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Stels [109]
3 years ago
5

In transverse waves, the medium moves perpendicular to the direction of energy transport? True or false?

Physics
1 answer:
saul85 [17]3 years ago
8 0

Answer:

This statement is true

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These the flow of electrons (the current) and where some of the electrons' energy gets converted into heat. (Lessons 5.01-5.03)
liq [111]

Answer:

Conductors.

Explanation:

Conductors allow the flow of electrons i. e. the current and where some of the electrons energy gets converted into heat while some electrons energy converted into light. Conductors are the materials which has the ability to allow heat and electricity to flow through them. Metals such as gold, copper, Silver, Aluminum, Mercury, Steel, Iron and salty water etc are good conductors of heat and electricity.

3 0
3 years ago
A 3.6-volt battery is used to operate a cell phone
Marina CMI [18]

Explanation :

It is given that,

Potential energy, V=3.6\ V

Power dissipated,  P=0.064\ Watt

We know that the power dissipated is given by :

P=VI

I is the current passing through the phone.

I=\dfrac{P}{V}

I=\dfrac{0.064\ W}{3.6\ V}

I=0.017\ A

or

I = 0.018 A

Hence, the current that passes through the phone is (1) 0.018 A.

3 0
3 years ago
Read 2 more answers
a 2.0-mole sample of an ideal gas is gently heated at constant temperature 330 k. it expands from initial volume 19 l to final v
shutvik [7]
Isothermal Work =  PVln(v₂/v₁)

PV = nRT =  2 mole * 8.314 J/ (k.mol) * 330 k = 5487.24 J

Isothermal Work =  PVln(v₂/v₁)            v₂ = ? v₁ = 19L, 

1.7 kJ = (5487.24)In(v₂/19)

1700 = (5487.24)In(v₂/19)

In(v₂/19) = (1700/5487.24) = 0.3098

In(v₂/19) = 0.3098

(v₂/19) = e^{0.3098}


v₂  =  19* e^{0.3098}

v₂ = 25.8999

v₂ ≈ 26 L        Option b.
6 0
3 years ago
Determine the empirical formula for a compound that is 36.86% n and 63.14% o by mass.
olga2289 [7]

Answer:

NO_{1.499}

Explanation:

Let assume that 100 kg of the compound is tested. The quantity of kilomoles for each element are, respectively:

n_{N} = \frac{36.86\,kg}{14.006\,\frac{kg}{kmol} }

n_{N} = 2.632\,kmol

n_{O} = \frac{63.14\,kg}{15.999\,\frac{kg}{kmol} }

n_{O} = 3.946\,kmol

Ratio of kilomoles oxygen to kilomole nitrogen is:

n^{*} = \frac{3.946\,kmol}{2.632\,kmol}

n^{*}= 1.499

It means that exists 1.499 kilomole oxygen for each kilomole nitrogen.

The empirical formula for the compound is:

NO_{1.499}

8 0
3 years ago
Read 2 more answers
You launch a ball at an angle of 35 degrees above the horizontal with an initial velocity of 38 m/s. What is the time the ball w
gayaneshka [121]

Vf=Vi+at

0=38+(-9.8)(?)

?=38-0+(-9.8)

?=28.2 s

5 0
3 years ago
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