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3 years ago

Which layer in the Earth has a composition similar to the granite shown in this photograph?

Physics

1 answer:

Vedmedyk [2.9K]3 years ago

4 0

Answer:

Earth crust and specifically the continental crust.

Explanation:

If we examine the earth crust there is mostly the granite and basalt and most of the granite is present in the continental crust part which is less thicker and denser. That's why we say that the continental crust has the composition similar to that of granite.

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Which literary device portrays a character's reflections?

DochEvi [55]

Of the literary devices you provided to choose from, snap hots and thought shots are not even devices so they fall off the list.

sensory language can explain the senses the character is using - helping figure out his reflections,

but figuratve language can helop the best. the character says something figuratively and we can analyze what they meant to find out what they were thinking

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3 years ago

which form of energy does a battery-powered flashlight receive as an input? chemical A) potential energy B) radiant energy C) so

geniusboy [140]

I believe that the best answer among the choices provided by the question is the second choice ,B) radiant energy


Hope my answer would be a great help for you. If you have more questions feel free to ask here at Brainly.

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3 years ago

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Complete this sentence. Average speed is worked out from dividing _________ by ________

Dmitry [639]

Average speed is worked out from dividing distance by time.

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3 years ago

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At a certain instant, the earth, the moon, and a stationary 1160 kg spacecraft lie at the vertices of an equilateral triangle wh

Afina-wow [57]

Answer:

$W = 1.22 \times 10^9 J$

Explanation:

Initial potential energy of the given spacecraft is given as

$U = -\frac{GM_e m}{r} - \frac{GM_m m}{r}$

so we have

$U = - \frac{Gm}{r}(M_e + M_m)$

so we have

$M_e = 5.98 \times 10^{24} kg$

$M_m = 7.35 \times 10^{22} kg$

$m = 1160 kg$

$r = 3.84 \times 10^8 m$

$U = - \frac{(6.67 \times 10^{-11})(1160)}{3.84 \times 10^8}(5.98 \times 10^{24} + 7.35 \times 10^{22})$

$U = -1.22 \times 10^9 J$

now total work done to move it to infinite is given

W = 0 - U

$W = 1.22 \times 10^9 J$

6 0

3 years ago

Topic Gravitational force amd firld strength.. help me please

I am Lyosha [343]

The gravitational force between m₁ and m₂ has magnitude

$F_{1,2} = \dfrac{Gm_1m_2}{x^2}$

while the gravitational force between m₁ and m₃ has magnitude

$F_{1,3} = \dfrac{Gm_1m_3}{(15-x)^2}$

where x is measured in m.

The mass m₁ is attracted to m₂ in one direction, and attracted to m₃ in the opposite direction such that m₁ in equilibrium. So by Newton's second law, we have

$F_{1,2} - F_{1,3} = 0$

Solve for x :

$\dfrac{Gm_1m_2}{x^2} = \dfrac{Gm_1m_3}{(15-x)^2} \\\\ \dfrac{m_2}{x^2} = \dfrac{m_3}{(15-x)^2} \\\\ \dfrac{(15-x)^2}{x^2} = \dfrac{m_3}{m_2} = \dfrac{60\,\rm kg}{40\,\rm kg} = \dfrac32 \\\\ \left(\dfrac{15-x}x\right)^2 = \dfrac32 \\\\ \left(\dfrac{15}x-1\right)^2 = \dfrac32 \\\\ \dfrac{15}x - 1 = \pm \sqrt{\dfrac32} \\\\ \dfrac{15}x = 1 \pm \sqrt{\dfrac32} \\\\ x = \dfrac{15}{1\pm\sqrt{\dfrac32}}$

The solution with the negative square root is negative, so we throw it out. The other is the one we want,

$x \approx 6.74\,\mathrm m$

5 0

3 years ago