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4 years ago

1. Round off 2553 N to three significant figures.

Engineering

1 answer:

Marizza181 [45]4 years ago

8 0

Answer:

(1) 2553 N = 2550 N

(2) 58342 m = 58300 m

(3) 68.534 s = 68.5 s

Explanation:

To round off a number to any significant number start from the last digit, round it off to 1 if the number is up to 5 and to 0 if the last digit is less than 5. Add this 1 or 0 to the preceding digit and continue the process until you are left with three non zero digits, if you are rounding off to three significant figures.

(1) Round off 2553 N to three significant figures.

= 2550 N

(2) Round off 58342 m to three significant figures.

= 58300 m

(3) Round off 68.534 s to three significant figures.

= 68.500 s (zero after decimal point is insignificant)

= 68.5 s

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Explain how you could transmit two independent base-band information signals by using SSB on a common carrier frequency.

Goshia [24]

Answer:

Using Hilbert Transformation, we can transmit two independent base-band information signals by using SSB on a common-carrier frequency.

Explanation:

In SSB modulators, we pre-process a real signal by Hilbert Transform filter to form another real signal
The signal has the same spectral amplitude but has 90° phase shift at each frequency relative to its input signal.
The ordered pair gets almost all of its negative components cancelled
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4 years ago

A car engine with a thermal efficiency of 33% drives the air-conditioner unit (a refrigerator) besides powering the car and othe

fgiga [73]

Answer:

The rate of fuel required to drive the air conditioner $Q_h = 6.061 kW$

The flow rate of the cold air is $\r m = 0.30765 kg/s$

Explanation:

From this question we are told that

The efficiency is $\eta = 33$ % = 0.33

Temperature for the hot day is $T_h = 35^oC = 308 K \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (35+273)$

Temperature after cooling is $T_c = 5^oC = 278K$

The input power is $P_{in} = 2kW$

The rate of fuel required to drive the air conditioner can be mathematically represented as

$Q_h = \frac{P_{in}}{\eta}$

$= \frac{2}{0.33} = 6.061 kW$

From the question the air condition is assumed to be half as a Carnot refrigeration unit

This can be Mathematically interpreted in terms of COP(coefficient of performance) as

$\beta_{air} = 0.5 \beta$

where $\beta$ denotes COP and is mathematically represented as

$\beta = \frac{Q_c}{P_{in}}$

= > $Q_c = \beta P_{in}$

Where $Q_c$ is the rate of flue being burned for cold air to flow

Now if the COP of a Carnot refrigerator is having this value

$\beta_{Carnot } = \frac{T_c}{T_h - T_c}$

$= \frac{278}{308-278}$

$\beta_{Carnot} = 9.267\\$

Then

$\beta_{air} = 0.5 * 9.2667$

$= 4.6333$

Now substituting the value of $\beta$ to solve for $Q_c$

$Q_c = \beta P_{in}$

$= 4.6333 *2$

$9.2667kW$

The equation for the rate of fuel being burned for the cold air to flow

$Q_c = \r mc_p \Delta T$

Making the flow rate of the cold air

$\r m = \frac{Q_c}{c_p \Delta T}$

$= \frac{9.2667}{1.004}* (308 - 278)$

$= 0.30765 kg/s$

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3 years ago

I am trying to create a line of code to calculate distance between two points. (distance=[tex]\sqrt{ (x2-x1)^2+(y2-y1)^2}) My li

k0ka [10]

Answer:

point_dist = math.sqrt((math.pow(x2 - x1, 2) + math.pow(y2 - y1, 2))

Explanation:

The distance formula is the difference of the x coordinates squared, plus the difference of the y coordinates squared, all square rooted. For the general case, it appears you simply need to change how you have written the code.

point_dist = math.sqrt((math.pow(x2 - x1, 2) + math.pow(y2 - y1, 2))

Note, by moving the 2 inside of the pow function, you have provided the second argument that it is requesting.

You were close with your initial attempt, you just had a parenthesis after x1 and y1 when you should not have.

Cheers.

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3 years ago

A mercury thermometer has a cylindrical capillary tube with an internal diameter of 0.2 mm. If the volume of the thermometer and

vova2212 [387]

To solve this problem we will proceed to calculate the specific volume from the area of the cylinder and the sensitivity. Later we will calculate the volumetric coefficient of thermal expansion and finally we will be able to calculate the volume through the relation of the two terms mentioned above. Our values are

$\text{Sensitivity}= 2mm/\°C$

$\text{Internal diameter } d= 0.2mm$

$\text{Differential expansion of Hg } \lambda_L = 1.82*10^{-4}/\°C$

Let's start by calculating the specific volume which is given by

$v = \pi (\frac{d}{2})^2 \gamma$

Here,

d = Diameter

$\gamma$ = Sensitivity

Replacing our values we have

$v = (\frac{\pi}{4})(0.2mm)^2(2mm/\°C)$

$v = 0.0628mm^3 /\°C$

Now we will obtain the value of the volumetric coefficient of thermal expansion of mercury through the differential expansion coefficient of Hg whic is three times, then

$\lambda_V = 3\lambda_L$