Physics questions. I sincerely ask you, brothers and sisters. The requirements are detailed.

Please put the picture up.

This one is too abstract.

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Physics 1 Question Physics teacher helps.

Hello, I am the co-teacher Gu Lin who asks a question, good at primary and secondary school homework guidance tremor, school selection, college entrance examination school selection and major selection, postgraduate examination adjustment guidance, now has been engaged in the education industry for more than ten years, I am happy to serve you. I have seen your question, and now I am sorting out the answer, it is expected to take about five minutes, please wait a while.

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Items. What does the main item mean.

Can you explain the problem clearly?

How to say that the main term transformation you search the Internet for <>

What do you ask for a solution to this question.

Look at it.

(1) The magnitude of the work done to overcome friction is equal to the work done by gravity minus the total resistance Joule heating, i.e. (2) The amount of electricity passing through the resistor r is equal to the total amount of BLVT R, that is, BL*3R. (3) Since the rod is moving at a uniform velocity, the Joule heat generated is equal to the power multiplied by time, i.e., b 2 * l 2 * v (.

The energy consumed by the two frictions is the same, q=1 2mv 2-1 2m(1 2v) 2, and in the second time, 1 2mv 2-q=1 2(m m)v 2, mv=(m m)v. m m = 3

The common denominator of the two modes of motion before and after is that the heat energy consumed by the bullet in the motion of the wooden block is the same (q=1 2mv 2-1 2m(1 2v) 2), so there is the kinetic energy equation of the second motion: 1 2mv 2-q=1 2(m+m)v 2 where the above equation can be obtained by the conservation of momentum mv=(m+m)v to solve the above equation.

Downstairs is correct and admits mistakes.

The heat energy consumed twice is the same, and if the velocity of the second uniform motion is a, mv 2-m(v 2) 2=mv 2-(m+m)a 2mv=(m+m)a

Substituting a mv (m+m) into the equation yields.

m/(m+m)=1/4

So, m m = 3

So, the ratio of the mass m of the block to the mass of the bullet m is 3

The heat energy consumed twice is the same, and if the velocity of the second uniform motion is x, mv 2-m(v 2) 2=mv 2-(m+m)x 2mv=(m+m)x

Substituting x mv (m+m) into the equation yields.

m/(m+m)=1/4

So, m m = 3

So, the ratio of the mass m of the block to the mass of the bullet m is 3

The decrease in the kinetic energy of the bullet is the sum of the increase in kinetic energy of the block and the energy expended by friction. The frictional energy consumed by both times is the same, both are q=δek=3 8mV2. So:

Q+1 2mv 2=1 2mv 2-1 2mv 2,(m+m)v=mv, where v is the common velocity, and m m=3 is solved

By Newton's second law g (component in the direction of the inclined plane) f (frictional force) = mag (component force in the direction of the inclined plane) = mgsin30

f (friction) =

Bringing in the data, the acceleration a=5 times the fifth of a quarter, the root number three is determined by the displacement formula s=1 2

a (t squared).

Substituting gives s=10 (5/5 of the root number 3).

The gravitational acceleration component on the inclined plane is GSIN30=5

The frictional acceleration component is 0 on the inclined plane

25 gcos30 = five-fourths of the root number three.

The combined acceleration is the component of gravitational acceleration on the inclined plane, the component of frictional acceleration on the inclined plane = 5, five quarters times the root number three.

By the displacement formula s=1 2

a (t squared).

s = 1 2 (5 fifths of a quarter of a root number three) (2 squared) = ten minus five times the root number three.

Solution: It is known that m=10kg, =30°, =,t=,g=10m s 2, let the acceleration a, the length of the inclined plane l, according to the topic, get.

l=1/2at^2

By force analysis, obtained.

mgsinθ-μmgcosθ=ma

Lianli solution. a=，l=

Finally, you can test the answer! )

Explain why mechanical energy is conserved.

The ruler only has a strong effect with the rope and not with the ball. (Of course, the gravitational force between the ruler and the ball is not taken into account here).

Therefore, the following only needs to analyze whether the ruler does work on the rope and whether the rope does work on the ball.

1. The ruler only has an effect on the rope at the point of contact. Because the point of contact is stationary, the work done by the ruler on the rope is 0

2. From beginning to end, the direction of movement of the ball is perpendicular to the direction in which the ball receives the pull force of the rope, so the work done by the rope on the ball is 0

"Thin line", implicitly, means that the quality of the rope is small and negligible. So the work required for the kinetic energy change of the rope can be seen as zero.

Moreover, the object of study of this problem is the small ball, the rope does not do work on the small ball, and only gravity does the work in the whole process, so the mechanical energy is conserved.

The team will answer for you.

The ruler is parked there.

There is no displacement in the force.

Don't do the work thankyou

In fact, the problem is very simple, when the force is applied to the p point, his shock arm becomes shorter and the amplitude becomes larger, why this is happening, it is because the mechanical energy is conserved.

The ball is not affected by an external force in the process of motion, and the ruler changes the radius of the circular motion of the ball, and does not provide an external force for the ball.

c Because the energy is only transformed, it does not disappear, so it is conserved.

Because the options are all about the ball, and this force only acts on the cycloid, and does not do work on the ball.

So the ball still continues to oscillate only under the work done by gravity, so the mechanical energy is conserved. Hope it works for you.

Option C, there is no loss of mechanical energy in this process, just the transformation of gravitational potential energy and kinetic energy.

There needs to be a relative displacement in the definition of work done by an external force. It seems to seem. The ruler has no displacement.

Take a step back. Even if you do the work. Seemingly. The object is not a ball.

c, energy neither disappears nor is generated out of thin air! A to B to C——— gravitational potential energy – kinetic energy – gravitational potential energy! The mechanical energy of the Mocha Force is gradually decreasing! It's not a good question.

The force on the rope did not do the work... It only plays a role in changing the state of motion, and there is no displacement in the direction of the force...

Clamping force = pressure x piston area = p r

Braking torque = clamping force x arm = rp r