Rescue 1 high school physics questions, help 1 high school physics problems emergency

It is conserved by mechanical energy.

If the mechanical energy is MGL initially, then when the gravitational potential energy is kinetic energy, the velocity at this time is also obtained from the formula.

The vertical component of velocity at this time is analyzed by plotting and analyzing vy=sin60°*v;

So in summary, the answer can be solved.

When the kinetic energy and potential energy are equal for the first time, there is the formula mgh = mv 2 2, and v = 2gh under the root number, so the instantaneous power of gravity p = fv = mg * 2 gh under the root number

It is at 1 2 in the vertical direction, 30 degrees horizontally downward, the kinetic energy is 1 2mgl, and the velocity is g under the root number by e=1 2mv squared, that is, 1 2mv square 11 2mg, and then the vertical velocity is the actual velocity multiplied by the cosine of 30 degrees, so that gravity is multiplied by this vertical velocity is the answer.

In the first process, a = (25-5) 10 = 2m s2

The second process, a2=(0-25) 2=

The direction is opposite, the first one is the same as the direction of motion, and the second is the opposite direction of motion.

a(Acceleration)=(25-5) 10=2m s 2 direction: Same as the initial velocity direction. (You can also set the initial velocity direction to the positive direction first).

a(deceleration)=25 2= direction: opposite to the direction of initial velocity.

The acceleration at acceleration is: 25-5 10 = direction is the direction of velocity (direction of travel).

When decelerating, 25 2 = direction is the reverse direction of speed (travel in the opposite direction).

How do you go to school Toes know.

Let the acceleration be a

Then the first segment: velocity at the end of time 1s v1 = a*1 = 1a displacement x1 = second segment: velocity at the end of time 2s v2 = a*(1+2) = 3a displacement x2 = v1*2+

The third stage: the velocity at the end of time 3s v3 = a * (1 + 2 + 3) = 6a displacement x3 = v2 * 3 +

So. v1：v2：v3=1：3：6

Displacement x1:x2:x3=1:8:27

If you read the information given on the V-T diagram directly, you can see that the initial velocity of 1 is 5m s and the initial velocity of 2 is 0, and the resultant velocity is 5m s along the x-axis

So A is one of the correct options.

Looking at the y direction, the velocity is 5m s when t=5, that is to say, the acceleration in the y direction is 1m s -2, and the resultant external force in the y direction is 1n from f=ma, so b is also one of the correct options.

Looking at option C, when t=5, the velocity in the Y direction is 5m s, and the velocity in the X direction is also 5m s, according to the parallelogram rule of vector sum, the resultant velocity should be 45 degrees to both the x and y axes, so the C option is also correct.

And finally, d. According to the information given in the V-T diagram, the area enclosed by the straight line t=0, t=5, v=0 and curve 1 and curve 2 can be calculated, and the uniform linear motion x=vt, uniform acceleration linear motion x=1 2*vt, x=1 2*at 2 can be calculated, and the displacement of the ball in the x direction is greater than the displacement in the y direction, and the vector sum of the displacement is not 45 degrees with the coordinate axis, and d is wrong.

Answer: ABC

a. The initial velocity is the velocity when t=0, and the x-axis velocity is 5m s and the y-axis velocity is 0 when t=0, so the initial velocity is 5m s

b. The acceleration of a uniform acceleration motion in the y-axis direction can be found as 1m s in the graph, f=ma=1n

C. When T=5S, the Y-axis velocity is 5m s, and the X-axis velocity is 5m s, both of which are equal in size, and the parallelogram rule is used to synthesize the velocity, and the angle is 45°

d. In 0 5s time, the x-axis displacement is 5*5=25m, and the y-axis displacement is at=, the two are not equal, and the resulting angle of displacement will not be 45 °

ABC correct.

From the figure, it is clear that the graph line 1 is a uniform motion in the x-axis direction, the velocity is 5m s, and the velocity in the y direction is 0 at time 0, so the initial velocity is 5m s, and a is correct.

Looking at the graph line 2, we can see that the velocity in the y direction reaches 5m s at 5s, so the acceleration is 1m s 2 constant force f=ma=1n, b is correct.

At 5s, the velocity in the xy direction is equal, and according to the parallelogram rule, the angle between the combined velocity and the horizontal direction is 45°, and c is correct.

The angle between the displacement direction of the ball and the x-axis at a certain time is always less than the angle between the instantaneous velocity direction of the ball and the x-axis, so the angle between the displacement direction of the ball and the x-axis is less than 45°, and d is wrong.

In summary, the answer is ABC

Hope it helps.

The ABC1 line represents the X axis direction with a constant linear motion velocity of 5m s, and when t=5, the 2 line vy is also equal to 5m s

Therefore, the angle between the direction of velocity and the direction of initial velocity is 45°

The ball is subjected to a constant force on the y-axis with Newton's second law has.

f=maa=1（m/s）

There is f=1 (n).

ABCA does not explain.

b a = 5 m s divided by 5s = 1 m s 2cd see figure.

The figure was done in a hurry, let's take a look.

According to Lenz's law, the ampere force is related to the speed of the change of magnetic flux per unit time, and the faster the change of magnetic flux, the greater the ampere force, and vice versa.

During the period from the midpoint of t=0 to m'n' to the point n (let t=1 at this time), the magnetic flux increases and changes more and more rapidly.

So the answer is chosen in A and B.

During the period from t=1 to p'q' to m'n' (let t=2 at this time), the magnetic flux is maximum and constant.

During the period from the midpoint motion from t=2 to p'q' to the n-point (let t=3 at this time), the magnetic flux decrease shear change becomes slower and slower.

So the answer is b