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2 years ago

12

Isabella deja caer accidentalmente un bolígrafo desde su balcón mientras celebra que resolvió satisfactoriamente un problema de

física Al suponer que la resistencia del aire es baja cuántos segundos tarda el bolígrafo en alcanzar una rapidez de 19.62 m/s

1 answer:

jok3333 [9.3K]2 years ago

8 0

Answer:

i cant read spanish lol

Explanation:

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How does energy from the sun affect the motion of molecules in a gas compared to molecules in a liquid?

AnnZ [28]

Molecules in a gas move faster than in a liquid.

hope it helps.

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1 year ago

Read 2 more answers

Two sinusoidal waves travel along the same string. They have the same wavelength and frequency. Their amplitudes are ym1 = 2.5 m

Nimfa-mama [501]

Answer:

0.5 m

Explanation:

Givens:

ym1 = 2.5 mm

ym2 = 4.5 mm

Ф_1=π / 4

Ф_2=π / 2

We have 2 ways to solve this problem. The first one given that the 2 waves have the frequency then we know that the resultant wave amplitude is

Ym = (ym1 + ym2)cos(Ф_2/2)

By substitution we have

Ym= (0.025 + 0.045)cos(π/4) = 0.496 m

The second one is it treat them as Phasors where the phase between them is Ф_2=π / 2 Therefore

Ym^2=(ym1^2+ym2^2)

So we have Ym=√0.025^2+0.045^2

= 0.5 m

7 0

2 years ago

The inductor in a radio receiver carries a current of amplitude 200 mA when a voltage of amplitude 2.40 V is across it at a freq

White raven [17]

Answer:92

Explanation:

3 0

2 years ago

3) 4 electrons are placed - one electron per corner - at the corners of a square of side 1 meter. One fixed proton is placed in

Eduardwww [97]

Explanation:

3

i believe that they are all going at 3.2 meters each, I did 4 times 0.8

4 0

2 years ago

Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.75 A out of the jun

AlexFokin [52]

Answer:

number of electrons = 2.18*10^18 e

Explanation:

In order to calculate the number of electrons that move trough the second wire, you take into account one of the Kirchoff's laws. All the current that goes inside the junction, has to go out the junction.

Then, if you assume that the current of the wire 1 and 3 go inside the junction, then, all this current have to go out trough the second junction:

$i_1+i_3=i_2$ (1)

i1 = 0.40 A

i2 = 0.75 A

you solve the equation i3 from the equation (1):

$i_3=i_2-i_1=0.75A-0.40A=0.35A$

Next, you take into account that 1A = 1C/s = 6.24*10^18

Then, you have:

$0.35A=0.35\frac{C}{s}=0.35*\frac{6.24*10^{18}e}{s}=2.18*10^{18}\frac{e}{s}$

The number of electrons that trough the wire 3 is 2.18*10^18 e/s

3 0

2 years ago