Physics winter vacation homework, summer vacation homework physics

Put a steel ball into a graduated cylinder containing 100ml of water, and the water surface rises to 160ml. The balance is used to weigh the mass of the ball to 234g, is the steel ball hollow or solid? If it is hollow, fill the hollow part with kerosene, then what is the total mass of the steel ball?

Steel=, Kerosene= Analysis: Combined with the requirement of filling the hollow part with kerosene in the second question, it is better to choose the method of comparing the volume to identify the hollow. According to the rising height of the water surface in the graduated cylinder, the volume of the steel ball is 60cm3, and the volume of the steel is found to be compared with the volume of the ball, if it is the same, it is solid, and if it is less than 60cm3, it is hollow.

Solution: The volume of the ball: v ball = v2-v1 = 160-100 = 60cm3 The volume of steel:

v steel = m = 234 so the steel ball is hollow. m total = m steel oil (v ball - v steel) = 234

1. 234 grams divided by the density of steel, the volume of steel is about cubic centimeters, that is, if it is a solid steel ball, cubic centimeters of water can be discharged, and the actual discharged water is 60 grams, and its volume is 60 cubic centimeters, which means that there is about cubic centimeters of space inside the steel ball.

2. Multiply the density of kerosene by grams, this part of the space can be filled with grams of kerosene, plus the mass of the steel balls, and the total mass is grams. (Done).

You will receive a system reward of 20 (Fortune + XP) + 20 (Fortune + XP) for solving within 15 minutes [14 minutes and 26 seconds until the end].

The formula for acceleration due to gravity: g = gm r 2 (where g is a constant).

So g fire: g earth = m fire r fire 2: m earth r land 2 = (m fire m earth) (r fire: r earth) 2

p q 2 This is to memorize the formula.

The apple falls straight down from the tree, which means that the direction of gravity is straight down, and the earth has gravity.

Peeled eggs plug the mouth of the bottle and fall when the cotton is extinguished.

Explanation: Igniting the cotton, the oxygen burns out, the pressure inside the bottle is small, the pressure outside is strong, and the atmospheric pressure pushes the egg down.

Find the relation: v1=gt1

v2 = displacement is represented by v-t plot area: (1 2) (v1 + v2) (t2-t1) = 125

Three equations, v1, t1, t2 three unknowns, can be solved.

The height from the ground when the athlete leaves the aircraft is S1+125m, and after leaving the aircraft, the time of reaching the ground is T2

When the friction force is 2N, the horizontal tensile force is 4N, the resultant external force is 2N, according to F=MA, the mass m can be found to be 1kg, when the tensile force is 10N, because the friction force is unchanged, the combined external force is 8N, or according to F=MA, a can be found to be 8M S2

It may be a homogeneous addition or subtraction. It should be multi-solution. 1: 1 kg, 8 m s2.

2: 3 kg, 4 m s2.

Use Newton's second law to solve the problem. f-f=ma, i.e.

4-2=2m to obtain m=1kg

10-2=ma yields a=8m s 2

By voltage invariant rli=2 3i(30+rl).

The solution is rl=60 ohms.

Since this circuit is a series circuit, (set the supply voltage to U), there is:

When the sliding rheostat is at terminal A: ia=u rl; B-side: ib=u (rl+30): and ib=2 3ia

The answer is r=60 euros.

Solution: Landing speed: v1= (2gh1)**Off-grid speed: v2= (2gh2).

The solution is v1=8 v2=10

v1-（-v2）=at

The solution is a=15

f = ma = 60 times 15 = 900 N.

f bomb = f + mg = 1500 N answer ......

v1 is the speed of the catenary when the athlete falls, v2 is the speed of the athlete** off the net, and v0 is the initial velocity of the athlete.

From v1 2-v0 2=2gx1 x1=v1=v1=8m s direction, straight downward.

v0* 2-v2 2=-2gx2 x2=get v2=10m s

A=(v2-v1) t yields a=15m s2 by f=ma+g f=1500n

The magnitude of this force is 1500N

Landing speed: v1= (2gh1) h1= h2= t=

**Off-grid speed: v2= (2gh2).

ft=mv2-mv1 f=

Break down the movement into 3 processes.

The first process: the athlete falls from the meter, and the speed v1 2=2*10* can get v1=8 meters per second.

The second process: after the force of the net on the athlete, ft=mv1--(mv2) the third process; The athlete does a uniform deceleration movement of 5 meters upward, v2 2=2*10*5 to obtain v2=10 meters per second.

In summary: f=900n