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Dimas [21]
3 years ago
15

Nesse texto, um argumento usado para defender a ideia de que o filme A Caminho da Lua cumpre o papel de tocar e de entreter enco

ntra-se no trecho:
“O filme foca em Fei Fei, garota de 12 anos que enfrenta dificuldades em superar o luto pela mãe e aceitar a nova noiva de seu pai.”. (1º parágrafo)
“... a menina decide construir um foguete para viajar até o satélite...”. (1º parágrafo)
“O que faz o longa se destacar é seu cenário mergulhado na cultura chinesa e a delicadeza com que o luto e sua superação são tratados.”. (2º parágrafo)
“... o foco quase exclusivo na garota prejudica todos os outros personagens...”. (3º parágrafo).

ajudem_me
Physics
1 answer:
sammy [17]3 years ago
3 0

Answer:O que faz o longa se destacar é seu cenário mergulhado na cultura chinesa e a delicadeza com que o luto e sua superação são tratados.

Explanation:

Por que se você pensar ela estava sempre buscando superar a morte da mãe.

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Which of the following is an example of a chemical change?
tankabanditka [31]

b. burning paper

hope it helps

3 0
2 years ago
The critical angle for water is 49°. If a ray of light
Sonja [21]

Answer:

Snell's Law states

Ni sin i = Nr sin r

Judging from the question the source of the ray is in the water (directed up)

or NI = 1 / sin 49      Ni = 1.325 deg     the critical angle

From inside the pond:

Nr = 1.325 * sin 45 / 1 = 94 deg  

So refraction can occur  outside the pond and you do not have total internal refection.

 

3 0
3 years ago
PLS HELP FAST!!!!!!!
Elan Coil [88]

Answer:D

Explanation:

6 0
1 year ago
Read 2 more answers
calculate the diameter of a silver wire of length 75cm , which is extended by 1.85mm when a 10kg mass is suspended from it's end
sdas [7]

Answer:0.8\ mm

Explanation:

Given

length of wire l=75\ cm

change in length \Delta l=1.85\ mm

mass of wire m=10\ kg

Young's modulus for silver E=7.9\times 10^{10}\ N/m^2

load on wire F=mg

F=10\times 9.8=98\ kg

change in length is given by

\Delta l=\dfrac{Pl}{AE}

Where A=area of cross-section

A=\dfrac{Pl}{\Delta lE}

A=\dfrac{98\times 0.75}{1.85\times 10^{-3}\times 7.9\times 10^{10}}

A=\dfrac{73.5}{14.615\times 10^{7}}

A=5.029\times 10^{-7}\ m^2

also wire is the shape of cylinder so cross-section is given by

A=\dfrac{\pi d^2}{4}=5.029\times 10^{-7}\ m^2

\Rightarrow d^2=\dfrac{5.029\times 10^{-7}\times 4}{\pi }

\Rightarrow d^2=64.02\times 10^{-8}

\Rightarrow d=8\times 10^{-4}\ m

\Rightarrow d=0.8\ mm

4 0
3 years ago
As a rock sinks deeper and deeper into water of constant density, what happens to the buoyant force on it? As a rock sinks deepe
Sidana [21]

Answer:

It remains constant

Explanation:

As we know that buoyant force on an object given as

Fb = ρ Vd g

ρ= Density of fluid

Vd=Volume displace by body

g=10 m/s²

Fb =buoyant force

So from above we can say that buoyant force does not depends on the depth. It only depends on the fluid density and volume displace by body.

So when rock gets deeper and deeper the buoyant force will remain constant.

It remains constant

3 0
3 years ago
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