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3 years ago

________ is characterized by the separation of light- and dark-colored minerals into thin layers or bands.

Physics

1 answer:

erastova [34]3 years ago

6 0

Answer:

Granitic gneiss is characterized by the separation of light- and dark-colored minerals into thin layers or bands.

Explanation:

Intense heat and pressure can also metamorphose granite into a banded rock known as "granite gneiss." This transformation is usually more of a structural change than a mineralogical transformation. Granite gneiss can also form through the metamorphism of sedimentary rocks.

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A 600W toaster, 1200W iron and a 100W bulb are all connected to household 120V circuit. a) find the current drawn by each applia

Advocard [28]

Answer:

(a) %a, 10 A, 0.833 A

(b) 12 ohm

Explanation:

Power of toaster, P1 = 600 W

Power of iron, P2 = 1200 W

Power of bulb, P3 = 100 W

V = 120 V

As they are in household circuit, so they are connected in parallel, so the voltage is same for all.

(a) Use the formula P = V x i

Current in toaster, i1 = P1 / V = 600 / 120 = 5 A

Current in iron, i2 = P2 / V = 1200 / 120 = 10 A

Current in bulb, i3 = P3 / V = 100 / 120 = 0.833 A

(b) Resistance of heating element of iron is R2.

V = i2 x R2

120 = 10 x R2

R2 = 12 ohm

= (600 + 1200 + 100) x 24 x 5 = 228000 Wh = 228 KWh

Cost of 1 KWh = $0.1

Cost of 228 KWh = 0.1 x 228 = $22.8

8 0

3 years ago

Read 2 more answers

Two slits separated by a distance of d = 0.190 mm are located at a distance of D = 1.91 m from a screen. The screen is oriented

Svetlanka [38]

Answer:

$\theta = 0.195^0$

Explanation:

wavelength $\lambda = 648 nm \ = 648*10^{-9}m$

d = 0.190 mm = 0.190 × 10⁻³ m

D = 1.91 m

By using the formula:

$dsin \theta = n \lambda\\\\\theta = sin^{(-1)}(\frac{n \lambda}{d})\\\\\\\theta = sin^{(-1)}(\frac{1*648*10^{-9}}{0.190*10^{-3}})$

$\theta = 0.195^0$

The first maximum will appear at an angle $\theta = 0.195^0$ from the beam axis

3 0

3 years ago

Coherent light with wavelength 598 nm passes through two very narrow slits, and the interference pattern is observed on a screen

Stella [2.4K]

Answer:

1.196 μm

Explanation:

D = Screen distance = 3 m

$\lambda$ = Wavelength = 598 m

y = Distance of first-order bright fringe from the center of the central bright fringe = 4.84 mm

d = Slit distance

$tan\theta=\frac{y}{D}\\\Rightarrow \theta=tan^{-1}{\frac{y}{D}}\\\Rightarrow \theta=tan^{-1}{\frac{4.84\times 10^{-3}}{3}}\\\Rightarrow \theta=0.09243\ ^{\circ}$

$sin\theta=\frac{\lambda}{d}\\\Rightarrow d=\frac{\lambda}{sin\theta}\\\Rightarrow d=\frac{598\times 10^{-9}}{sin0.09243}\\\Rightarrow d=0.00037066\ m$

For first dark fringe

$dsin\theta=\frac{\lambda'}{2}\\\Rightarrow \lambda'=2dsin\theta\\\Rightarrow \lambda'=2\times 0.00037066\times sin0.09243\\\Rightarrow \lambda'=1.196\times 10^{-6}\\\Rightarrow \lambda'=1.196\ \mu m$