Tomorrow will take the exam and ask a few physics questions Hurry up Thank you

1.The centripetal acceleration provided by the gravitational force is: a=gm r 2

So: a1:a2=gm1 r1 2 : gm2 r2 2 = (m1 m2)*(r2 2 r1 2), substitute your data but your r1:r2 value is not given!

Because the centripetal acceleration a=gm r 2=r(2 t) 2, so t = under the root number (4 2r 3 (gm)).

Therefore, the ratio of the period is: t1:t2 root number ((r1 3*m2) (r2 3*m1)), and you can substitute your data!

2.As can be seen from the question, on this planet, the gravitational acceleration is one-tenth of that of the Earth (60 600 = 1 10), i.e., g'=

For a satellite in low orbit on the surface of the planet (the radius of motion is the radius of the planet), gravity provides the centripetal force: mg=gmm r 2, r is the radius of the planet, and the substitution relation: gm=r 2g

For the Earth: gm(ground) = r(ground) 2*g

For planets: gm(rows) = r(rows) 2*g'

Compared with the upper and lower about g, substituting the data can get:

r(row) r(ground)=8 1

3.The first cosmic velocity is the velocity of the moon (mass m) mentioned above, thus

mg=mv 2 r, where r is the radius of the planet. Get: v = root number (rg).

In this topic:

v = under the root number (rg).

g=mg The two equations are solved:

r=mv^2/g

Come to an end! I hope you do well in the exam!

f=g(mm) r squared.

This is the formula is the gravitational force formula.

g = -11 power.

a=vsquared r=wsquare*r

This is the circular motion formula.

1 Raise an object weighing 200n by 1m with a movable pulley, and the tensile force used is 125n, then the useful work done by the movable pulley w is useful = 200 J, and his mechanical efficiency = self-percentage). If friction is not taken into account, the mechanical efficiency of the pulley block increases when the weight of the object to be lifted increases.

2. Under the action of the horizontal tensile force F of 50N, the object weighing 800N moves in a uniform straight line along the horizontal ground, and the sliding friction between the object and the ground is 120N, then the mechanical efficiency of the pulley block is; If the velocity of the object is, then the work done by the tensile force in 1 min is 1200 J

3 Xiaohong then used the pulley block to do the test of "measuring the mechanical efficiency of the pulley block", and used the spring dynamometer to pull the object weighing 3N vertically at a constant speed, the object was raised, the indication of the spring dynamometer was 3 N, the useful work done to the object during the experiment was J, and the mechanical efficiency of the pulley block was.

1 The action of the man, the wall, the force, the force, and the force is mutual.

2 c Rope, because you take the rope in your hand, and the pull of the bucket acts on the rope, and the hand is pulled by the rope.

3 b A square wooden stool, 20n is about 2kg, which can be easily lifted with one hand. A blackboard erases less than 300g, a desk 6 10kg, and middle school students 50kg

Person. Wall. Forced. Apply force. The force action is reciprocal

1. The answers are man, wall, stillness, and motion;

2、c3、b

1. When rollerblading, push the wall hard, but the person is pushed away by the wall, which shows that the person is also affected by the force of the wall. At this time, if the human being is the object of study, the human being is the force object, and the wall is the force object, this phenomenon shows that the force is reciprocal

2. Tie the bucket with a rope, hold the rope to lift water from the well, and the hand is subjected to a vertical downward tension, and the force object of this tension is (c) a. Earth b bucket c rope d hand.

3. The object that can be lifted with a force of about 20n may be (b) a blackboard eraser b a square wooden stool c a desk d a middle school student.

1. The force applied to the human wall The effect of the force between the objects is mutual.

2、c3、b

The knowledge of this place is focused on understanding and hope to accept this answer.

1.If Xiao Ming wants to measure the width of a physics textbook, he should choose a scale with a graduation value (the minimum scale is millimeters), if the actual measurement result is, the accurate value is (, and the estimated value is (

1. Change: In order to reduce the error, the method of (averaging multiple measurements) should be adopted.

Two variations: If the results of the three measurements are , then the width of the textbook is (2How do you roughly measure the distance from the classroom to the school gate?

Measure the distance you walk in one step m0;

Record the number of steps taken from the classroom to the school gate;

Calculate the distance from the classroom to the school gate: nm0

The particle moves in a uniform and circular motion in the electric field, and the particle is released at the position of the y-axis away from the point d at the point o, and the velocity entering the magnetic field is v, and the radius of the particle's circular motion in the magnetic field is r

In the electric field qed=1 2mV

In the magnetic field qvb=mv r

The particle moves in a semi-circular motion in a magnetic field n times to reach the point m, then.

l=2nr Missing reed calendar

d=b l q (8n me) (n=1,2,3...).)<

After entering the magnetic field, the particles move in a uniform circular motion.

In this case, the radius is only related to velocity.

Make l=2r.

The first question of the first question, you can draw a V-T diagram, use the area to represent the distance, calculate, list a system of equations, the first question and the second question are the same as above, that is, the diagram is a triangle on the line, note: when setting t, you can set two, one is 4 times the other one, the inverse ratio of speed The second question, first calculate the length of the trunk 14,!! Then think backwards, put the car still, move people, and calculate the time it takes for people to travel 56 to 70.

The third question, because of uniform deceleration, still draw a V-T diagram. An RT triangle with an area of 3 to 4 when remaining time.

It's easy to solve.

1.(1) The time elapsed to accelerate to VM t1=vm A1, and the time elapsed to deceleration to zero t2=vm A2

Area on the V-T image 130 vm=1600+then,vm=

2) Acceleration is directly followed by deceleration. Look at the area of the triangle in the v-t diagram 1600 = solution vm = 64m s

t=t1+t2=64/