Two single choice or multiple choice questions in junior high school physics require you to write ou

1.a Correct.

Because when using the movable pulley, the movable pulley must also be lifted when lifting the heavy object, which does more work. Under the condition that the useful work remains unchanged, the total work increases, so the mechanical efficiency decreases. On the contrary, when it is used as a fixed pulley, the extra work is reduced, and the mechanical efficiency is bound to increase.

B False Obviously, the fixed pulley is not labor-saving, and the movable pulley can be labor-saving.

c False The formula for calculating power is p w t, and from the formula, it is the size and time of the work that determines the power, so it has nothing to do with how the pulley is used.

d Correct. Needless to say.

2.a False. When an arrow pierces the bull's-eye, it is the kinetic energy that is converted into internal energy. (similar to frictional thermogenesis).

b Yes. Elastic potential energy is converted into kinetic energy.

c False. The kinetic energy of the wind is transferred to the kinetic energy of the windmill. (Note: It's a transfer, not a conversion).

d False. Electrical energy is converted into kinetic energy.

Additional question: Kinetic energy: Determined by the mass and velocity of the object. Gravitational potential energy: Determined by the mass of the object and the dispatch of the lifted.

When an object falls from a height, the height decreases and the velocity increases, so, the potential energy is converted into kinetic energy.

When the object moves upwards, the height becomes greater and the velocity decreases, so, the kinetic energy is converted into potential energy.

Regarding this supplementary question, I think it was written by the landlord incorrectly, or the writing you read was wrong.

I majored in physics in college, and I have been teaching physics for almost 20 years. You can't be wrong with these two simple questions.

a.The fixed pulley is used to lift the object (not counting friction), while the moving pulley has a certain amount of work, and the passive pulley itself is consumed by the rise. So choose.

b.Let the object weight g movable pulley f1 = 1 2g fixed pulley f2 = g so do not choose.

The object is the same as the reference, and the power is the same, so it is not selected. d.Choose.

a.Kinetic energy is converted into frictional potential energy, not selected.

b.Elastic potential energy is converted into kinetic energy selection.

c.Kinetic energy is converted into potential energy.

d.Electrical energy is converted into kinetic energy.

The last one, I think, should be purely the conversion of potential energy into kinetic energy.

Above, the first time I came here to answer the question.

I majored in physics in college and am now working in human resources. So it's rusty, alas.

First of all, it is necessary to analyze this question, and the key to this question is to analyze the two separate connection methods: because the question tells that the resistance value of L1 is greater than the resistance value of L2 and remains unchanged, and in the first connection of the inform, the two lights emit light normally, so it can be seen that it is connected in series (the basic circuit in series, I will not draw a diagram to help you analyze), and the second time it must be in parallel [according to: in the series circuit, the current is the same, and the voltage is different; The reverse is true for parallel connections].

After analysis, you can draw two connection diagrams, and then analyze the simple circuit diagram to obtain: l1=u p1

Then analyze the condition in the question "the actual power of L1 is 4 9 of the rated power", and the equation UB L1=4 9P1 can be listed, and the generation L1=U P1 can be obtained by entering the above formula, and ub P1 9=4 9P1 can be simplified to UB = 4V, then UB = 2V

When L1 is connected in parallel with L2, L2 emits light normally, so U2=2V When L1 is connected in series with L2, L2 emits light normally, so U2=2V So, at this time, I2=I1=I Total =P2 U2=1W 2V=So, P1=UI=3V

After my inspection, it's completely correct, although the process is a little simpler, but I also think about how long, the landlord will give points.

1 should be facing the window.

Why: In order to see yourself in the mirror, you need to have a good contrast between your face and your background.

When facing away from the window, the face in the mirror is dim, but the background (outside the window) is bright, so the face cannot be seen clearly; On the other hand, when facing the window, the face in the mirror is bright, but the background (inside the house) is dim, so the face is very clear.

2 The farther the slide projector is from the screen, the larger the image, because the slide projector is an inverted and magnified real image, and the larger the image distance, the larger the image.

I specifically found a light path diagram of the slide projector, and I added the red part. As you can see, the screen moves back and the image becomes larger. Of course, in order to make the image clear, the slide position also has to be moved, that is, slightly to the right.

The face should be facing the window, so that the light can reach the face, and after the mirror reflection, the person can see clearly.

If the image on the screen is larger, the object will be closer to the convex lens, the object distance will be smaller, and the image distance will be larger, so the slide projector will be farther away from the screen.

1. Facing the window, the face imaging process is that the light on the face is reflected into the mirror, and then reflected into the eye through the mirror reflection. The reason why you want to face the window is because the face is diffuse, and you should face the window to strengthen the reflected light and make the image clearer. 2. Farther away, because the light emitted by the slide is scattered, the light is scattered by a point around the image, and the farther the imaging interface is from the light source, the greater the imaging.

It doesn't matter if the window is a window or not, if you want to see the face clearly, you only need to make enough light on the face, and at the same time, don't let the mirror reflect and stab the eyes; The face should be facing the window, but the mirror should not block out the light.

If you want to make the image bigger, how can there be a mobile projector, there is nowhere to use it, and there is no room for it, in fact, the lens should be adjusted to be closer to the slide, rather than the entire mobile slide projector. It's common sense, haven't you ever used a slide projector in class? The so-called "object advances and retreats", because the closer the object is to the convex lens, the greater the dispersion angle of light (not the so-called scattering angle), and the image will of course become larger.

As for how to move the slide projector, it is a question that does not need to be asked, and someone actually answers, it is too impractical, I am very disappointed in this phenomenon, too many people answer questions in order to answer questions, without thinking, answering questions not only with the brain, but more importantly, with the heart, the motherland needs fewer and fewer people the most, and I am feeling very sad. However, it is good to have the spirit of studying.

I think I should make it clear.

If you really can't do it, you will report the Q number.

Positively, the reflection of light.

Imaging of far, light and graphs.

1. Solution: The resistance of the small bulb: R lamp = U forehead 2 p forehead = (6V)2 3w = 12 When the slide moves to the far left, the bulb emits light normally, and the power supply voltage can be known: u = u forehead = 6V

When the slider moves to the far right, the bulb is connected in series with the sliding rheostat, and the current in the circuit at this time: i= u (r lamp + r slip) = 6v (12 +20) =

Therefore, the actual voltage at both ends of the bulb is: u real = i r lamp = x 12 = 2, and it is not clear to see.

In the second question, 2 resistors are connected in parallel, and the voltage u is equal.

r1 = 10 ohms, power p1 = u 10

The power of r2 is p2=

p1+p2=

According to the above 3 equations, we can know that u=6

R1 power p1 = u 10 = watts.

Known: P total = 600W, P1 = 200W, P2 = 1200W, P3 = 1200W

Seeking: i=? Solution: i=p u=(600w+200w+1200w+1200w) 220v=

Answer; A 10 (20)A ammeter should be selected in this circuit.

So c