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mrs_skeptik [129]
2 years ago
9

सत्य से अधिक उपयोगी एवं आज्ञापालन से श्रेष्ठ क्या है ?answer fast plz​

Engineering
1 answer:
PilotLPTM [1.2K]2 years ago
5 0

Answer:

I hope this helps you

Explanation:

किसी की जान बचाने के लिए झूठ बोलना सत्य से अधिक उपयोगी है । अन्यायपूर्ण आज्ञा का विरोध आज्ञापालन से श्रेष्ठ है ।

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When asked about favorite Thanksgiving leftovers, 9% of the people said turkey and 7100 said mashed potatoes. Which food is more
irakobra [83]

Answer:

how many people were asked though

Explanation:

8 0
2 years ago
Set up the following characteristic equations in the form suited to Evanss root-locus method. Give L(s), a(s), and b(s) and the
Sunny_sXe [5.5K]

Answer:

attached below is the detailed solution and answers

Explanation:

Attached below is the detailed solution

C(iii) : versus the parameter C

The parameter C is centered in a nonlinear equation, therefore the standard locus will not apply hence when you use a polynomial solver the roots gotten would be plotted against C

4 0
3 years ago
Match the military operation to the category of satellite that would perform it.
SOVA2 [1]

Answer:

1. Location of enemy ground troops  - EARTH OBSERVING.

Using earth observing satellite imagery, the military can observe vast expanses of land and in so doing, find the location of enemy ground troops.

2. Routine reconnaissance of an unfamiliar climate  - WEATHER

In other to find out more about the climate of an area, a weather satellite can be used to observe the areas and its changing weather patterns.

3. Analyze waterways in an unfamiliar location  - NAVIGATION

Using navigation satellites, navigation conduits such as roads and waterways can be observed.

4. Provide warning of an attack - COMMUNICATION.

Communications satellites enable people to communicate over great distances and so can be used by the military to warn of an impending attack.

5 0
3 years ago
1. A glass window of width W = 1 m and height H = 2 m is 5 mm thick and has a thermal conductivity of kg = 1.4 W/m*K. If the inn
emmasim [6.3K]

Answer:

1. \dot Q=19600\ W

2. \dot Q=120\ W

Explanation:

1.

Given:

  • height of the window pane, h=2\ m
  • width of the window pane, w=1\ m
  • thickness of the pane, t=5\ mm= 0.005\ m
  • thermal conductivity of the glass pane, k_g=1.4\ W.m^{-1}.K^{-1}
  • temperature of the inner surface, T_i=15^{\circ}C
  • temperature of the outer surface, T_o=-20^{\circ}C

<u>According to the Fourier's law the rate of heat transfer is given as:</u>

\dot Q=k_g.A.\frac{dT}{dx}

here:

A = area through which the heat transfer occurs = 2\times 1=2\ m^2

dT = temperature difference across the thickness of the surface = 35^{\circ}C

dx = t = thickness normal to the surface = 0.005\ m

\dot Q=1.4\times 2\times \frac{35}{0.005}

\dot Q=19600\ W

2.

  • air spacing between two glass panes, dx=0.01\ m
  • area of each glass pane, A=2\times 1=2\ m^2
  • thermal conductivity of air, k_a=0.024\ W.m^{-1}.K^{-1}
  • temperature difference between the surfaces, dT=25^{\circ}C

<u>Assuming layered transfer of heat through the air and the air between the glasses is always still:</u>

\dot Q=k_a.A.\frac{dT}{dx}

\dot Q=0.024\times 2\times \frac{25}{0.01}

\dot Q=120\ W

5 0
3 years ago
A continuous random variable, X, whose probability density function is given by f(x) = ( λe−λx , if x ≥ 0 0, otherwise is said t
Ganezh [65]

Answer:

a) F(x) = \lambda \int_0^{\infty} e^{-\lambda x} dx= -e^{-\lambda x} \Big|_0^{\infty} = 1- e^{-\lambda x} \

b) P(10 < X

Explanation:

Previous concepts

The cumulative distribution function (CDF) F(x),"describes the probability that a random variableX with a given probability distribution will be found at a value less than or equal to x".

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution".

Part a

Let X the random variable of interest. We know on this case that X\sim Exp(\lambda)

And we know the probability denisty function for x given by:

f(x) = \lambda e^{-\lambda x} , x\geq 0

In order to find the cdf we need to do the following integral:

F(x) = \lambda \int_0^{\infty} e^{-\lambda x} dx= -e^{-\lambda x} \Big|_0^{\infty} = 1- e^{-\lambda x} \

Part b

Assuming that X \sim Exp(\lambda =0.1), then the density function is given by:

f(x) = 0.1 e^{-0.1 x} dx , x\geq 0

And for this case we want this probability:

P(10 < X

And evaluating the integral we got:

P(10 < X

4 0
3 years ago
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