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3 years ago

Which of the following is a chemical property of a mineral?

Physics

2 answers:

zzz [600]3 years ago

6 0

Hardness, color , and luster.

But since you can pick one, I would say hardness

zaharov [31]3 years ago

5 0

The chemical properties of a mineral are:-Hardness,Luster and Color.

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What happens to mechanical energy if the potential energy of an object decreases?

OleMash [197]

D. Mechanical energy remains unchanged.

Mechanical energy = Potential energy + Kinetic energy

So if potential energy decreases, then kinetic energy must increase so that the system is balanced.

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3 years ago

An airplane flies 250.0 km at 300 m/s. How long does this take?

givi [52]

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3 years ago

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What units would you use to measure time that has elapsed?

dedylja [7]

You would use seconds as seconds is a unit of time

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3 years ago

A metal block has a density of 5000 kg per cubic meter and a mass of 15,000 kg. What is its volume?

Naily [24]

Taking into account the definition of density, the volume of the metal block is 3 m³.

<h3>What is density</h3>

Density is defined as the property that matter, whether solid, liquid or gas, has to compress into a given space.

In other words, density is a quantity that allows us to measure the amount of mass in a certain volume of a substance.

Then, the expression for the calculation of density is the quotient between the mass of a body and the volume it occupies:

$density=\frac{mass}{volume}$

From this expression it can be deduced that density is inversely proportional to volume: the smaller the volume occupied by a given mass, the higher the density.

<h3>Volume of the metal block</h3>

In this case, you know that:

Density= 5000 $\frac{kg}{m^{3} }$
Mass= 15000 kg
Volume= ?

Replacing in the definition of density:

$5000 \frac{kg}{m^{3} } =\frac{15000 kg}{volume}$

Solving:

volume×5000 $\frac{kg}{m^{3} }$ = 15000 kg

volume= $\frac{15000 kg}{5000 \frac{kg}{m^{3} }}$

<u><em>volume= 3 m³</em></u>

In summary, the volume of the metal block is 3 m³.

Learn more about density:

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4 0

2 years ago

The inclined plane in the figure above has two sections of equal length and different roughness. The dashed line shows where sec

saveliy_v [14]

The static friction exerted on the block by the incline is $\mu _ s _1 Mgcos \ \theta$ .

The given parameters;

<em>mass of the block, = M</em>
<em>coefficient of static friction in section 1, = </em> $\mu_s_1$ <em />
<em>angle of inclination of the plane, = θ</em>

<em />

The normal force on the block is calculated as follows;

Fₙ = Mgcosθ

The static friction exerted on the block by the incline is calculated as follows;

$F_s = \mu_s F_n\\\\F_s = \mu _s_1(Mg cos\ \theta)\\\\F_s = \mu _s_1 Mgcos\ \theta$

Thus, the static friction exerted on the block by the incline is $\mu _ s _1 Mgcos \ \theta$

Learn more here:brainly.com/question/17237604

3 0

3 years ago