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3 years ago

Explain moment of inertia

Engineering

1 answer:

DochEvi [55]3 years ago

6 0

Answer:

The moment of inertia is basically a physical quantity which describes how easy it is for a body to rotate itself around a given axis.

In easier wording: Moment of Inertia is a body and when it starts rotating or moving, it will keep doing so until we stop it by force.

Explanation:

For example, a car that is moving and active will continue to move even if you were to switch the engine off.

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An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500

Assoli18 [71]

Answer: The exit temperature of the gas in deg C is $32^{o}C$ .

Explanation:

The given data is as follows.

$C_{p}$ = 1000 J/kg K, R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

$P_{1}$ = 100 kPa, $V_{1} = 15 m^{3}/s$

$T_{1} = 27^{o}C = (27 + 273) K = 300 K$

We know that for an ideal gas the mass flow rate will be calculated as follows.

$P_{1}V_{1} = mRT_{1}$

or, m = $\frac{P_{1}V_{1}}{RT_{1}}$

= $\frac{100 \times 15}{0.5 \times 300}$

= 10 kg/s

Now, according to the steady flow energy equation:

$mh_{1} + Q = mh_{2} + W$

$h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}$

$C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}$

$(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}$

$(T_{2} - T_{1})$ = 5 K

$T_{2}$ = 5 K + 300 K

$T_{2}$ = 305 K

= (305 K - 273 K)

= $32^{o}C$

Therefore, we can conclude that the exit temperature of the gas in deg C is $32^{o}C$ .

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4 years ago

How to code the round maze in CoderZ?

dlinn [17]

Answer:

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3 years ago

When encountering low visibility from rain or fog, you should use your ____.

kirza4 [7]

Answer:

Explanation:

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3 years ago

Write a function named "read_prices" that takes one parameter that is a list of ticker symbols that your company owns in their p

ohaa [14]

Answer:

import pandas pd

def read_prices(tickers):

price_dict = {}

# Read ingthe ticker data for all the tickers

for ticker in tickers:

# Read data for one ticker using pandas.read_csv

# We assume no column names in csv file

ticker_data = pd.read_csv("./" + ticker + ".csv", names=['date', 'price', 'volume'])

# ticker_data is now a panda data frame

# Creating dictionary

# for the ticker

price_dict[ticker] = {}

for i in range(len(ticker_data)):

# Use pandas.iloc to access data

date = ticker_data.iloc[i]['date']

price = ticker_data.iloc[i]['price']

price_dict[ticker][date] = price

return price_dict

7 0

3 years ago

Design a 7.5-V zener regulator circuit using a 7.5-V zener specified at 10mA. The zener has an incremental resistance of rZ = 30

hram777 [196]

Answer:

The answer is given in the explanation.

Explanation:

The circuit is as indicated in the attached figure.

From the analytical description the zener voltage is given as

$V_z=V_z_o+I_zr_z$

Here

Vzo is the voltage at which the slope of 1/rz intersects the voltage axis it is equal to knee voltage.

The equivalent model is shown in the attached figure.

From the above equation, Vzo is calculated as

$V_z_o=V_z-I_zr_z$

Here Vz is given as 7.5 V

Iz is given as 10 mA

rz is given as 30 Ω

Thus the Vzo is given as

$V_z_o=V_z-I_zr_z\\V_z_o=7.4-30*10*10^{-3}\\V_z_o=7.5-0.3\\V_z_o=7.2 V$

The value of I_L is given as 5 mA

Now the expression of current is as

$I=I_z+I_L\\I=10mA+5mA\\I=15 mA$

Now the resistance is calculated as

$R=\dfrac{V-Vo}{I}\\R=\dfrac{10-7.2}{15*10^{-3}}\\R=186.66$

So the value of resistance is 186.66 Ω.

Considering the supply voltage is increased by 10%

V is 10-10%*10=10+1=11 so the

$R=\dfrac{V-Vo}{I}\\186.66=\dfrac{11-V_o}{15*10^{-3}}\\V_o=8.2 V$

Considering the supply voltage is decreased by 10%

V is 10-10%*10=10-1=9 so the

$R=\dfrac{V-Vo}{I}\\186.66=\dfrac{9-V_o}{15*10^{-3}}\\V_o=6.2 V$

Now if the supply voltage is 10% high and the value of Load is removed i.e I=Iz only which is 10mA

$R=\dfrac{V-Vo}{I'}\\186.66=\dfrac{11-V_o}{10*10^{-3}}\\V_o=9.13 V$

Now the largest load current thus that the supply voltage is 10% low and the current of zener is knee current thus

$V_z_o=V_z-I_zr_z\\V_z_o=7.5-30*0.5*10^{-3}\\V_z_o=7.5-0.015\\V_z_o=7.485 V$

$R=\dfrac{V-Vo}{I'}\\186.66=\dfrac{9-7.485}{I}\\I=10.71 mA$

The load voltage is 7.485 V

8 0

3 years ago

Explain moment of inertia​

Explain moment of inertia