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2 years ago

6

A student sorted mineral samples into two groups: dull and shiny. Which of the following properties did the student use to sort

the mineral samples into groups? a
streak
b
luster
c
cleavage
d
color free brainliest

2 answers:

Katyanochek1 [597]2 years ago

7 0

Answer:

luster

Explanation:

xenn [34]2 years ago

4 0

The answer would be luster

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The tuning fork has a frequency of 426.7 is found to cause resonance in a closed column of air measuring 0.186 (the diameter of

lidiya [134]

Answer:

The velocity of the tuning fork sound is 317.4648 m/s

Explanation:

The given parameters are;

The frequency of the fork, f = 426.7 Hz

The length of the closed air column, L = 0.186 m

The diameter of the tube, d = 0.15 m

At the fundamental frequency in a closed tube, we have;

λ = 4·L

Where;

λ = The wavelength of the wave

L = The length of the tube

From the equation for the velocity of a wave, we have;

v = f·λ

Where;

v = The velocity of the (sound) wave

f = The frequency of the wave = 426.7 Hz

λ = 4·L = 4 × 0.186 m = 0.744 m

∴ v = 426.7 Hz × 0.744 m = 317.4648 m/s

Therefore, the velocity of the sound produced by the tuning fork, v = 317.4648 m/s

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3 years ago

How efficient are the small and large scale solar-power systems used in individual homes and industrial settings? What is the en

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Answer:

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Explanation:

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3 years ago

The part of the Electromagnetic (Light) Spectrum used to see through the dust

dimaraw [331]

Answer:

A or C

Explanation:

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3 years ago

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A garden hose attached with a nozzle is used to fill a 10‐ gal bucket. The inner diameter of the hose is 2 cm, and it reduces to

nadya68 [22]

Answer:

Explanation:

Given

Volume of bucket $V=10\ gallon$

Time taken to fill the bucket $t=50\ s$

so volume flow rate is $\dot{V}=\frac{10}{50}=0.2\ gal/s$

1 gal is equivalent to $0.133\ ft^3$

$\dot{V}=0.0267\ ft^3/s$

mass flow rate $\dot{m}=\rho \times \dot{V}$

$\dot{m}=62.4\times 0.0267$

$\dot{m}=1.668\ lbs$

(b)Average velocity through nozzle exit

$\dot{V}=Av_{avg}$

$v_{avg}=\dfrac{0.0267}{\frac{\pi}{4}\times (0.0262)^2}$

$v_{avg}=49.51\ ft/s$

8 0

3 years ago

When a man returns to his well-sealed house on a summer day, he finds that the house is at 35°C. He turns on the air conditioner

Paul [167]

Answer:

1353.38 Watt

Explanation:

T₁ = Initial temperature of the house = 35°C

T₂ = Final temperature of the house = 20°C

Δt = Time taken to cool the house = 38 min = 38×60 = 2280 s

m = mass of air in the house = 800 kg

Cv = Specific heat at constant volume = 0.72 kJ/kgK

Cp = Specific heat at constant pressure = 1.0 kJ/kgK

Heat removed

q = mCvΔT

⇒q = 800×720×(35-20)

⇒q = 8640000 J

Average rate of hear removal

$Q=\frac{q}{\Delta t}\\\Rightarrow Q=\frac{8640000}{2280}\\\Rightarrow Q=3789.47\ W$

$COP=\frac{Q}{W}\\\Rightarrow W=\frac{Q}{COP}\\\Rightarrow W=\frac{3789.47}{2.8}\\\Rightarrow W=1353.38\ W$

∴ Power drawn by the air conditioner is 1353.38 Watt

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2 years ago

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