Ask a high school physics question and explain it in detail.

If the kinetic energy of b m increases by 1 3mgh and the acceleration is 1 3g, then it can also be regarded as the downward resultant force is 1 3g, then it is the same as the acceleration g hypothetical = 1 3g actual, in the absence of air resistance and other effects, then the kinetic energy increases = the potential energy decreases, w = gh = mg assumes h = 1 3mg actual h

The gravitational potential energy of a m decreases by 1 3mgh and the gravitational potential energy decreases = gh=mgh so it is wrong.

The mechanical energy of c m decreases by 1 3mgh and the kinetic energy increases by 1 3mgh and the potential energy decreases by mgh then the decrease in mechanical energy should be 2 3mgh so it is also false.

d Gravity work 1 3mgh gravity work = gh = mgh is also wrong.

So the right one is only B m with an increase in kinetic energy of 1 3 mgh

bYour understanding is not quite correct. It should not be thought of as a situation where the gravitational force decreases due to being too far from the earth.

Think of it this way: the object is actually subjected to a vertical upward tension of 2 3mg at the same time as the mg.

So you can understand.

If you think so, acceleration will be a quantity that changes at any time, which does not match the topic.

It's not that you can't do it, it's that you don't understand it correctly.

That is, it falls vertically from rest with an acceleration of 1 3g, why not g? This is not due to its position being too high, but rather to the fact that it has received a resistance of 2 3 mg

Okay, I'm sure you'll do it, right?

Hello, dear <>

Based on the questions you provided, Gensen will answer the following questions for you: Hello, question: A small ball with mass m, moving in the horizontal direction at a velocity of v, touches a large stationary ball with a mass of m, the ball bounces back, and the velocity is sold as v', how far does the big ball travel?

Analysis: Since the momentum of the ball is conserved before and after the collision, there is: mv = mv'+MV where V is the velocity of the ball before it collides, V'is the velocity of the small ball after collision, and m is the mass of the large ball.

And because the velocity of the ball is reversed after collision, this sock has: v'=v substituting the above equation into the momentum conservation equation yields: mv = mv + mv simplification yields:

mv = 2mv Therefore, the distance traveled by the big ball is: s = vt = mv) (m 2) t = 2mv (t m) where t is the time of the big ball motion.

For the physics topic of the first year of high school, I believe you have already learned kinematics.

It's the same as solving for unknowns. Let its acceleration be a

v1-v2=at= v (velocity in the back minus velocity in front) Directional acceleration is the amount of change in the velocity of an object per unit of time.

East is the positive direction.

The initial velocity is 3 meters per second, and the final velocity is -3 meters per second.

Obtained by v=at.

3 meters per second - 3 meters per second = a*3

a = -2 meters per second.

It's okay to know 2m s!!

If you really want the process, you can't help it:

Let East be the positive direction.

The initial velocity is 3 meters per second, and the final velocity is -3 meters per second.

Obtained by v=at.

3 meters per second - 3 meters per second = a * 3 = 6

a = -2 meters per second.

Quack, give more points Frog call.........: quack,

Let the two forces be f1 and f2 respectively, where f1 is greater than f2 and f1-f2 is obtained from the figure f1-f2=4

f1+f2=12

So the solution is f1=8n

f2 = 4n and if f2 is greater than f1 then f2 = 8n

f1=4n

The sum of the time for the stone to fall to the ground and the sound to Xiao Ming is 6 seconds, and the time for the stone to fall to the ground is t

Then the depth is h=ut+1 2at 2=

The time of the sound wave is 5t 2 340

Then t+5t2 340=6

5t^2+340t-2040=0

The solution yields t=h=154m

Hope you are satisfied :)

According to the inscription, if the depth of the cliff is h, the time of stone movement is t1, and the time of sound propagation is t2, then there is.

t=t1+t2

h=1/2gt1^2

h = v sound * t2

t2=t-t1

1/2gt1^2=2040-340t1

t1^2+68t1-408=0

That's what you did, I don't know if your data was wrong!! Or that's it, you're looking at it yourself, you won't be asking me!!