Physics around you, thank you, there s a physics problem

The fan is actually a low-power pressurizer, which draws air from the back and pressurizes the impeller to send air out from the front.

After the air is sucked into the fan from behind, the velocity increases, according to Bernoulli's equation, since the total pressure of the air remains the same, the total pressure of the air is composed of dynamic pressure and static pressure, the dynamic pressure increases, the static pressure decreases, so a negative pressure zone will be formed behind the fan, you put the paper behind the fan, because there is a pressure difference between the front and back, so the paper is sucked in.

If you put the paper behind an open fan, the air behind the fan is pumped away, so that the air pressure near the end of the fan becomes smaller, so the paper will be covered"Suction"Past.

That's related to the direction of the fan's sector.

Electric fans can't produce their own air. If the front is out of the air, the back is the inlet air, it just gives the flow of the wind direction. Otherwise, the air will be blown out from both sides, and the fan used in the kitchen will be useless.

I don't think you can suck it up, you can't suck it in the general fan, the wind is relatively small and the wind can also enter from around the fan, if there is only one inlet and one outlet for the air, it will also produce uneven air, sometimes it will vibrate, and the force will be uneven and fall down.

The air in front flows fast, the pressure is small, and there is a pressure difference between the front and rear.

The fan makes the air flow, and where there is air outlet, there must be a place where air comes in.

This is explained in fluid mechanics.

I can't tell you clearly.

There are three non-extensible insulated light wires with a length of 1m, one end of two of them is fixed at the O point on the ceiling, and the other end is tied to the charged balls A and B with a mass of m=<-2>kg, their electricity is +Q and -Q respectively, Q=<-7>C, there is a third wire connected between A and B, and there is a uniform electric field of size E=<6>N C in the space, and the direction of the field strength is horizontal to the right, and ABO is in a regular triangle when equilibrium, A is on the left, and B is on the right. Now the line between the ob is burned, due to the air resistance A, the ball B will finally reach a new equilibrium position, and when the ball finally reaches equilibrium, the rope will be subjected to tension?

The electrostatic force of interaction between two charged spheres is not counted).

Calculate the volume v=

Then find the mass m = density * v

Then you do the math yourself.

Believe you can do it.

People are supported by gravity, friction and inclined planes during the upward slide.

The component of gravity along the inclined plane is mgsin37°, and the component of the perpendicular inclined plane is mgcos37°

Let the supporting force of the inclined face be n, then there is n=mgcos37° friction f= n= mgcos37°

During the sliding process, the net force along the inclined plane is f1=f+mgsin37°, and the acceleration is a and f1=ma

The solution is a=g*(sin37°+ cos37°)=8m s and the initial velocity is v0=8m s, and the maximum distance along the inclined plane is set to be s after t1 time

Then there is s=v0t1-1 2at1

v0-at1=0

Solve s=4m

The second question: the net force of the person in the process of writing f2=mgsin37°-f=mgsin37°- mgcos37°

Let the acceleration be a1 and slide to the starting point after t2 time.

then there is f2=ma1

1/2at2²=s

Solve t2=sqrt2 (root number 2).

a=4m/s²

Then the terminal velocity v1=at2=4sqrt2m s (4 times the root number 2).

1 Formula for conservation of energy.

1/2mv^2-μmgcosθ ×h/sinθ=mgh;H2 is conserved by energy.

mgh- mgcos h sin =1 2mv2 2 can be solved to obtain v2;

Both of these problems are conservation of energy problems: initial energy - expenditure = end energy here the energy includes kinetic energy and potential energy, while for the two cases of this problem there is only kinetic energy or potential energy (relative to the ground).

1.The gravity of the person is decomposed downward along the slope and perpendicular to the slope downward, and the direction of kinetic friction is downward along the slope, which can be listed:

mgsin37+μmgcos37=ma

The time it takes for the athlete's speed to be reduced to zero (because the deceleration a is negative): vt=v0+at, distance to slide: s=v0t+1 2at 2

Calculated s=2When gliding in the opposite direction, the direction of dynamic friction changes to upward along the slope, which can be listed

mgsin37-μmgcos37=ma'

It is calculated as: vt=

I take sin37=, cos37=

,p=f s=360 (,f=pvg=,the key gravity of a thin plank g=f=

The redundant step is that the actual density of the cork is greatly dissipated.

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A uniformly charged ring of radius r has a charge line density of , the center of the circle is at point o, and there is a point p on the axis perpendicular to the center of the circle and the torus, po=r, and the potential is zero point at infinity, then the potential up of point p is 0.

The answer is b

Work: 100N*5M+10N*5M=550J Power: 550J 10S=55W

Efficiency: 500J 550J=91%.

Total work: w = g total h = 100n * 5m + 10n * 5m = 550j useful work: w has = g h = 100n * 5m = 500j power:

The power of the total work p=w total t = 550j 10s = 55w efficiency: w has w total = 500j 550j = 91%.