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3 years ago

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An octave is the Question 3 options: a) absolute frequency difference between two notes in the same interval. b) musical distanc

e between two chords. c) amplitude of a piece of music. d) frequency range of a particular piece of music. e) interval between two sound frequencies having a ratio of 2:1.

1 answer:

Scorpion4ik [409]3 years ago

6 0

Answer:

e) interval between two sound frequencies having a ratio of 2:1.

Explanation:

Have a great day! ^w^

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How often do spring and neap tides occur

iragen [17]

Answer:

Without thought to the season, Spring tides happen [two times] every [lunar] month the whole year. When the Sun and the Moon are quadratic, Neap tides also happen two times during a month.

Here is a picture to help show you:

(With the Spring tide, the moon is toward the "left" or "right" of the Earth. With the Neap tide, the moon is "above" or "below/under" the Earth. It's kind of hard to explain).

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3 years ago

Convert 9.065 m to<br> cm<br> 90.65<br> 906.5<br> 9065<br> 0.0965<br> SUBMIT ANSWER

DedPeter [7]

Answer:

906.5 cm

Explanation:

1 m = 100 cm

9.065 m = 100 cm * 9.065 m = 906.5 cm

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3 years ago

The amount of energy that remains after the amount of energy used for respiration is subtracted from the total is called

lawyer [7]

Answer:Gibb's free energy

Explanation:

The Free energy change describes the amount of energy that is available in any system to do work. It is often designated with the symbol G

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4 years ago

A metal ring 5.00 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpend

labwork [276]

Answer:

Ein: 2.75*10^-3 N/C

Explanation:

The induced electric field can be calculated by using the following path integral:

$\int E_{in} dl=-\frac{\Phi_B}{dt}$

Where:

dl: diferencial of circumference of the ring

circumference of the ring = 2πr = 2π(5.00/2)=15.70cm = 0.157 m

ФB: magnetic flux = AB (A: area of the loop = πr^2 = 1.96*10^-3 m^2)

The electric field is always parallel to the dl vector. Then you have:

$E_{in}\int dl=E_{in}(2\pi r)=E_{in}(0.157m)$

Next, you take into account that the area of the ring is constant and that dB/dt = - 0.220T/s. Thus, you obtain:

$E_{in}(0.157m)=-A\frac{dB}{dt}=-(1.96*10^{-3}m^2)(-0.220T/s)=4.31*10^{-4}m^2T/s\\\\E_{in}=\frac{4.31*10^{-4}m^2T/s}{0.157m}=2.75*10^{-3}\frac{N}{C}$

hence, the induced electric field is 2.75*10^-3 N/C

8 0

3 years ago

A piston having 7.23 g of steam at 110°C increases its temperature by 35°C. At the same time it expands from a volume of 2.00 L

My name is Ann [436]

Answer : The value of $q,w,\Delta U\text{ and }\Delta U$ is 505 J, -599 J, -94 J and -693 J respectively.

Explanation : Given,

Mass of steam = 7.23 g

Initial temperature = $110^oC$

Final temperature = $(110+35)^oC=145^oC$

Initial volume = 2 L

Final volume = 8 L

External pressure = 0.985 bar

Heat capacity of steam = 1.996 J/g.K

First law of thermodynamic : It states that the energy can not be created or destroyed, it can only change or transfer from one state to another state.

As per first law of thermodynamic,

$\Delta U=q+w$

First we have to calculate the heat absorbed by the system.

Formula used :

$Q=m\times c\times \Delta T$

or,

$Q=m\times c\times (T_2-T_1)$

where,

Q = heat absorbed by the system = ?

m = mass of steam = 7.23 g

$C_p$ = heat capacity of steam = $1.966J/g.K$

$T_1$ = initial temperature = $110^oC=273+110=383K$

$T_2$ = final temperature = $145^oC=273+145=418K$

Now put all the given value in the above formula, we get:

$Q=7.23g\times 1.966J/g.K\times (418-383)K$

$Q=505J$

Now we have to calculate the work done.

Formula used :

$w=-p_{ext}dV\\\\w=-p_{ext}(V_2-V_1)$

where,

w = work done = ?

$p_{ext}$ = external pressure = 0.985 bar = 0.985 atm (1 bar = 1 atm)

$V_1$ = initial volume of gas = 2.00 L

$V_2$ = final volume of gas = 8.00 L

Now put all the given values in the above formula, we get :

$w=-p_{ext}(V_2-V_1)$

$w=-(0.985atm)\times (8.00-2.00)L$

$w=-5.91L.atm=-5.91\times 101.3J=-599J$

conversion used : (1 L.atm = 101.3 J)

Now we have to calculate the change in internal energy of the system.

$\Delta U=q+w$

$\Delta U=505J+(-599J)$

$\Delta U=-94J$

Now we have to calculate the change in enthalpy of the system.

Formula used :

$\Delta H=\Delta U+P\Delta V$

$\Delta H=\Delta U+w$

$\Delta H=(-94J)+(-599J)$

$\Delta H=-693J$

Therefore, the value of $q,w,\Delta U\text{ and }\Delta U$ is 505 J, -599 J, -94 J and -693 J respectively.

4 0

4 years ago