PhysicsPhysicsPhysics Urgency There must be a problem solving process

Sorry I haven't learned yet.

1.The circuit current i=u (r1+r2)=5 16 w1=i 2*r1=i 2*r2=w2 is the same.

Ohms r2 = u 2 p2 = 484 ohms.

i=u (r1+r2)=220 2420a=1 11ap1=i 2*r1=16w p2=i 2*r2=4w, so u 2 r=

p3=(2u)^2/r=4*

1.Just as big

Because they are connected in series, the current of the two resistors is equal, and the resistance is also equal, and the power gain is equal according to p=i i r.

2 1500revs (kWh) means that 1500 revolutions of the meter are the energy consumed in kilowatt hours w 1 1000w 3600s

Time t=100s

So p w t=24w

3 R1 1936 Euro and R2 484 Euro according to r u u p.

According to i u (r1 r2) i 5 88a is obtained by p i i r: p1 6 25 watts p2 3 125 watts.

4 p1 u u r p2 (u 3) (u 3) rp1 p2 (8u u 9) r 0 8w, so u u r 0 9w

So 2u 2u r 4u u r 4 0 9w 3 6w

The first question is how to feel the bright ...... of low power in a series circuit

r (resistance of the soldering iron); r (the resistance of the bulb);

U square r square = 40 (the electric soldering iron is rated power when the switch is closed) to solve the ru square * r (r + r) square = 40 4 (when the switch is disconnected, the electric soldering iron is 1 4 of the rated power).

Just solve r lao ( o number is more difficult to calculate, I'm lazy hehe.

The resistance of the soldering iron.

r branding = (uu) amount p amount.

220v×220v）/40w

When the electric power of the soldering iron is 10W, the voltage at both ends of the soldering iron is U branding, (Uu) branding = P branding R branding.

10w×1210ω

12100vv

U branding = 110V

The voltage at both ends of the lamp.

U light = u-u branding.

220v-110v

The current of the 110V series circuit is equal, and the distribution of voltage is proportional to the resistance.

U lamp u branding = (IR lamp) (IR branding).

R lamp R branding.

Because U light = U branding, so R light = R branding = 1210

The bulb resistance is 1210 ohms.

Specific proof: because the rated power of the electric network iron is 40W, and the rated voltage is 220V, the resistance is.

220V) divided by 40 watts is equal to 1210 ohms, because the electric network iron belongs to the state of insulation, the power at this time is 1 4 of the rated power, so they divide the voltage 110

So the bulb resistance is also 1210 ohms.

This question is very simple, the minimum must not be less than 500 meters, that is, when the limit is 500 meters, the falling speed of the aircraft is just 0. In the state of overweight with a maximum of two times the gravity, that is, the resultant force is upward to double the gravitational force, that is, the deceleration is -gIt is equivalent to 25s of free fall, and then decelerates with -g for 25s to 500 meters in the air, and the speed is just 0.

That is, the height is at least 500 + 2 (g * 25 2 2) = 500 + 6250 = 6750 meters.

The descent of the aircraft is divided into two processes: one. 25s of weightlessness. Two. Deceleration with acceleration g (2g-g) with a final velocity of 0 is 500m above the ground.

s1=s=2s1+500=6750m

There are two processes to this problem: first it falls freely, and then it decelerates to zero with twice the acceleration of gravity.

Momentum is conserved, m1*v1=m2*v2

v1=m2*v2/m1=

That is, the backward speed of the machine gun.

Momentum theorem: the speed at which the machine gun retreats = the mass of the bullet * the speed of the bullet The mass of the machine gun =

a=g 2So, the actual acceleration is.

a0=4a/5

v=√（2*

40√2（m/s）

t=√（2*400/

10√2（s）

Glide acceleration a=g*sin30-g*sin30*; (Multiplying both sides of the equation by m at the same time is better understood, and I can do this if I'm proficient).

v^2=2as=3200;

t=v/a;

I didn't play the straw paper at hand, so the landlord will count it himself.

According to mg=mv2r, obtained.

The velocity of the ball through the highest point of the arc orbit is at least v= (gr) and is conserved by mechanical energy:

mgh=mg2r+

h=Hope the above answer can be helpful to you