this the reason?

The orbit of the moon around the earth is close to a circle with a radius of 3.84 × 10 (to the fifth power), and the revolution period is 27.3 days. What is the centripetal acceleration of the moon around the earth?
I think it should be done with the quadratic power of (2 π / T) × R
But the number is too big. Is the centripetal acceleration so big?
Process after 27.3 days of conversion

To be specific, the centripetal acceleration can't be very large, its size is only equal to the acceleration produced by the earth's gravity here, right? How much gravity can there be so far away from the earth!

It is calculated with the centripetal acceleration formula. The topic is as follows: a car runs on a circular track with a radius of 120m, the track inclines inward at an angle of 22 degrees, the weight of the car is 850kg, and the friction coefficient between the tire and the road is 0.7. It is a bit unclear to find out the maximum speed of the car. If the faster the speed is, the more the car inclines to the inside of the track, or the outside of the track? What will the friction change with it?

In the speed range, there is a speed, which is just the size of the car will not slide left and right
mV^2/r=mgtanθ
V=√(grtanθ)
This is the best speed for turning
If the actual speed of the car is greater than the optimal speed, the centripetal force required by the car will increase, mgtan θ can not meet the centripetal force demand, the car will slide outward, and the friction force will point inward to the center of the circle to help provide centripetal force
If the speed of the car is lower than the optimal speed, the centripetal force required by the car will be smaller, and mgtan θ will meet the centripetal force needs. If there is still something left, the car will slide inward, and the friction force will deviate from the center of the circle to help offset some mgtan θ

As shown in the figure, the arc track AB is a 14 circle with radius r in the vertical plane, and the tangent line of the track at point B is horizontal. A particle begins to slide down from the static point a, regardless of the friction and air resistance between the slider and the track. When the speed of the slider reaches B is ν, the acceleration of the particle just reaching point B is______ The acceleration just after sliding over point B is______ ．

When the particle just reaches point B, there is f = mv2r = MA1. According to the kinetic energy theorem: Mgr = 12mv2, the solution is: A1 = 2G. When sliding over point B, the resultant force of gravity and supporting force is zero. According to Newton's second law, the acceleration is zero, that is, A2 = 0. So the answer is: 2G; 0

Please write down the detailed steps and ideas:
The earth can be regarded as a perfect sphere with a radius of 6400km, rotating around an axis connecting the north and south poles. There are two identical objects moving freely in the earth's poles and equator respectively. The acceleration difference of them in the free fall is estimated
I first calculate the angular velocity of the earth rotation = 1.99 E-7 rad / s, and then calculate the centripetal acceleration = 2.54 E-7 M / S ^ 2, but the difference seems too small

V=2pi*R/T,R=6400km=6.4x10^6 m,T=24h=24*3600s=8.64x10^4 s
a=V^2/R=4*pi^2 *R/T^2
=4*3.14*3.14*6.4*10^6/(8.64*8.64*10^8)
=252.41/7464.96
=0.0338m/s^2

Forget the formula of centripetal acceleration
a=rω^2=v^2/r=4π^2r/T^2
On the basis of these formulas, the evolution formula is used to find the physical quantity formula with rotational speed n and frequency f, which believe in acceleration
I forgot to dial it

1. Linear velocity v = s / T = 2 π R / T
2. Angular velocity ω = Φ / T = 2 π / T = 2 π f ω × r = v
3. Centripetal acceleration a = V2 / r = ω 2R = (2 π / T) 2R
4. Centripetal force fcenter = MV2 / r = m ω 2R = MR (2 π / T) 2 = m ω v = f
5. Period and frequency: T = 1 / F
6. Relationship between angular velocity and linear velocity: v = ω R
7. Relationship between angular velocity and rotational speed ω = 2 π n (here the meaning of frequency and rotational speed is the same)
8. Main physical quantities and units: arc length (s): m; angle (Φ): rad; frequency (f): Hz; period (T): s; rotational speed (n): R / S; radius (R): m; linear velocity (V): M / S; angular velocity (ω): rad / S; centripetal acceleration: M / S2
Note:
(1) The centripetal force can be provided by a specific force, a resultant force or a component force, and its direction is always perpendicular to the direction of velocity and points to the center of the circle;
(2) The centripetal force is equal to the resultant force of an object in uniform circular motion, and the centripetal force only changes the direction of velocity, not the size of velocity, so the kinetic energy of the object remains unchanged, the centripetal force does not do work, but the momentum changes constantly