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3 years ago

Que hace falta para descubrir hasta que punto todas y todos somos diferentes y ver que eso, más bien es valioso y positivo

Physics

1 answer:

lara [203]3 years ago

5 0

Answer:

Para descubrir que las diferencias entre las personas son valiosas y positivas para la creación de sociedades más diversas y enriquecidas culturalmente, es necesario valorar las condiciones personales de cada persona de modo positivo, abandonando posturas discriminatorias como el racismo o la xenofobia, para así aceptar la diversidad social. Además, debe fomentarse el respeto por la opinión y la visión del otro, para que las personas no tomen al que piensa diferente como un adversario al que hay que confrontar. Solo así podrá lograrse una sociedad más diversa y respetuosa de las diferencias, sean culturales, étnicas o intelectuales.

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A 100kg couch is being pushed with 196N of force. As it slides along the ground it experiences a coefficient of friction of 0.1.

QveST [7]

Answer:98

Explanation:hope this helps!

4 0

3 years ago

Which of the following has high specific heat capacity?

Molodets [167]

Option D - Water (why does it needs to be 20 words long)

4 0

3 years ago

A thick, spherical shell made of solid metal has an inner radius a = 0.18 m and an outer radius b = 0.46 m, and is initially unc

KonstantinChe [14]

(a) $E(r) = \frac{q}{4\pi \epsilon_0 r^2}$

We can solve the different part of the problem by using Gauss theorem.

Considering a Gaussian spherical surface with radius r<a (inside the shell), we can write:

$E(r) \cdot 4\pi r^2 = \frac{q}{\epsilon_0}$

where q is the charge contained in the spherical surface, so

$q=5.00 C$

Solving for E(r), we find the expression of the field for r<a:

$E(r) = \frac{q}{4\pi \epsilon_0 r^2}$

(b) 0

The electric field strength in the region a < r < b is zero. This is due to the fact that the charge +q placed at the center of the shell induces an opposite charge -q on the inner surface of the shell (r=a), while the outer surface of the shell (r=b) will acquire a net charge of +q.

So, if we use Gauss theorem for the region a < r < b, we get

$E(r) \cdot 4\pi r^2 = \frac{q'}{\epsilon_0}$

however, the charge q' contained in the Gaussian sphere of radius r is now the sum of the charge at the centre (+q) and the charge induced on the inner surface of the shell (-q), so

q' = + q - q = 0

And so we find

E(r) = 0

(c) $E(r) = \frac{q}{4\pi \epsilon_0 r^2}$

We can use again Gauss theorem:

$E(r) \cdot 4\pi r^2 = \frac{q'}{\epsilon_0}$ (1)

where this time r > b (outside the shell), so the gaussian surface this time contained:

- the charge +q at the centre

- the inner surface, with a charge of -q

- the outer surface, with a charge of +q

So the net charge is

q' = +q -q +q = +q

And so solving (1) we find

$E(r) = \frac{q}{4\pi \epsilon_0 r^2}$

which is identical to the expression of the field inside the shell.

(d) $-12.3 C/m^2$

We said that at r = a, a charge of -q is induced. The induced charge density will be

$\sigma_a = \frac{-q}{4\pi a^2}$

where $4 \pi a^2$ is the area of the inner surface of the shell. Substituting

q = 5.00 C

a = 0.18 m

We find the induced charge density:

$\sigma_a = \frac{-5.00 C}{4\pi (0.18 m)^2}=-12.3 C/m^2$

(e) $-1.9 C/m^2$

We said that at r = b, a charge of +q is induced. The induced charge density will be

$\sigma_b = \frac{+q}{4\pi b^2}$

where $4 \pi b^2$ is the area of the outer surface of the shell. Substituting

q = 5.00 C

b = 0.46 m

We find the induced charge density:

$\sigma_b = \frac{+5.00 C}{4\pi (0.46 m)^2}=-1.9 C/m^2$

3 0

3 years ago

A cue ball initially moving at 3.4 m/s strikes a stationary eight ball of the same size and mass. After the collision, the cue b

romanna [79]

Answer:

speed of eight ball speed after the collision is 3.27 m/s

Explanation:

given data

initially moving v1i = 3.4 m/s

final speed is v1f = 0.94 m/s

angle = θ w.r.t. original line of motion

solution

we assume elastic collision

so here using conservation of energy

initial kinetic energy = final kinetic energy .............1

before collision kinetic energy = 0.5 × m× (v1i)²

and

after collision kinetic energy = 0.5 × m× (v1f)² + 0.5 × m× (v2f)²

put in equation 1

0.5 × m× (v1i)² = 0.5 × m× (v1f)² + 0.5 × m× (v2f)²

(v2f)² = (v1i)² - (v1f)²

(v2f)² = 3.4² - 0.94²

(v2f)² = 10.68

taking the square root both

v2f = 3.27 m/s

speed of eight ball speed after the collision is 3.27 m/s

5 0

3 years ago

Once you start pulling your object with less force than friction, what should you expect your object to do? What about when your

forsale [732]

#Case -1

If Pulling force is less than frictional force the object won't move .

#Case-2

If Pulling force is greater than frictional force then object will be .

In order to calculate friction force you need Limiting friction first .

$\\ \sf\longmapsto F_L=\mu sN$

u s is coefficient of static friction and N is normal reaction

$\\ \sf\longmapsto F_L=\mu smg$

As N=mg

8 0

3 years ago

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