Help with a physics problem!! Finding a Physics Problem!!

The first time there was no pull, so it was subjected to static friction, which was equal to 200N

The second wooden box moves at a constant speed, and according to the equilibrium conditions, it is known that the amount of friction it is subjected to is equal to the magnitude of the horizontal tensile force, that is, 500N

1.Because the acceleration of an object is 1 meter and no second square, it can be obtained from Newton's law that the net force experienced by the object is 1*1000kg=1000n. In the first two seconds, the displacement of the object can be found using the formula 1 2*a*t =s, i.e., 2 meters.

Therefore, the work done by the crane on the cargo is w=f*s=(1000N+g)*2m=22000JAnd at the end of two seconds, the velocity of the object is a*t=2m s, so its kinetic energy is e=1 2mv =2000j.

2.Because the inclination angle of the inclined plane is 30 degrees, the height is sin30*1= and the inclined plane is smooth, when the object slides down, the gravitational potential energy is completely converted into kinetic energy, and the mechanical energy is not lost, so the kinetic energy at the moment of sliding out of the inclined plane is e=g*h=m*g*, so the velocity of the object is the root number (g). And the coefficient of dynamic friction is, so the acceleration a = produced by friction on the object, from the formula 1 2 * a * t = s can be pushed out s s = v 2a = 5m, so its displacement is 5 meters.

3.Let the height be h, so the bevel plane has a long root number (2)*h, and if the object is in free fall, then the velocity v = root number (2*h*g), and the velocity of the object sliding down the inclined plane to the bottom is, so the kinetic energy e=, the energy difference. It is the heat energy lost by friction.

From the formula w=f*s, the frictional force is the root number (2)*m*g 10, and the pressure of the object on the inclined plane is the root number (2)*m*g 2, so the dynamic friction coefficient is.

Hehe, some symbols can't be played, but you can understand it, sorry, sorry. ^

The electrical energy consumption of the microwave oven is e p*t search 1100*60 66000 joules local leakage included.

The mass of 250 ml of water is 250 grams of kilograms.

The heat absorbed by water is q cm* t Tongxiao 60900 joules.

The heating efficiency is Qe 60900 66000*100%.

w stove sippp = pt = 66000 (j).

W water = cmt = 60900 (j).

Jiajia Zhichang thermal efficiency teasing jujube = 60900 66000 =

The electrical energy consumed by the microwave oven is W pt 1100W*60s, and the mass of the water is m water = PV = stove type stuffy bend 3 = 250 grams kg.

The heat absorbed by water is q cm(t-t0).

The heating efficiency of this microwave oven is w

(1) At this time, the sliding rheostat is connected in parallel with the bulb by: the small bulb L is marked with 3V It can be known that the rated current is When the maximum resistance of the sliding rheostat is R0 (the size is unknown), the current flowing through the sliding rheostat is also The maximum resistance and the resistance of the small bulb are equal, that is, 6 ohms.

2) At this time, the sliding rheostat is short-circuited by p=u2 r, p=6*6, 6=6w (according to the first question, the power supply voltage can be known to be 6v).

The mass of the OA side: 2*OA, the force arm OA 2

The mass of the ob side: ob-oa, the force arm (ob-oa) 2 is balanced according to the moment, oa 2 = (ob-oa) 2 2 so oa : ob-oa) = 1 sqrt(2)oa:

ob = 1 : sqrt(2) +1] where sqrt stands for the root number.

Solution: oa, w1=oa(for the density of the rod).

ob,w2=ob.ρ

According to the leverage balance conditions:

2 oa.ρ.i/2oa=i/2(ob-oa)^2.Finishing: (oa ob) 2+2oa ob=1(oa ob) 2+1=2

Solution: OA ob = root number 2-1

Let the time interval between each drop of water be t, and it can be seen from the known conditions that the time for the first drop of water to fall from the eaves to the ground is 3t, so t=?

The distance between each and every two drops of water can then be found.

Force analysis:

Gravity = frictional force.

The weight of the anteroposterior person does not change. So c

c is when the two people weigh the same weight, and the question does not say. However, it is estimated that the two weigh the same.

Choose c, gravity = friction.

In this way, students should not have weight changes (such as going to the toilet in the middle, etc.), and the friction is the same twice.

3 root No. 30 3.

You first decompose the gravity g, a1 = gsin30 = 5m s 2 along the inclined plane, and then decompose the a1 along the inclined plane again into the reverse and vertical chute directions along the chute. Along the direction of the chute is the acceleration of the slide of the particle. There is a trigonometric relationship, and it can be found that along the direction of the chute a2=a1sin =3m s 2.

Then, according to 1 2A2*T 2=s, find the slip time t = root number 30 3s.

Solution: Increased internal energy e=(23-3)*

The external force does the work, and the friction generates heat.

If it is substituted sin = 1 and the maximum progression calculated km = 10, then the maximum number of bright stripes that can appear on the screen is 9 levels. When the diffraction angle is 90 degrees, the outgoing rays are already parallel to the screen and cannot appear on the screen, so the maximum series should be less than 10 steps.