Physics questions in junior high school that challenge IQ should be a specific process

Solution: 840m-230m=610m

So h=610m

Then the actual work distance is 610m (while the horizontal motion of 1200m does not overcome resistance and friction, so no work is done).

g (top) = 60kg 15 = 900kg

g (down) = 60kg 5 = 300kg

g (cable car) = 600kg

W(Upper) = F(Upper) S(Upper) + F(Gondola) S (Gondola) G(Upper) H + G(Gondola) H

900kg×610m+600kg×610m549000j+366000j

915000j

W(down) = F(down) S(down) + F (cable car) S (cable car) G(down) H + G (cable car) H

300kg×610m+600kg×610m183000j+366000j

549000j

W (Total) = PT 60%.

45000w×（7×60）s×60%

18900000j×60%

11340000j

w (useful) = 549000J + 915000J1464000J

w (useful) w (total) 100%.

1464000j 11340000j 100% Answer: Then the mechanical efficiency of the cable car at this time is about.

My process is very specific, I checked it a few times and it shouldn't be wrong, so refer to it.

Altitude difference h=610 m.

The mass of the mountain m1 = 900 kg + 600 kg.

The mass of the descent m2 = 480 kg + 600 kg.

Running time t=420 seconds.

The actual power of the motor p = watts.

Useful work refers to the work done by the system to overcome gravity, therefore.

w1=(m1-m2)gh

Total work w=pt

Mechanical efficiency =w1 w=22%.

For the first time, the ground is facing the support force of the worker n1, then the rope is like the pulling force of the stalker f1 = g man - n1 = 1 5 (g goods 1 + g), n1 = g people - 1 5 (g goods 1 + g).

Second, if the ground is facing the worker's support force n2, then the pulling force of the rope f2 = g person - n2 = 1 5 (g goods 2 + g), n2 = g people - 1 5 (g goods 2 + g).

The ratio of N1 to N2 is 15:19, then G person -1 5 (G goods 1 + g): G people -1 Shelter 5 (G goods 2 + g) = 15:19

i.e. 70*:70*:19

The total gravity g of the sum of the slag scale of the cargo box and the movable pulley is obtained g=

You're okay with the topic, why can't I come?

Upstairs how do you know that it can be all submerged. Is the density of an object greater than that of water?

Load the column container into a height sufficient to submerge the pendant, note the scale v1 into the pendant to completely submerge, note this moment v2, get the pendant volume v=v2-v1 Fill the column container with enough water, put the plastic box into it, and after it is stable, write down this moment v3.

Put the pendant into the locket, at this time the locket can not touch the bottom of the cylinder container, after stabilization, note this moment v4

The mass of the pendant is m=p(v4-v3).

p(pendant)=m v=p(v4-v3)] (v2-v1).