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Anonymous users2024-02-11Considering that the topic is two balls chasing, the direction of movement of the two balls in the horizontal direction should be the same, that is, both move to the left. If the velocity of ball A after reaching the bottom of the inclined plane b exceeds the velocity of ball B, then ball A can always catch up with ball B in any case; On the other hand, if the velocity of ball A after reaching the bottom of the inclined plane b is equal to or less than the speed of B, then ball A can never catch up with ball B. Take the inclined plane as the gravitational potential energy 0, let the mass of the small ball A be m, the gravitational acceleration is 10m s 2, the gravitational potential energy of the small ball A is m * 10* to the bottom of the inclined plane, the kinetic energy of the small ball is m 2 2 = m10 *, it is not difficult to find v = 2m s, that is to say, after the ball slides down, the speed is 2m s, and the speed of ball B can only exceed 2m s in order not to be caught up by A.
Considering that the inclined plane is 30°, and the length of the opposite side of the 30-degree angle is half of the hypotenuse, when the small ball A rolls on the inclined plane, the acceleration in the direction of the inclined plane is 5m s 2, and the velocity changes from 0m s to 2m s, 2 5=, that is to say, the small ball A slides on the inclined plane for seconds, and then catches up with B in seconds, and in the following seconds, the small ball A moves in a straight line with a speed of 2m s on the plane, and the displacement is, that is to say, the displacement of the ball B in the whole 1 second is, and the velocity of the ball B is It can be seen that the speed of ball B is indeed very slow, otherwise why would A be caught up so quickly.
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Anonymous users2024-02-10First, A is 10cm away from the horizontal plane, and the final velocity of A on the horizontal plane can be calculated, because it is smooth, the speed is unchanged, and the speed of A is the minimum velocity of B. Second, let the velocity of B be v0, then the distance of B from the bottom of the inclined plane after one second is v0 + the time for A to move from a to the bottom, the square of = find t, and find the velocity v of A by one, that is, v(1-t) = v0 + v t can be found, and v0 can be found. The phone called, but I was tired, ah.
Supplement: A 20cm from the horizontal plane and 40cm from the bottom of the inclined plane. Follow-up:
I'm also on my phone. Help to put the first question in detail: two, squared = supplement:
A slides down from point A, since it is smooth, there is no energy loss, the inclined plane and the horizontal plane are connected by a smooth arc, and there is no energy loss, so the loss of A's gravitational potential energy from A to the bottom is all converted into the kinetic energy of A, mgh = 1 2mv squared, you can find the speed of A, the horizontal plane is smooth, and the speed of B is greater than or equal to the speed of A, so A can not catch up with B.
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Anonymous users2024-02-09According to the title, 15km h refers to the average speed of the typhoon moving northwest for a period of time or a period of Shenmo displacement, while 25m s refers to the instantaneous speed at the time of the typhoon's landfall.
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Anonymous users2024-02-08The lower left arc and the upper right straight bar are a lever.
The upper left arc and the lower right straight rod part are also a lever, take this part as an example, the f in the lower right corner up is the force that makes the lever rotate counterclockwise, when the workpiece is pulled out to the left, the workpiece has a leftward friction force at one end of the lever circle, it is also the force that makes the lever rotate counterclockwise, so the pressure on the workpiece increases, so it is difficult to pull out, and AD is correct.
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Anonymous users2024-02-07The answer is shown in the diagram below.
I can't read 10 questions clearly, and 4 of the 11 questions are incorrect.
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Anonymous users2024-02-06The whole scroll will not be a piece.
Then you should think of other ways.
How can a whole paper be asked?
You do it yourself first, and you won't ask again.
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Anonymous users2024-02-05Friends, individual topics will not be available for communication here, but they should not be all exercises. Beginners may not have started physics yet, so they should practice more. "Do it yourself, have enough food and clothing" personal opinion, for reference only. I wish you success in your studies.
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Anonymous users2024-02-0435. It can be seen from Figure B that when r1 = 10 ohms, the power it consumes is p1=, and the current in the circuit is i1= p1 r1=
When r1*=40 ohms, it consumes power p2=. The current in the circuit i2= p2 r1*=
Minimum current ix=
If the power supply voltage U does not change, then there is.
u=(ro+r1)i1=(ro+10) （1）u=(ro+r1*)i2=(ro+40)
Solution: (1) u=
ro=(u-5) (3) substitute (3) into (2).
u=u=10v。
When the slide P is moved to the B-terminus, the RO consumes the most power.
ro=(10-5) ohm.
pm=u²/ro=10²/10=10w。
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Before the line is broken, the three objects move together at V1 speed, according to the conservation of momentum, there is MV0=3MV1, so there is V1 (1 3)V0, when the line is broken, the spring has impulse at both ends, so it can be regarded as the total momentum is still unchanged, let C leave the spring, the speed of AB is V2, the direction is to the right, then there are 2MV2 + MV0 MV0, V2 0, that is, when C leaves the spring, A and B are stationary. Before the collision, the kinetic energy of the system e0=(1 2)m(v0) 2;The kinetic energy of the system before the line is broken is e1 (1 2)*3m*(v1) 2=(1 6)m(v0) 2;When c leaves the spring, the kinetic energy of the system is e2 (1 2)*2m*(v2) 2+(1 2)*m*(v0) 2 (1 2)*m*(v0) 2. We can see that before the line is broken after the collision, the total kinetic energy of the system is reduced to e1 = (1 6) m (v0) 2, and after the broken wire c leaves the spring, the kinetic energy e e2 = (1 2) m (v0) 2, and the system kinetic energy returns to the initial kinetic energy, indicating that after the spring ** is broken, the kinetic energy is added to the system, and the potential energy of the spring is e, then there is e e2 e1 (1 2) m(v0) 2 (1 6) m(v0) 2=(1 3) m(v0) 2. >>>More