1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
allochka39001 [22]
3 years ago
12

Magnetism, reactivity, and fluorescence are three special properties used to identify minerals. Please select the best answer fr

om the choices provided
True
or False
Physics
2 answers:
Dominik [7]3 years ago
6 0

Answer:true

Explanation: test

astraxan [27]3 years ago
3 0

Answer: True

Explanation: Following are the properties that are used to identify minerals:

(1) Color

(2) Hardness

(3) Magnetism : It is the property of mineral to attract or repel with other magnetic materials.

(4) Luster : It is the property that shows the surface is reflecting light or not.

(5) Reactivity: Reactivity is also responsible to identify mineral. How the mineral is reacting with acids, bases etc helps to identify type of minerals.

You might be interested in
What is quantum physics? ​
lara [203]

Answer:

A fundamental theory that provides a description of the physical properties of nature at the scale of atoms and subatomic particles.

Explanation:

5 0
2 years ago
What are the characteristics of the radiation emitted by a blackbody? According to Wien's Law, how many times hotter is an objec
jasenka [17]

Answer:

a) What are the characteristics of the radiation emitted by a blackbody?

The total emitted energy per unit of time and per unit of area depends in its temperature (Stefan-Boltzmann law).

The peak of emission for the spectrum will be displaced to shorter wavelengths as the temperature increase (Wien’s displacement law).

The spectral density energy is related with the temperature and the wavelength (Planck’s law).

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wave length of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

Explanation:

A blackbody is an ideal body that absorbs all the thermal radiation that hits its surface, thus becoming an excellent emitter, as these bodies express themselves without light radiation, and therefore they look black.

The radiation of a blackbody depends only on its temperature, thus being independent of its shape, material and internal constitution.

If it is study the behavior of the total energy emitted from a blackbody at different temperatures, it can be seen how as the temperature increases the energy will also increase, this energy emitted by the blackbody is known as spectral radiance and the result of the behavior described previously is Stefan's law:

E = \sigma T^{4}  (1)

Where \sigma is the Stefan-Boltzmann constant and T is the temperature.

The Wien’s displacement law establish how the peak of emission of the spectrum will be displace to shorter wavelengths as the temperature increase (inversely proportional):

\lambda max = \frac{2.898x10^{-3} m. K}{T}   (2)

Planck’s law relate the temperature with the spectral energy density (shape) of the spectrum:

E_{\lambda} = {{8 \pi h c}\over{{\lambda}^5}{(e^{({hc}/{\lambda \kappa T})}-1)}}}  (3)

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wavelength of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

It is need it to known the temperature of both objects before doing the comparison. That can be done by means of the Wien’s displacement law.

Equation (2) can be rewrite in terms of T:

T = \frac{2.898x10^{-3} m. K}{\lambda max}   (4)

Case for the object with the blackbody emission spectrum peak in the blue:

Before replacing all the values in equation (4), \lambda max (450 nm) will be express in meters:

450 nm . \frac{1m}{1x10^{9} nm}  ⇒ 4.5x10^{-7}m

T = \frac{2.898x10^{-3} m. K}{4.5x10^{-7}m}

T = 6440 K

Case for the object with the blackbody emission spectrum peak in the red:

Following the same approach above:

700 nm . \frac{1m}{1x10^{9} nm}  ⇒ 7x10^{-7}m

T = \frac{2.898x10^{-3} m. K}{7x10^{-7}m}

T = 4140 K

Comparison:

\frac{6440 K}{4140 K} = 1.55

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

4 0
3 years ago
A circuit is made of a battery, a light bulb, and a 2 resistor. The battery has a voltage of 3 volts. When connected, the ammete
Monica [59]

Answer:

3ohms

Explanation:

From Ohm's Law

V = IR

V is that voltage = 3volts

I = current = 1amp

R = resistance in ohms

Putting those values into the above formula.

3volts = 1amp×R

Making R the subject

R = 3/1

R = 3ohms

The resistance of the light bulb is 3ohms.

6 0
3 years ago
working alongside the pharmacist , one of the duties a medication reconciliation technician would perform is to a ) deal with va
Vanyuwa [196]

Answer:

okkiikkkkkkkkoiiiiiiii8iiii

5 0
2 years ago
Three charges lie along the x axis. The positive charge q1 = 15 μC is at x= 2.0 m and the positive charge q2 = 6.0μC is at the o
Delicious77 [7]

Answer:

x = 0.775m

Explanation:

Conceptual analysis

In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.

We apply Coulomb's law to calculate the electrical forces on q₃:

F_{23} = \frac{k*q_{2}*q_{3}}{x^2} (Electric force of q₂ over q₃)

F_{13} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (Electric force of q₁ over q₃)

Known data

q₁ = 15 μC = 15*10⁻⁶ C

q₂ = 6 μC = 6*10⁻⁶ C

Problem development

F₂₃ = F₁₃

\frac{k*q_{2}*q_{3}}{x^2} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (We cancel k and q₃)

\frac{q_2}{x^2}=\frac{q_1}{(2-x)^2}

q₂(2-x)² = q₁x²

6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)

6(2-x)² = 15(x)²

6(4-4x+x²) = 15x²

24 - 24x + 6x² = 15x²

9x² + 24x - 24 = 0

The solution of the quadratic equation is:

x₁ = 0.775m

x₂ = -3.44m

x₁ meets the conditions for the forces to cancel in q₃

x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel

The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.

6 0
3 years ago
Read 2 more answers
Other questions:
  • The gas tank of Dave’s car has a capacity of 12 gallons. The tank was 38 full before Dave filled it to capacity. It cost him $2.
    8·2 answers
  • A certain lightning bolt moves 40.0 C of charge. How many fundamental units of charge |qe| is this?
    13·2 answers
  • Because water molecules are polar and carbon dioxide molecules are non polar A water has a lower boiling point than carbon dioxi
    14·2 answers
  • Three forces act on an object. a 3 n force acts due east and a 8 n force acts due north. if the net force on the object is zero,
    14·1 answer
  • A box is pulled 6 meters across the ground at a constant velocity by a horizontally applied force of 50 newtons. At the same tim
    11·1 answer
  • The condition of a country’s depends on its people’s ability to exchange money for goods and services.
    15·2 answers
  • mass is tied to spring and begins periodically . the distance between its highest and its lowest position is 48cm. what is the a
    15·1 answer
  • A river with a flow rate of 7 m3/s discharges into a lake with a volume of 9,000,000 m3. The concentration of a particular VOC i
    9·1 answer
  • Protective equipment and protective measures help keep all types of workers safe on the job.
    7·2 answers
  • Explain how you would find the volume of a regular-shaped cube in comparison to an irregular shaped object such as a rock.
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!