Physics problems in the first year of high school. Physics problems in the first year of high school

When f=6n, a=2m s2 f=ma=4 so friction f=6-4=2 f=12 resultant force=12-2=10 so a=f m=10 2=5

Since they are on the same horizontal plane, the coefficient of friction between the object and the plane does not change, that is, the friction force does not change. Therefore, the right known condition deduces the friction force (f-f=ma, f=2n) and then finds a'（f'-f=ma',a'=5m/s2）

According to the formula f=ma because there is friction in the plane of pulling, so the formula is f-f=ma f=umg f is the force you pull m is the mass ah is the acceleration through the first known acceleration to find the friction factor between the object and the contact surface, you can find the acceleration when f = 12.

The magnitude of friction is 2 N, so a=5m s2

f-f=am

Substituting f=6n, a=2m s2 gives f=2n

Thus, when f=12n, a=5m s2

Because "when f=6n, a=2m s2", the frictional force f=f-ma=6-2*2=2n

When the file f=12,,a=(f-f) m=10 2=5m s 2

(1) If the magnitude of the two components that are 120° to each other is the same, then the direction of their resultant force is the direction of the angular bisector, and the magnitude is equal to the component force.

2) Three forces of equal magnitude at 120° to each other, their net force is zero.

(1) The magnitude of the resultant force is 10N, and the direction is on the angular bisector of the angle of F1F2.

2) The resultant force of any 2 forces is equal in magnitude and opposite to the other force. The resultant force of the three forces is 0

Rule: The direction of the resultant force of equal magnitude forces acting on the same object is on the bisector of the angles of these two forces.

n and n

The direction of the first question doesn't need to be said, it's very easy, the drawing takes 1cm as a unit to represent 2 bulls and 10n, that is, to draw five line segments, and then use the line segments to represent the force to draw it, oh, oh, it is the use of the parallelogram law There is a small knowledge in the physics book.

Let me answer the question: Why can't the picture be uploaded?

For example, the final velocity of the barrel v v 2=2gl v= 20*, if the maximum acceleration of the upward moving barrel is a, then v 2=2ah a=v 2 2h=4 2*

From Newton's second law, f-(m+m)g=(m+m)af=(m+m)g+(m+m)a =

In order for the keg to rise to the inlet, and the small stone p does not cross AB, the pull force is 2N

The first phase of the first state to accelerate the movement together:

a = f (m+m) -g = f collapse ( -10 = 5f-10 when the small stone p stops to gain initial velocity v = group base (2ah) =2(5f-10)* =8f-16).

h = v^2/(2g)≤l

8f-16)/(2*10)≤

8f-16≤4f≤

Agree with the solution upstairs to the brother collapse bucket process, but F still has a condition, that is, it can pull the barrel and stones, so F has a lower limit of value.

f is equal to (180+20)*

Therefore, the value range of f is: the number of shirts.

Hehe, add ha.

v+ This is the lack of judgment for this public feast.

v+v+a=v=

It takes time for the car to stop t=v a=18 6=3 s

So the velocity at the end of 4s is 0

The displacement is the displacement in 3s s=v 2 2a=27 m