What is the mass of an object with a density of 13.5 g/cm3 and a volume of 5cm3?

Helllllllppppp

disa [49]

Answer: c and a

Explanation:

3 0

3 years ago

The ________ method is duct sizing that considers the each duct section has the same air velocity.

Gennadij [26K]

The equal velocity approach for duct size assumes that the air velocity in each duct segment is the same.

How fast is the air moving through a duct?

The most common unit of air velocity (distance traveled in a unit of time) is feet per minute (FPM). The amount of air passing past a location in the duct per period of time may be calculated by multiplying the airflow by the area of the duct. The standard unit for volume flow is cubic feet per minute (CFM).

What happens when the size of ducts changes to the airflow?

Result for an image The equal velocity technique for duct size makes the assumption that air velocity is constant across the entire duct system.

The main lesson to be learned from this is that when air goes from a bigger to a narrower duct, its velocity rises. The velocity drops when it transitions from a shorter to a bigger duct. The flow rate or the amount of air passing through the duct in cubic feet per minute is the same in all scenarios.

Learn more about air velocity here:

brainly.com/question/3255148

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5 0

2 years ago

A musical chord consists of several notes, each formed by a sound wave of a different frequency. The three graphs show the patte

sweet [91]

Answer:

If a chord had notes with frequencies of 100, 1,000, and 6,000 Hz, the basilar membran would vibrate at multiple positions, with peaks at A, B, and C.

Explanation:

3 0

3 years ago

A 75-hp (shaft output) motor that has an efficiency of 91% is worn out and is replaced by a high efficiency 75-hp motor that has

ZanzabumX [31]

Answer:

The heat gain of the room due to higher efficiency is 2.84 kW.

Explanation:

Given that,

Output power of shaft = 75 hp

Efficiency = 91%

High efficiency = 95.4%

We need to calculate the electric input given to motor

Using formula of efficiency

$\eta=\dfrac{Work\ Output}{Work\ input}$

$Work\ Input =\dfrac{Work\ output}{\eta}$

$Work\ Input =\dfrac{75\times746}{0.91}$

$Work\ Input =61483.51\ W$

$Work\ Input = 61.48 kW$

We need to calculate the electric input

For, heigh efficiency

$Work\ Input_{inc} =\dfrac{75\times746}{0.954}$

$Work\ Input_{int} =58647.7\ W$

$Work\ Input_{int} = 58.64\ kW$

The reduction of the heat gain of the room due to higher efficiency is

$Q=Work\ Input-Work\ Input_{int}$

Put the value into the formula

$Q=61.48 -58.64$

$Q=2.84\ kW$

Hence, The heat gain of the room due to higher efficiency is 2.84 kW.

3 0

3 years ago

The spectrum from a hydrogen vapour lamp is measured and four lines in the visible light range are observed. These lines are the

Elan Coil [88]

Answer:

$10942249.24 m^{-1}$

Explanation:

Rydberg's formula is used to describe the wavelengths of the spectral lines of chemical elements similar to hydrogen, that is, with only one electron being affected by the effective nuclear charge. In this formula we can find the rydberg constant, knowing the wavelength emitted in the transcision between two energy states, we can have a value of the constant.

$\frac{1}{\lambda}=Z^2R(\frac{1}{n^2_{1}}-\frac{1}{n^_{2}^2}})$

Where $\lambda$ it is the wavelength of the light emitted, R is the Rydberg constant, Z is the atomic number of the element and $n_{1} n_{2}$ are the states where $n_{1}$ .

In this case we have Z=1 for hydrogen, solving for R:

$R=\frac{1}{\lambda}*(\frac{1}{n^2_{1}}-\frac{1}{n^_{2}^2}})^{-1}\\R=\frac{1}{658.9*10^{-9}m}*(\frac{1}{2^2}-\frac{1}{3^2}})^{-1}\\R=1.52*10^6m^{-1}*(\frac{36}{5})=1.09*10^7 m^{-1}=10942249.24m^{-1}$

This value is quite close to the theoretical value of the constant $R=10967758.34 m^{-1}$

5 0

4 years ago