Thank you for the high school physics problems that need to be solved urgently

The inclination angle is 30°, which is long, so the initial package height is h=l*sin30°=conservation according to energy: ek-0=δep=mgh=3j.

So p= (2mek)=2 3kg m s and horizontal momentum: px=p*cos30°3kg m s.

Conservation of momentum in the horizontal direction at the moment of entering the trolley: that is, the momentum in the horizontal direction at the moment of falling into the trolley is px and the friction force f= (m+m)g, and the distance of the trolley is s, then the energy is conserved: fs=δek=px 2 2(m+m).

So s=px 2 2(m+m) f=

The whole process has gravity and friction to do the work.

mgh=fs

s=mgh f=mg(l*sin30) u(m+m)g= m 1m.

First, use the ox two or the conservation of mechanical energy to find out the velocity at the end of the bottom of the inclined plane.

mgh=mv^2/2

v = 2gh under the root number

Here's the key.

Just now this speed is down the inclined plane, after falling into the car, the vertical component of this speed is lost, only the horizontal component remains, so this speed multiplied by cos30 is a useful horizontal velocity.

It is equal to 2GH under the cos30* root number

Then this velocity and the car hit, applying the conservation of momentum.

2 kg * cos30 * root number under 2gh = 15 kg * v to solve this v is the horizontal initial velocity of the two of them.

Then you can use Niu Er to find the distance.

This is a collision problem, first find the velocity at the beginning of the collision.

v = root number 3m s

After the collision, the car and the object reach the same speed v'= The collision time and displacement are negligible, pay attention to the velocity decomposition).

Let's do it.

2 root number 2 times v 2 g

Brother, you have too many topics.

(1) If the angle is 45 degrees, then at the end of one second, horizontal velocity = vertical velocity = gt = 10m s2 vertical velocity at the end of the second = g2 = 20m s; The horizontal velocity is unchanged = 10 m s, so the velocity size at the end of two seconds is 10v5 m s

2) The initial velocity of the fall in the vertical direction is 0, which can be understood as the opening of the square = 15s from the height of 1125m free fall time t = 2 * 1125 10

Horizontal distance = 50v3 * 15 = 750 v3 m5 vertical velocity at the end of seconds = g * 5 = 50m s; Horizontal velocity = 50v3 m s, the velocity magnitude is 100m s, the direction is tana = 50, 50v3 = 1 v3, that is, it is 30 degrees with the horizontal.

4) 45 degrees, 20m

5） s ;45 degrees horizontally.

Anyway- I remember that there is a formula, what is tangent than what can be set of formulas - - formula I forgot - -

Since the geomagnetic pole and the north and south poles do not overlap, it can be considered that the magnetic inductance line of the northern hemisphere is deflected downward with the ground level, and the right hand rule, the thumb points to the east, and the palm of the hand is upward, then the four point to the north, because the aircraft does not constitute a closed loop, there is no electric current, at this time, there is only induced electromotive force and no charge movement, and the potential difference cannot be formed, so the potential energy of the left wing and the right wing (only the electric potential energy and gravitational potential energy are considered in high school) are equal.

I just talked about it in a round of review, and I thought that the left wing was high, because I felt that the charge was running over there, but as long as the charge moved, there would be an electric current, and this was ironclad and there was no current, so there was no electric field line, so there was no work from the left end to the right end, so there was no potential difference...

In addition, I checked and found that the knowledge of the physics competition was used, which is summarized as follows: in a closed loop in a uniformly varying magnetic field, because there is an electromotive force everywhere, the electric potential rises and the electric potential decreases cancel, so the electric potential is equal everywhere, and the voltage exists between any two points, otherwise there will be no current.

In the case of an airplane, there is neither a potential difference, nor a voltage, nor an electric current.

I opted for the chemistry competition, not a physics major, so I can't say it well, please bear with me

The right-hand rule. Beijing - Northern Hemisphere.

The direction of the magnetic inductance line is from south to north, and there is a downward component.

The left wing has high potential energy.

Southern hemisphere or northern hemisphere.

From left to right facing the computer.

The set resistance is R1R2R3

The analysis of the meaning of the question can be known:

V1 is the voltage at both ends of the resistor R1 and R2.

V2 is the voltage at the r@ and R3 of the resistor.

The resistance that can crack the old to find r2 is:

u=ir=oh.

In fact, the circuit is a circuit with 3 empty orange resistors in series, so the ammeter measures the branch current) Then you can know the voltage at both ends of R1.

That's 6-1=5

It is also possible to know that the current of R3 is 9-1=8

Since the voltage of the series circuit is equal to the sum of the voltages of the individual resistors.

So the total voltage = 1 + 5 + 8 = 13V

From left to right, the resistors are R1, R2, R3

As can be seen from the figure. The burn-up measured by V1 is the stool blind R1, and the total voltage of R2 is 6V, while V2 measures R2, and the total voltage of R3 is 9V

Because 3 resistors are connected in series, ur2 = multiplied by 2 = 1v and ur1 + ur2 + ur3 = ur1 + ur2 + ur2 + ur3 - ur2 = 6 + 9-1 = 14v