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3 years ago

Which two options describes behaviors of particles that are related to the chemical properties of the materials

Physics

1 answer:

jarptica [38.1K]3 years ago

3 0

Answer:

The two correct answers are B.) reacting quickly with water, and D.) forming bonds with other atoms.

Explanation:

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A 265 g mass attached to a horizontal spring oscillates at a frequency of 3.40 Hz . At t =0s, the mass is at x= 6.20 cm and has

lara [203]

Answer:

The phase constant is 7.25 degree

Explanation:

given data

mass = 265 g

frequency = 3.40 Hz

time t = 0 s

x = 6.20 cm

vx = - 35.0 cm/s

solution

as phase constant is express as

y = A cosФ ..............1

here A is amplitude that is = $\sqrt{(\frac{v_x}{\omega })^2+y^2 }$ = $\sqrt{(\frac{35}{2\times \pi \times y})^2+6.2^2 }$ = 6.25 cm

put value in equation 1

6.20 = 6.25 cosФ

cosФ = 0.992

Ф = 7.25 degree

so the phase constant is 7.25 degree

5 0

3 years ago

Calculate the orbital period for Jupiter's moon Io, which orbits 4.22×10^5km from the planet's center (M=1.9×10^27kg) .

Verdich [7]

According to the <u>Third Kepler’s Law of Planetary motion</u> “<em>The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $a$ of its orbit.

This Law is originally expressed as follows:

Where;

$G$ is the Gravitational Constant and its value is $6.674(10^{-11})\frac{m^{3}}{kgs^{2}}$

$M=1.9(10^{27})kg$ is the mass of Jupiter

$a=4.22(10^{5})km=4.22(10^{8})m$ is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

$T=\sqrt{\frac{4\pi^{2}}{6.674(10^{-11})\frac{m^{3}}{kgs^{2}}1.9(10^{27})kg}(4.22(10^{8})m)^{3}}$

$T=\sqrt{\frac{2.966(10^{27})m^{3}}{1.268(10^{17})m^{3}/s^{2}}}$

$T=\sqrt{2.339(10^{10})s^{2}}$

Then:

Which is the same as:

Therefore, the answer is:

The orbital period of Io is 42.482 h

7 0

3 years ago

Words a child uses are called the spoken or ___ vocabulary

Vaselesa [24]

The answer is: Expressive vocabulary

Expressive vocabulary refers to the combination of all the words that a person has acquire throughout his/her life and can be used in various type of situations.

This would include all words in the child vocabulary, starting from the child's written language, spoken language or even the child's manually signed words.

4 0

3 years ago

Electricity costs 6 cents per kilowatt hour. In one month one home uses one megawatt hour of electricity. How much will the elec

Luda [366]

Answer:

Cost of 1000 kilowatt hour = 6000 cents

Explanation:

Given that

Electricity cost is 6 cents per kilowatt hour.

And we have to found out the cost for one megawatt hour

We know that

1 kilowatt = 1000 watt

1 megawatt = = 1000000 watt

1 megawatt = 1000 kilowatt

1 megawatt hour = 1000 kilowatt hour

Given that cost of 1 kilowatt hour = 6 cents

So the cost of 1000 kilowatt hour = 6 x 1000 cents

Cost of 1000 kilowatt hour = 6000 cents

3 0

3 years ago

When a magnetic field is first turned on, the flux through a 20-turn loop varies with time according to Φm=5.0t2−2.0t, where Φm

goldfiish [28.3K]

Answer:

A) ( - 200t + 40 ) volts

B) b) anticlockwise , c) anticlockwise , d) clockwise , e) clockwise

Explanation:

Given data:

magnetic flux (Φm) = 5.0t^2 − 2.0t

number of turns = 20

<u>a) determine induced emf </u>

E = - N $\frac{d\beta }{dt}$

= - N ( 10t - 2 ) = - 20 ( 10t - 2 )

= - 200t + 40 volts

<u>b) Determine direction of induced current </u>

i) at t = 0

E = - 0 + 40 ( anticlockwise direction )

ii) at t = 0.10

E = -20 + 40 = 20 ( anticlockwise direction )

iii) at t = 1

E = - 200 + 40 = - 160 ( clockwise direction)

iv) at t = 2

E = -400 + 40 = - 360 ( clockwise direction )

8 0

3 years ago