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Ipatiy [6.2K]
3 years ago
10

Considering the symmetry of the charge distribution, determine the symmetry of the electric field and choose one of the followin

g options as the most appropriate choice of Gaussian surface to use in this problem. Considering the symmetry of the charge distribution, determine the symmetry of the electric field and choose one of the following options as the most appropriate choice of Gaussian surface to use in this problem. A cube with one of its edges coinciding with the axis of the rod A cube whose center lies on the axis of the rod and with two faces perpendicular to the rod axis A sphere whose center lies on the axis of the rod A finite closed cylinder whose axis coincides with the axis of the rod An infinite cylinder whose axis coincides with the axis of the rod
Physics
1 answer:
BaLLatris [955]3 years ago
4 0

Answer:

my name jeff hahahhahahahahahhahhahhahhahhahhah the correct a

Explanation:

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Please answer this question for me and explain why.
horsena [70]

Answer:

D.None of these

Explanation:

The derivation of acceleration formula:

Let us call the 5kg mass m_2 and the 4kg mass m_1. If the tension in the string is T then for the mass m_2

(1). T-m_2g=-m_2a <em>(the negative sign on the right side indicates that acceleration is downwards)</em>

And for the mass m_1

(2). T-m_1g =m_1a<em> (the acceleration is upwards, hence the positive sign)</em>

Solving for T in the 2nd equation we get:

T =m_1a+m_1g,

and putting this into the 1st equation we get:

m_1a+m_1g-m_2g=-m_2a\\\\m_1a+m_2a = m_2g-m_1g\\\\a(m_1+m_2)= (m_2-m_1)g

\boxed{a= \dfrac{(m_2-m_1)}{(m_1+m_2)} g}

Back to the question:

Using the formula for the acceleration we find

a= \dfrac{(5kg-4kg)}{(5kg+4kg)} g

a = \dfrac{g}{9},

which is the acceleration that none of the given choices offer. Also, the acceleration of the two blocks is the same, because if it weren't, the difference in the instantaneous velocities of the objects would cause the string to break. Therefore, these two reasons make us decide that none of the choices are correct.

7 0
3 years ago
What happens to the image when u make the hole bigger in a pinhole camera?​
marta [7]

Answer:

more light enters and disturbs the formation of the image.

7 0
2 years ago
Read 2 more answers
You are driving your car along a country road at a speed of 27.0 m/s. as you come over the crest of a hill, you notice a farm tr
Bezzdna [24]

speed of the car = 27 m/s

speed of truck ahead = 10 m/s

relative speed of car with respect to truck

v_r = 27 - 10 = 17 m/s

relative deceleration of car

a_r = -7 m/s^2

now the distance before they stop with respect to each other is given by

v_f^2 - v_i^2 = 2 a d

0 - 17^2 = 2 *(-7)*d

d = 20.6 m

so it will come at the same speed of truck after 20.6 m distance and hence it will not hit the truck as the distance of the truck is 25 m from car

Part b)

Distance traveled by car before it stops is given by

v_f^2 - v_i^2 = 2 a s

0^2  - 27^2 = 2 * (-7)* s

s = 52.1 m

so it will stop after it will cover total 52.1 m distance

Part c)

time taken by the car to stop

v_f - v_i = at

0 - 27 = (-7) * t

t = 3.86 s

now the distance covered by truck in same time

d = 3.86 * 10 = 38.6 m

now after the car will stop its distance from the truck is

D = 25 + 38.6 - 52.1 = 11.5 m

<em>so the distance between them is 11.5 m</em>

6 0
3 years ago
Un avión vuela a 10000m de altura y otro a 33300 pies, si un pie equivale a 30.48cm ¿Cuál vuela a mayor altura?
user100 [1]

Answer:

Avion A (10000 meters).

Explanation:

Deje que la altura de los aviones sea A y B respectivamente.

Dados los siguientes datos;

Altura A = 10000 metros

Altura B = 33300 pies

Para encontrar el avión que voló más alto, tendríamos que hacer alguna conversión de unidades.

Conversión:

Metros a centímetros;

1 metro = 100 cm

10000 metros = 100 * 10000 = 1.000.000 centímetros.

Por lo tanto, la altura A en cm = 1,000,000 centímetros

Pies a centímetros;

1 pie = 30,48 centímetros

33300 pies = 33300 * 30,48 = 1014984 centímetros.

Por lo tanto, la altura B en cm = 1014984 centímetros.

De los cálculos anteriores, podemos deducir que el avión A voló más alto.

6 0
3 years ago
New 5G networks utilize millimeter-wave radiation. Millimeter-wave radiation refers to electromagnetic waves with frequencies in
seraphim [82]

Answer:

It corresponds to 1mm-10 mm range.

Explanation:

  • Electromagnetic waves (such as the millimeter-wave radiation) travel at the speed of light, which is 3*10⁸ m/s in free space.
  • As in any wave, there exists a fixed relationship between speed, frequency and wavelength, as follows:

        v = \lambda * f  (1)

  • Replacing v= c=3*10⁸ m/s, and the extreme values of f (which are givens), in (1) and solving for λ, we can get the free-space wavelengths that correspond to the 30-300 GHz range, as follows:

       \lambda_{low} = \frac{c}{f_{high}}  = \frac{3e8m/s}{300e9Hz} = 1 mm (2)

      \lambda_{high} = \frac{c}{f_{low}}  = \frac{3e8m/s}{30e9Hz} = 10 mm (3)

4 0
3 years ago
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