Physics junior high school life answers, junior high school physics questions and answers

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First, the large ** that occurs in mountainous areas often forms a lot of dammed lakes, which may burst at any time, seriously threatening the lives of the people downstream. If the water depth at the bottom of the embankment of a dammed lake has reached 55 meters, the density of the water is 10 kg

1) Find the pressure of water at the bottom of the embankment in the above case.

2) Assuming that the maximum pressure of the water that the bottom of the embankment and the parts of the embankment can withstand is 6x10 to the power of 5 pa, how many meters can the water surface rise be allowed?

Solution: 1: p=pgh, (p is the density of water).

10n kg 55 meters.

Handkerchief. 2：p=pgh

6x10 to the power of 5 = 10 n kg h

h = 60 m.

60-55 = 5 meters.

Therefore, it is necessary to lift 5 meters.

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3. A middle school research study group drove a boat to Dongchang Lake. They tied the stone with a thin wire, and immediately measured its gravity with a spring meter, and then put the stone into the lake, and the spring dynamometer showed it. When Shilu Ai touched the bottom of the lake, they made a mark on the surface of the water on the western line and measured the length of the line from the stone to the mark.

g takes 10n|kg, the density of water is cubic).

1) The buoyancy of the stones in the lake.

2) The pressure of the lake water on the bottom of the lake.

3) The density of the stone.

Solution: 1) The buoyancy of the stones in the lake water: f buoyancy =

2) The pressure of the lake water on the bottom of the lake here: p= gh=1000*10*3) The mass of the stone m=g g=

The volume of the stone v stone = v row = f float ( water g) = cubic meter.

Density of stones: = m stone v stone = cubic meter = 2500kg cubic meter.

Left ob and od balance lever f1*l1=f2*l2 will not.

Solution: Because the OA segment and the OB segment are elastic deformations, it can be seen by observing the image that the maximum tensile force that A can withstand is 4N, and the maximum tensile force that B can withstand is 6N. When the tensile force received by A is 4N, the spring elongation is 6cm, when the tensile force received by B is 8N, the spring elongation is 4cm, indicating that the tensile force of A is small, but the elongated length is long

So the answer is: B; a

The first one chooses B, and the second one chooses A.

B can measure to 8N, while A can only measure to 4N, so B's range is large (to clarify: the range is based on the condition that the spring is not broken, that is, the elastic deformation stage).

According to f=k*l, you can find their respective k, and the higher the ** clear accuracy of k, so the second one is selected.

B, A's reason is: B can measure to 8N, while A can only measure to 4N, so B's range is large (to explain: the range is based on the condition that the spring is not broken, that is, the elastic deformation stage).

According to f=k*l, you can find their respective k, and the higher the ** clear accuracy of k, so the second one is selected.

If B is selected with a large range, the elongation of B is shorter when subjected to the same amount of force; High precision nail selection, under the action of the same force, the elongation is longer, and the index value can be more accurate.

The first is B, and the second is A.

When the switches S1 and S2 are closed, R2 is short-circuited, R1 and R3 are connected in parallel, and the voltmeter measures the power supply voltage, i.e.:

u=u1=u3=12v

i1=u1/r1=12v/24ω=

The current indication measures the dry circuit current, i.e., i=

The current in r3 i3=i-i1=

r3=u3/i3=12v/

When the switches S1 and S2 are disconnected, R1 and R2 are connected in series, and the current in the ammeter circuit R total = R1 + R2 = 24 + 8 ohms = 32

The indication of the ammeter is i=ur=12v32=

When switches S1 and S2 are closed, the circuit R1 and R3 are connected in parallel to both ends of the power supply. The current in r1 is i1=12 24=

The current in r3 is i2=i-i1=

r3=u/i2=12/

When the switches S1 and S2 are disconnected, the circuit is R1 and R2 connected in series to both ends of the power supply.

i=u/(r1+r2)=12/(24+8)=

When S1 and S2 are closed, R1 and R3 are connected in parallel, R2 is short-circuited, and the voltage at both ends of R1 and R3 is 12V

r1 = 24 ohms.

Current through R1 i1 = U R1 = 12V 24 ohms = Current through R3 i3 = i a i1 = one.

Then r3 = u i3 = 12v ohms

I can only use Chinese...

1. Only consider the movement in the vertical direction (free fall). h=gt; 2 2 g is taken to obtain t=

2. The time of whereabouts is known. . . Only horizontal motion (uniform linear line) is considered: s=vt=22*

It's simple.

Let the power supply voltage be U R1=U 2 36 R2=U 2 18 when connected in parallel

In series, p total = u 2 r total = u 2 r1+r2 and then substitute the data u 2 to be eliminated.

It is concluded that p total = 12w

To calculate the power of r1, use p=i 2*r=u*r1 (r1+r2) and then substitute r1 r2 to calculate p=4w

The above are all typed by yourself, if you don't understand, you can ask me.

Solution: When R1 and R2 are connected in parallel, the voltages are equal.

p1=u r1 =36w p2=u r2=18w p1 p2=r2 r1=2 1 r2=2r1 When r1 and r2 are connected in series, the total voltage is u, then the voltage of r1 is u1=u 3 r string=r1+r2=3r1

P string = u 3r1 = 12w

p1’=u1²/r1=u²/9r1=36w/9=4w

Suppose the voltage is U, for R1 electrical power = U2 R1, so R1=U2 36, for R2=U2 18

When connected in series, the current i=U (R1+R2) Power consumed = i 2*R1+i 2*R2 = 12W

The power consumed on R1 accounts for 2 3 of the total power, so it is 8W

It's too simple, my brother doesn't bother to do it.