Senior 1 Physics Questions, Senior 1 Physics Questions.

1) With energy conservation, let the velocity of the throwing at the highest point of the arc be v, then there is a calculated velocity v = root number (2gr).

The initial velocity is a flat tossing motion of 2gr.

We can use the equation using the same motion time t in the horizontal and vertical directions

Let the height be h, then:

vt=This equation shouldn't be difficult, so let's solve it yourself.

2) The first question is to calculate the relative acceleration, and after putting the slider, the force analysis of the trolley:

f-umg=a-car*m, so a-car

For small wooden blocks: Amu ug 2

So the relative acceleration: a phase 2

So the time t v a phase is 3 seconds).

The second question is, the wooden block does a uniform acceleration linear motion, then s wood wood 3) because in time t, the volume of water passing through each cross-section is the same, we take two extremes, the first is the cross-section of the outlet of the water pipe, and the second is the cross-section of the floor.

Let the area to the floor section be s1, the velocity is v1, and the time from the water to the ground is t, then:

h=vt+substitution: h=,g=10,v=1 solution t seconds.

So v1=v+gt=4

From the equal volumes, it can be obtained:

So s1=vs v1=

We found that since the water reaches the ground in seconds, then we only need to calculate the volume of water flowing through the first section in this time to get the volume of water in the air a.

a Cubic metres.

1.The problem description is not clear.

2.There is something wrong with the question itself. No solution.

Isn't it the original question?

Horizontal: ff=fsin30

Vertical direction: fn+fsin60=g

f=200n g=500n

It's easy to find fn and ff.

First, the time taken to brake and accelerate is xand y. Then there's y=2x, x+y=12, and then you solve them to eight and four, and the acceleration is just one and two, and the second question is eight and six.

Question 2: It should be noted that the plane is already stationary for ten seconds, and the distance is calculated as five meters by v=2as. Three questions:

s one-half a t square. a ＝2.So s t squared, and the displacement of the last two seconds is equal to the average velocity multiplied by time, to emphasize, the average velocity of the last two seconds is the instantaneous velocity in the middle of the last two seconds, so (2t 2) 2 three-quarters s s three-quarters t squared.

Four-thirds or four-thirds of the solution, of course, four-thirds have to be discarded. So s t squared 16m. Finished.

Set the deceleration time t1 and accelerate t2

t1+t2=12

t1=2t2

t1=8 t2=4

According to a=(v2-v1) t

Deceleration a=(2-10) 8=-1

Acceleration a=(10-2) 4=2

Others need to tell me QQ, like it, thank you.

v=gt=t=s/v=

2. Set the length of the inclined plane to l, yes.

l=v0t+at^2/2

L-7L 15=V0(T-1)+A(T-1) 2 2 The two formulas are combined, which can solve the bevel length l and the motion time t, and then substitute t into the formula:

v(t)=v0+at to find the speed of the ball to the bottom, I will not calculate the number.