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melamori03 [73]
3 years ago
8

Basic Question please help

Engineering
1 answer:
Dvinal [7]3 years ago
7 0
1(A)
2(B)
3(E)
4(C)
5(D)
6(B)
7(C)
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What is an Algorithm? *
barxatty [35]

Answer:

a

Explanation:

4 0
4 years ago
A centimeter is Viooth of a meter, while a kilo-
tatuchka [14]

Answer:

b. False

Explanation:

A centimeter is a hundredth of a meter. This means 1/100 ,1 cm = 0.01 m

A kilo is 1000 grams.

The first answer is false because a hundredth isnot written well.

The second statement is false because the comparison given is of different units of measure. For distance is meters where as for weight is kilograms.

8 0
4 years ago
Find: factor of safety (n)for point A and B by using both MSS and DE (you can neglect shear stress due to shear force and also n
gladu [14]

Answer:

Hello your question is incomplete attached below is the complete question

Answer : Factor of safety for point A :

i) using MSS

(Fos)MSS =  3.22

ii) using DE

(Fos)DE = 3.27

Factor of safety for point B

i) using MSS

(Fos)MSS =  3.04

ii) using DE

(Fos)DE = 3.604

Explanation:

Factor of safety for point A :

i) using MSS

(Fos)MSS =  3.22

ii) using DE

(Fos)DE = 3.27

Factor of safety for point B

i) using MSS

(Fos)MSS =  3.04

ii) using DE

(Fos)DE = 3.604

Attached below is the detailed solution

8 0
3 years ago
Steam at 1 MPa, 300 C flows through a 30 cm diameter pipe with an average velocity of 10 m/s. The mass flow rate of this steam i
stealth61 [152]

Answer:

\dot m = 2.74 kg/s

Explanation:

given data:

pressure 1 MPa

diameter of pipe  =  30 cm

average velocity = 10 m/s

area of pipe= \frac[\pi}{4}d^2

                 = \frac{\pi}{4} 0.3^2

A = 0.070 m2

WE KNOW THAT mass flow rate is given as

\dot m = \rho A v

for pressure 1 MPa, the density of steam is = 4.068 kg/m3

therefore we have

\dot m = 4.068 * 0.070* 10

\dot m = 2.74 kg/s

7 0
4 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
4 years ago
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