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I'm sorry, I'm an elementary school student, Emmer can wish

It's the picture I drew like this:

From the figure, we can get v2=v1*cosa

V1 does not change, cosa is less than 1, so when V2 is far away, the A angle decreases, Cosa increases, V2 increases, so the object accelerates. Select: c

bc, right?! A first of all, no, because the trolley is in motion, so the angle of the horizontal plane is getting smaller and smaller, and the object is always changing, so it cannot be a constant velocity.

d Because f>mg so the object starts to be subjected to an upward force, do the acceleration motion of variable acceleration a (the angle is changing, f is changing), and the car motion how f-mg will not be 0 so there is an acceleration a because t does not know so bc is possible.

My physics is not very good, don't blame me for the wrong analysis. If you have a question about biology, ask me that I'm pretty good at biology!

Choose C, the car is subject to rope tension, gravity and friction.

The car is affected by friction, and the pulling force is equal, and gravity does not participate in the horizontal motion of the object and the car.

The car is subjected.

The resultant force is necessarily less than the tensile force on the object.

So choose C

It depends on how the rope is pulled, and pulling it horizontally is not the same as pulling it obliquely.

a Let the speed of the car pull the rope be v0, because the vertical distance between the car and the pulley is constant, v0 is proportional to v1 and v0 = v1

From (1) and (2) we know that (-1,0) and (1,0) are used, so -1+1=-(2 a) of the relation between the root and the coefficient, and the quadratic function a is not equal to 0, so it cannot be (1) and (2).

From (3) and (4) to know the image (0,0), substitute the function to get a = 1 or -1 from (3) to know, the opening up a >0, get a = 1, substitute it into the function to get y=x2+2x, at this time the intersection of the function and the x-axis is (0,0) and.

1,0), which does not match the known image, so it cannot be (3) from (4) to know, the opening is up a<0, and a=-1 is obtained, and it is substituted into the function to obtain y=-x2+2x, and the intersection of the function and the x-axis is obtained as (0,0) and.

1,0), which is consistent with the known image, so it can only be (4) and the image is (4), a=-1

Suppose f(x) passes through the origin.

a^2-1=0

a = plus or minus 1, axis of symmetry 2 (-2a), >0

A<0 so choose B

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Solution: A proportional sequence of positive numbers, then q 0, and the square of a 3=a2a4=1, a3=1 0;

and the square of s3 = a1 + a2 + a3 = 1 q + 1 q = 7, that is, 6q2-q-1 = 0, and the solution gives q=1 2, or q=-1 3

If it doesn't fit the topic, give it up.

then an=a3 q to the power (n-3) = (1 2) to the power (n-3);

a1=4；s5 = 4 x [1 - (1 fifth of 2)] 1-1 2 = 31 4

So the answer is 31 4

Solution: According to the question, we know that f(x)=x+4 x is a subtraction function on (0,2], an increasing function on [2,+, and at x=2, there is a minimum value of 4

f(x) defines the domain as [1,m].

1) When m<=2, f(x) decreases monotonically on [1,m] f(m)<=f(x)<=f(1)=5

The value range is [4,5].

m=22) When m>=2, f(x) decreases monotonically on [1,2] and increases monotonically on [2,m].

So when f(m)=5 there is m=4

In summary, the value range of m is [2,4].