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3 years ago

1) Una carga de 10 uC colocada en un campo eléctrico experimenta una fuerza de 60x104N ¿Cuál es la magnitud del campo eléctrico?

Physics

2 answers:

Dmitrij [34]3 years ago

5 0

Answer:

pls translate in English

wolverine [178]3 years ago

4 0

Answer:

yeah it'd be okay if you translate to English

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They compress or expand depending on amount of pressure or depending on the temperature

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A cantilever beam with a width b=100 mm and depth h=150 mm has a length L=2 m and is subjected to a point load P =500 N at B. Ca

DerKrebs [107]

Answer:

Explanation:

Given that:

width b=100mm

depth h=150 mm

length L=2 m =200mm

point load P =500 N

Calculate moment of inertia

$I=\frac{bh^3}{12} \\\\=\frac{100 \times 150^3}{12} \\\\=28125000\ m m^4$

Point C is subjected to bending moment

Calculate the bending moment of point C

M = P x 1.5

= 500 x 1.5

= 750 N.m

M = 750 × 10³ N.mm

Calculate bending stress at point C

$\sigma=\frac{M.y}{I} \\\\=\frac{(750\times10^3)(25)}{28125000} \\\\=0.0667 \ MPa\\\\ \sigma =666.67\ kPa$

Calculate the first moment of area below point C

$Q=A \bar y\\\\=(50 \times 100)(25 +\frac{50}{2} )\\\\Q=250000\ mm$

Now calculate shear stress at point C

$=\frac{FQ}{It}$

$=\frac{500*250000}{28125000*100} \\\\=0.0444\ MPa\\\\=44.4\ KPa$

Calculate the principal stress at point C

$\sigma_{1,2}=\frac{\sigma_x+\sigma_y}{2} \pm\sqrt{(\frac{\sigma_x-\sigma_y}{2} ) + (\tau)^2} \\\\=\frac{666.67+0}{2} \pm\sqrt{(\frac{666.67-0}{2} )^2 \pm(44.44)^2} \ [ \sigma_y=0]\\\\=333.33\pm336.28\\\\ \sigma_1=333.33+336.28\\=669.61KPa\\\\\sigma_2=333.33-336.28\\=-2.95KPa$

Calculate the maximum shear stress at piont C

$\tau=\frac{\sigma_1-\sigma_2}{2}\\\\=\frac{669.61-(-2.95)}{2} \\\\=336.28KPa$

6 0

3 years ago

Vlad1618 [11]

Answer:

a) 16m/s b) 192m

Explanation:

v1=32m/s a=-2m/s^2 t=8s v2=? d=??

a) I will use this equation v2= v1 + a*t

v2= 32m/s + -2m/s^2 * 8s

v2= 32m/s + -16m/s

v2= 16m/s

b) v2^2=v1^2 + 2ad

rearranging

v2^2-v1^2=2ad

v2^2-v1^2/2= a d

v2^2-v1^2/2a=d

16m/s^2 - 32m/s^2/ 2 x-2m/s^2 =d

d=192m

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NISA [10]

By the Newtons Second Law:

F = ma

F = (47 kg)(15 m/s²)

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3 years ago

180 J of work is done when lifting a box up to a shelf that is 3 m high. What is the mass of the box?

ale4655 [162]

Explanation:

Work done= MGH

180=m×10×3

180/30=m

m=6kg

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3 years ago