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3 years ago

The diagram below shows a portion of the rock cycle.

Physics

1 answer:

SCORPION-xisa [38]3 years ago

8 0

Referring to the attached image below which represents the complete rock cycle, we will find that a marble can be formed when hot magma is cooled and crystallized.

Applying this to the given diagram, we will find that marble is most likely formed in position C

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If you drive at 50 km/hr for 8 hours, how far will you go?

bonufazy [111]

50km/hr* 8hr = 400 km

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3 years ago

Runner x starts 50 feet in front of runner z and moves with a velocity of 8.0 m/s. At what speed would runner z have to run to c

Mariulka [41]

Answer:

15.625

Explanation:

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3 years ago

Light shines through a single slit whose width is 5.7 x 10-4 m. A diffraction pattern is formed on a flat screen located 4.0 m a

lilavasa [31]

Answer:

$\lambda = 570\ nm$

Explanation:

Given,

Width of slit, W = 5.7 x 10⁻⁴ m

Distance between central bright fringe, L = 4 m

distance between central bright fringe and first dark fringe, y = 4 mm

Diffraction angle

$tan \theta = \dfrac{y}{L}$

$tan \theta = \dfrac{4}{4\times 10^3}$

$\theta = 0.0572$

Now.

$W sin \theta = m \lambda$

m = 1

$5.7 \times 10^{-4} \times sin (0.0572) = 1 \times \lambda$

$\lambda = 569.99 \times 10^{-9}\ m$

$\lambda = 570\ nm$

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3 years ago

What color is the sky?

alexdok [17]

A clear cloudless day-time sky is blue because molecules in the air scatter blue light from the sun more than they scatter red light. When we look towards the sun at sunset, we see red and orange colours because the blue light has been scattered out and away from the line of sight.

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4 years ago

Read 2 more answers

Infer When one object heats another, does the temperature increase of one object always equal the temperature decrease of the ot

scoray [572]

Answer:

Explanation:

When one object heats another, the amount of heat transferred by the hotter object ( $Q_h$ ) to the colder object is equal to the amount of heat absorbed by the colder object ( $Q_c$ ):

$Q_h=Q_c$

The two amounts of heat can be written as:

$Q_h=m_h C_h \Delta T_h$

where

$m_h$ is the mass of the hotter object

$C_h$ is the specific heat capacity of the hotter object

$\Delta T_h$ is the change in temperature of the hotter object

And

$Q_c=m_c C_c \Delta T_c$

where

$m_c$ is the mass of the colder object

$C_c$ is the specific heat capacity of the colder object

$\Delta T_c$ is the change in temperature of the colder object

So we can write

$m_h C_h \Delta T_h = m_c C_c \Delta T_c$

$\frac{\Delta T_h}{\Delta T_c}=\frac{m_c C_c}{m_h C_h}$

We see that this ratio is not always equal to 1, since the two objects can have different masses and specific heat capacities: therefore, the changes in temperature are not equal.

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4 years ago