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Vedmedyk [2.9K]
3 years ago
10

Un electrón en un tubo de rayos catódicos acelera desde el reposo con una aceleración constante de 5.33x10¹²m/s² durante 0.150μs

(1μs-10⁻⁶s). El electrón se mueve otros 0.200μs a velocidad constante y, finalmente se para con una aceleración de -2.67x10¹³m/s² ¿Qué distancia recorre el electrón?
Physics
1 answer:
Andre45 [30]3 years ago
3 0
Can you translate that in English ? I'll try to help you out with that..
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A rope of negligible mass passes over a uniform cylindrical pulley of 1.50kg massand 0.090m radius. The bearings of the pulley h
boyakko [2]

Answer:

Explanation:

mass of pulley, m3 = 1.5 kg

Radius of pulley, R = 0.09 m

mass of monkey, m2 = 4.5 kg

mass of banana bunch, m1 = 3 kg

Let a is teh acceleration ans T1 and T2 be the tension in the rope.

The moment of inertia of the pulley

I = 0.5 x m3 x R² = 0.5 x 1.5 x 0.09 x 0.09 = 0.006075 kgm²

According to Newton's second law

T1 - m1 g = m1 x a .... (1)

m2 g - T2 = m2 x a ..... (2)

(T2 - T1 ) x R = I x α    

where, α is the angular acceleration

α = a / R

(T2 - T1)R = 0.5 x m3 x R² x a / R

T2 - T1 = 0.5 x m3 x a ..... (3)  

from (1), (2) and (3)

a = \frac{m_{2}-m_{1}}{m_{1}+m_{2}+\frac{m_{3}}{2}}\times g

a = \frac{4.5-3}{3+4.5+0.75}\times 9.8

a = 1.78 m/s²

from equation (1)

T1 = m1 ( g + a) = 3 ( 9.8 + 1.78) = 34.77 N

from equation (2)

T2 = m2 (g - a) = 4.5 (9.8 - 1.78) = 36.13 N

4 0
3 years ago
Which phrase best describes the outer planets?
pav-90 [236]

Answer:

Astronomers have divided the eight planets of our solar system into the inner planets and the outer planets. The 4 inner planets are the closest to the Sun, and the outer planets are the other four – Jupiter, Saturn, Uranus, and Neptune. The outer planets are also called the Jovian planets or gas giants.

Explanation:

6 0
3 years ago
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A fire engine is moving south at 35 m/s while blowing its siren at a frequency of 400 Hz.
vodomira [7]

To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as

f_d= f_s \frac{(v+v_d)}{(v-v_s)}

Here,

f_d=frequency received by detector

f_s=frequency of wave emitted by source

v_d=velocity of detector

v_s=velocity of source

v=velocity of sound wave

Replacing we have that,

f_d = 400(\frac{(343+18)}{(343-35)})

f_d=422 Hz

Therefore the frequencty that will hear the passengers is 422Hz

8 0
3 years ago
Calculate the electric field at the center of a square
pantera1 [17]

Answer:

E_y=1175510.2\ N.C^{-1}

The Magnitude of electric field is in the upward direction as shown directly towards the charge q_1.

Explanation:

Given:

  • side of a square, a=52.5\ cm
  • charge on one corner of the square, q_1=+45\times 10^{-6}\ C
  • charge on the remaining 3 corners of the square,q_2=q_3=q_4=-27\times 10^{-6}\ C

<u>Distance of the center from each corners</u>=\frac{1}{2} \times diagonals

diagonal=\sqrt{52.5^2+52.5^2}

diagonal=74.25\cm=0.7425\ m

∴Distance of center from corners, b=0.3712\ m

Now, electric field due to charges is given as:

E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}

<u>For charge q_1 we have the field lines emerging out of the charge since it is positively charged:</u>

E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}

  • E_1=2938775.5\ N.C^{-1}

<u>Force by each of the charges at the remaining corners:</u>

E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}

  • E_2=E_3=E_4=1763265.3\ N.C^{-1}

<u> Now, net electric field in the vertical direction:</u>

E_y=E_1-E_4

E_y=1175510.2\ N.C^{-1}

<u>Now, net electric field in the horizontal direction:</u>

E_y=E_2-E_3

E_y=0\ N.C^{-1}

So the Magnitude of electric field is in the upward direction as shown directly towards the charge q_1.

8 0
3 years ago
Steam types of<br> forces in<br> nature
max2010maxim [7]

Answer:

Gravity, Weak, Electromagnetic and Strong.

7 0
3 years ago
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