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Anika [276]
4 years ago
6

these are the types of attractions found between molecules. their strength determines the state of the substance at room tempera

ture.
Physics
2 answers:
Elina [12.6K]4 years ago
7 0
<h2>Answer</h2>

The physical state of the elements depends upon the <u>attraction forces </u>and their <u>kinetic energy</u>.

<h2>Explanation</h2>

The elements or substances are fixed with each other with the help of different chemical forces including ionic bonding, covalent bonding, H- bonding etc. The strength of these forces is also one of the factors that affect their physical natures. For example, covalent or ionic bonds are the strongest bonds than all other bonds and metals that contain these forces are mostly in solid form. The kinetic motion of electrons in the element also affects the physical state of the element and potential of bonding.

insens350 [35]4 years ago
5 0

Answer:

Intermolecular forces

Explanation:

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Define absorption and explain why it makes light dimmer as it travels farther away
ale4655 [162]

Answer:

the process or action by which one thing absorbs or is absorbed by another.

Explanation:

bc it spreeds out more the farther away it gets

8 0
3 years ago
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An automotive paint sequence on a suspect's car is as follows: clear coat, silverstone ii metallic, primer. a paint smear is fou
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3 years ago
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a grinding wheel is initially rotating with an angular velocity 5500 rad/srad/s when its motor is suddenly turned off. it comes
kiruha [24]

The angle through which the grinding wheel rotates in the first second =  <u>5300 rad</u>

Angular velocity is, the time charge at which an object rotates, or revolves, about an axis, or at which the angular displacement between our bodies changes. within the discern, this displacement is represented via the angle θ among a line on one body and a line on the alternative.

The angular velocity is described as the charge of trade of the angular position of a rotating body. Linear speed is defined because the charge of change of displacement with respect to time whilst the item moves alongside a straight course.

Initial angular velocity of the grinding wheel = ω1 = 5500 rad/s

Final angular velocity of the grinding wheel = ω2 = 0 rad/s   (Comes to rest)

Time is taken by the grinding wheel to come to rest = T = 10 sec

Angular acceleration of the grinding wheel = α

2 = ω1 + αT

0 = 5500 + α(10)

α = - 400 rad/s2

Negative as it is deceleration.

The angle through which the grinding wheel rotates in the first second = θ

Time period = T1 = 1 sec

θ = ω₁T1 + αT1²/2

θ = (5500)(1) + (-400)(1)²/2

θ = 5300 rad

The angle through which the grinding wheel rotates in the first second =  <u>5300 rad</u>

Learn more about angular velocity here:-brainly.com/question/6860269

#SPJ4

5 0
1 year ago
A sample of an unknown gas has a density of 0.0009 g/cm3.
Gnoma [55]

Answer:   A

The option A is the most accurate description of unknown gas density relative to other gases in the table.

Observe the table clearly and compare it with given density of unknown gas i.e., <em>0.0009 g/cm³ . </em>

In the table, it is given that

1. Hydrogen  - 0.00009 g/cm³, which is less than 0.0009 g/cm³, because the value of hydrogen has four zeros after the decimal point on right side, (that is decimal zero places up to ten thousands) but when comparing the value of unknown gas, it has zeros up to thousands place only on right side.

Hence, while moving on right side of the decimal point, the number of zeros decreases the value of given number.

(0.00009 > 0.0009)

similarly  observe the remaining values

2. 0.0009  >  0.00018 (Helium)

3. 0.0009  <  0.00122 (AIr)

4. 0.0009  <  0.00142 (Oxygen)

<em>So, answer is A. </em>


8 0
3 years ago
Read 2 more answers
A series circuit has a capacitor of 0.25 × 10−6 F, a resistor of 5 × 103 Ω, and an inductor of 1 H. The initial charge on the ca
svetoff [14.1K]

Answer:

q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Explanation:

Given that L=1H, R=5000\Omega, \ C=0.25\times10^{-6}F, \ \ E(t)=12V, we use Kirchhoff's 2nd Law to determine the sum of voltage drop as:

E(t)=\sum{Voltage \ Drop}\\\\L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)\\\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+\frac{1}{0.25\times10^{-6}}q=12\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+4000000q=12\\\\m^2+5000m+4000000=0\\\\(m+4000)(m+1000)=0\\\\m=-4000  \ or \ m=-1000\\\\q_c=c_1e^{-4000t}+c_2e^{-1000t}

#To find the particular solution:

Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Hence the charge at any time, t is q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

6 0
4 years ago
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