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2 years ago

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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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As a rough approximation, the human body may be considered to be a cylinder of length L=2.0m and circumference C=0.8m. (To simpl

Brilliant_brown [7]

Answer:

Thermal Power = 460W

Explanation:

From Stephan-Boltzmann Law Formula;

P = єσT⁴A

Where,

P = Radiation energy

σ = Stefan-Boltzmann Constant

T = absolute temperature in Kelvin

є = Emissivity of the material.

A=Area of the emitting body

Now, σ = 5.67 x 10^(-8)

є = 0.6

Temperature = 30°C and coverting to kelvin = 30 + 273 = 303K

Area ; since we are to consider the sides of the human body as 2m and 0.8m,thus area = 2 x 0.8 = 1.6

Thus thermal power = 0.6 x 5.67 x 10^(-8) x303⁴ x 1.6 = 458. 8W

Normally, we approximate to the nearest 10W. Thus, thermal power is approximately 460W

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2 years ago

Which of the following represents a convex lens?<br> A. +f<br> B. -di<br> C. -f<br> D. +di

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F is the focal length of the lens. A positive f indicates a converging optical device, like a convex lens or a concave mirror.

Choice A

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2 years ago

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two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i

NARA [144]

A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 - 14.7t - 4.9t^2$

$4.9t^2 + 14.7t - 19.6 = 0$

$t^2 + 3t - 4 = 0$

$(t + 4)(t - 1) = 0$

$(t - 1) = 0$

$\boxed {t = 1 ~ second}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 + 14.7t - 4.9t^2$

$4.9t^2 - 14.7t - 19.6 = 0$

$t^2 - 3t - 4 = 0$

$(t - 4)(t + 1) = 0$

$(t - 4) = 0$

$\boxed {t = 4 ~ seconds}$

The difference in the two ball's time in the air is:

$\Delta t = 4 ~ seconds - 1 ~ second$

$\large {\boxed {\Delta t = 3 ~ seconds} }$

<h2>Question B:</h2><h3>First Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (-14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

<h3>Second Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 11.025 ~ m}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 25.725 ~ m}$

The difference in the two ball's height after 0.500 s is:

$\Delta h = 25.725 ~ m - 11.025 ~ m$

$\large {\boxed {\Delta h = 14.7 ~ m} }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

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2 years ago

You drop a 6.0x10^-2 kg ball from height of 1.0m above hard flat surface. Ball strikes surface and energy decreases by 0.14J, th

harina [27]

If you drop a <span>6.0x10^-2 kg ball from height of 1.0m above hard flat surface, and a</span>fter the ball had bounce off the flat surface, the kinetic energy of the ball would be mgh - 0.14 = 0.45.

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2 years ago

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an elastic cord 61 cm long when a weight of 75N hangs from it, but 85cm when a weight of 210N hangs from it. what is the spring

pishuonlain [190]

Answer:

560 N/m

Explanation:

F = kx

75 N = k (0.61 m − L)

210 N = k (0.85 m − L)

Divide the equations:

2.8 = (0.85 − L) / (0.61 − L)

2.8 (0.61 − L) = 0.85 − L

1.708 − 2.8L = 0.85 − L

0.858 = 1.8L

L = 0.477

Plug into either equation and find k.

75 = k (0.61 − 0.477)

k = 562.5

Rounded to two significant figures, k = 560 N/m.

3 0

1 year ago