High School Object Questions, Help? High School Physics Ask for help

1.Since the front and rear velocities are 0, v=at can be obtained.

v (maximum) = a (plus) t (plus) = a (minus) t (minus).

And because a(plus)=2a(minus), t(add)=(1 2)t(minus) b is correct.

2.Since the front and rear velocities are both 0, both sections can be seen as a uniform acceleration motion with an initial velocity of 0.

s(plus) = (1 2) a (add) t (add) 2 s (minus) = (1 2) a (minus) t (minus) 2

Because a(plus)=2a(minus), t(plus)=(1 2)t(minus).

So s(plus) = (1 2)s(minus) d is wrong.

3.From the previous step, we can see that s(plus)=4m s(minus)=8m

And because t(plus)=(1 2)t(minus), t(plus)=1s, t(minus)=2s

By s(plus) = (1 2) a (plus) t (plus) 2 s (minus) = (1 2) a (minus) t (minus) 2

A(plus) = 8m s2 a (minus) = 4m s2

Therefore, the maximum velocity during the glide process v (max) = a (plus) t (plus) = 8m s a error.

4.Let the upward friction force when accelerating be f (plus) and the upward friction force when decelerating is f (minus).

There is g-f(plus)=ma(plus) f(minus)-g=ma(minus).

So f(plus)=g-ma(plus)=600-480=120 n

f(minus) = g + a (minus) = 600 + 240 = 840 n

So f(plus):f(minus)=1:7 c is correct.

To sum up, B and C should be chosen

The acceleration when setting the deceleration is a, the acceleration at the time of acceleration is 2a, the deceleration time is t, and the acceleration time is 3-t

2a(3-t)=at

t=2s 3-t=1s

The deceleration time is 2s and the acceleration time is 1s

1/2 (2a)*1^2+1/2 a*2^2=12a=4m／s2 2a=8m／s2

a Maximum speed 8*1=8m s

b Time ratio 1:2

c Friction force f=g-m*8m s2=120n during acceleration f=g+m*4m s2=840n ratio 120:840=1:7 when decelerating

d When accelerating, s=1 2 *8*1 2=4m

When decelerating, s=1 2 *4*2 2=8m

Ratio 4:8 = 1:2

The answer is b c

A's 4ms is the average speed!!

D's should be 1:5

A is not true, if the maximum speed is 4, the acceleration distance is 2 meters, and the deceleration distance is 4 meters, which is not true.

b is right, because the acceleration is twice the deceleration, and the landing speed is 0c, the acceleration is found as: acceleration is 8 and deceleration is 4

The friction force is found to be upwards of 14N when accelerating and 84N when decelerating. So correct.

D is not true. Specific ideas: first find the acceleration, from the question, accelerate for 1 second, decelerate for 2 seconds, set the acceleration when decelerating, then the acceleration is 2a, 1 2 * (2a) * 1 square + 1 2a * 2 square = 12

Result A. It's all understood.

Draw an image of f-t, the area enclosed by the plot line represents the impulse of the force, and then solve it according to the kinetic absolute quantity theorem (the impulse of the resultant force on the object and the bond of the body is equal to the change in the momentum of the object!). ）

The problem of variable force is dusty, and points are used:

f(t) brother laugh = 2t + 2

m=2a(t)=f(t)/m=t+1

v(t)=v0+ (0,t)a(t)dt (0,t)(t+1)dt

t²/2+t]（0，t）

t²/2+t

v（2）=2²/2+2=4

10.There are so many words that are not useful Knowing the rated speed, the diameter of the wheel knows how to find the result, and the answer to the silver meaning of the high old speed refers to a few revolutions per minute. Result: Qingyan 508* choose C

3.There is acceleration only when there is an external force, so it only accelerates when kicking the ball, and it slows down after leaving the foot.

Average acceleration: 10

The magnitude of the acceleration is constant, and it is a linear motion after leaving the foot, which is a uniform deceleration motion, and the acceleration is -2. But the title is a bit problematic.

Average speed: (10+10-3*2) 2=7What did that 20m come from? Or the result is 20 3=

According to the topic you don't know a lot of algorithms:1Ignore that 20 to count 10-3*2=42I don't know the initial velocity.

3.It is superfluous for 3 seconds.

How is it possible to give a detailed process without giving a single point for several questions.

The ball A is thrown flat, B is thrown vertically and B does not reach the highest point to meet the two balls, so the algebraic sum of the vertical displacement of ball A and the vertical displacement of ball B is h, and let the meeting time be t, then there is: h = xa + xb = (1 2)gt 2 + v2t - 1 2)gt 2. Simplification gets:

h = v2t, then: t = h v2, a is false. Because the two balls meet T seconds after the two balls are thrown, the horizontal displacement of the A ball is equal to the horizontal distance of the two balls, and there is:

s = v1t = h (v1 v2), b pair. The velocity of the ball A when they meet is calculated by the Pythagorean theorem, and the sum of the square of the vertical velocity and the horizontal velocity is the second square, and c is wrong. Finally, just bring v2 in and count xa = (1 2)gt 2.

When thrown horizontally at a horizontal velocity of 1m s.

The motion is decomposed into a free-fall motion in the vertical direction and a uniform linear motion in the horizontal direction, and the final velocity v in the vertical direction satisfies 2gh=v 2 (displacement velocity formula) to obtain v 2=20 so the velocity of the field when it lands is (1 2+20) = 21m s

In the same way, when the throwing speed is 2m s, the landing speed is Li Yu shouting (2 2 + 20) = 2 6m s

If Q is positively charged, the gravity mg is straight downward.

The electric field force qe is horizontal to the right.

Then the Lorentz force is in the upper left to balance the three forces.

And mg = 2 10-5

qe=2×10-5

So the direction of the Lorentz force is 135 degrees upward with the direction of the electric field.

The size is 2 10-5 root number 2Equal to qvb

v = 10 root number 2

Negative electricity is the same.

The positive charge v is 60° to E and the negative charge V is 120° to E at 20mS

Let's tell you how to do it, and the specific process will be calculated by yourself.

1。When the rotor is at the highest point, according to the pressure of the machine on the ground is zero, there is a circular motion formula, you can find its velocity at the highest point (the pressure is zero, indicating that the centripetal force of the rotor is mg+mg), and then use energy conservation to calculate the speed of the rotor at the lowest point, find the centripetal force of the rotor again at the lowest point, and then force analysis can find the maximum pressure, when the rotor is at the lowest point, the pressure on the ground is the maximum.

Sorry, I didn't understand the first question.

2.First force analysis.

f=fcosα

When (mg-fsin) > fcos, the block no longer slides.

At this point, tan >1 (square multiplied by mg).

Nadir. Conservation of mechanical energy mgh=1 2mv 2 circular motion shows f-mg=mv 2 h

So f=3mg

f=mg+3mg

umg\cosa-usina