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3 years ago

11

Can anyone answer this question its a science question NO LINKS !!!!

1 answer:

liubo4ka [24]3 years ago

5 0

Answer: Change

Explanation:

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1 pts<br> How can two different nonmetals form a compound?

Molodets [167]

Explanation:

Covalent or molecular compounds form when elements share electrons in a covalent bond to form molecules. Molecular compounds are electrically neutral. Ionic compounds are (usually) formed when a metal reacts with a nonmetal (or a polyatomic ion). Covalent compounds are formed when two nonmetals react with each other.

3 0

4 years ago

Light traveling through air encounter a second medium which slows the light to 2.7 x 10^8. What is the index of the medium?

kozerog [31]

Answer:

1.11

Explanation:

The index of the medium can be calculated using below formula

V= c/ n ............eqn(1)

Where V= velocity of the light is reduced to while traveling through the second medium= 2.7 x 10^8 m/s

n= index of the medium

c= speed of light= 3 x 10^8 m/s

Substitute for the values in eqn(1)

2.7 x 10^8 = (3 x 10^8 m/s)/ n

Making " n" subject of the formula, we have

n= (3 x 10^8 )/(2.7 x 10^8)

n= 1.11

Hence, the index of the medium is 1.11

5 0

3 years ago

Electrons and protons are basic particles (and together make atoms) that have charge. What is the electric charge of an electron

trapecia [35]

Answer:

hi, here's the answer

Explanation:

the electric charge of an electron is negative (-ve).

pls mark this as the brainliest...

8 0

3 years ago

A solid, horizontal cylinder of mass 18.0 kg and radius 1.70.0 m rotates with an angular speed of 40 rad/s about a fixed vertica

Radda [10]

Answer:39.88 rad/s

Explanation:

Given

mass of cylinder m_1=18 kg

radius R=1.7 m

angular speed $\omega =40rad/s$

mass of $m_2=0.8 kg$ dropped at r=0.3 m from center

let $\omega _2$ be the final angular velocity of cylinder

Conserving Angular momentum

$L_1=L_2$

$\left ( \frac{m_1R^2}{2}\right )\omega =\left ( \frac{m_1R^2}{2}+m_2r^2\right )\omega _2$

$\left ( \frac{18\cdot 1.7^2}{2}\right )\cdot 40=\left ( \frac{18\cdot 1.7^2}{2}+0.8\cdot 0.3^2\right )\omega _2$

$26.01\times 40=26.082\times \omega _2$

$\omega _2=39.88 rad/s$

3 0

3 years ago

A 0.40 kg mass hangs on a spring with a spring constant of 12 N/m. The system oscillated with a constant amplitude of 12 cm. Wha

Vaselesa [24]

Answer:

The maximum acceleration of the system is 359.970 centimeters per square second.

Explanation:

The motion of the mass-spring system is represented by the following formula:

$x(t) = A\cdot \cos (\omega \cdot t + \phi)$

Where:

$x(t)$ - Position of the mass with respect to the equilibrium position, measured in centimeters.

$A$ - Amplitude of the mass-spring system, measured in centimeters.

$\omega$ - Angular frequency, measured in radians per second.

$t$ - Time, measured in seconds.

$\phi$ - Phase, measured in radians.

The acceleration experimented by the mass is obtained by deriving the position equation twice:

$a (t) = -\omega^{2}\cdot A \cdot \cos (\omega\cdot t + \phi)$

Where the maximum acceleration of the system is represented by $\omega^{2}\cdot A$ .

The natural frequency of the mass-spring system is:

$\omega = \sqrt{\frac{k}{m} }$

Where:

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass, measured in kilograms.

If $k = 12\,\frac{N}{m}$ and $m = 0.40\,kg$ , the natural frequency is:

$\omega = \sqrt{\frac{12\,\frac{N}{m} }{0.40\,kg} }$

$\omega \approx 5.477\,\frac{rad}{s}$

Lastly, the maximum acceleration of the system is:

$a_{max} = \left(5.477\,\frac{rad}{s})^{2}\cdot (12\,cm)$

$a_{max} = 359.970\,\frac{cm}{s^{2}}$

The maximum acceleration of the system is 359.970 centimeters per square second.

7 0

3 years ago