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gladu [14]
4 years ago
10

Which of the following are true about algorithms? (You may select more than one)

Engineering
2 answers:
Marysya12 [62]4 years ago
8 0

Answer: algorithms produce result

Algorithms have clear instructions

Algorithms make use of functions

Algorithms are used to do a number of calculations

Algorithms stop in a finite amount of time

Algorithms can be done by a computer

Algorithms have an order

Explanation: Algorithms are an ordered process of solving mathematical problems,they are made and splitter into functions.

Algorithms have the following characteristics

Finiteness (specific time and space)

Definite ( they must be clear and unambiguous)

Input( they must entered)

Output( result of the various inputs )

Effectiveness ( algorithms must give tangible/ correct results).

r-ruslan [8.4K]4 years ago
7 0
Have a orderf. Fr

R
F
F
D
Djeiiekee
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Technician A states that if someone is asking a question of you, then he or she is showing interest. Technician B states that si
eimsori [14]

Answer:

They are both correct

Explanation:

When someone is asking a question of you, it means he or she has either been paying attention, and is interested in you, what you're doing, or what you're saying. Also, silence when used properly can be golden in the sense that it can prevent unnecessary problems from arising, and can save one from a lot of unforeseen problem. Whatever is said cannot be taken back again, and some things should never be said at all, especially in a professional setting.

8 0
4 years ago
Molten metal is poured into the pouring cup of a sand mold at a steady rate of 400 cm3/s. The molten metal overflows the pouring
umka2103 [35]

Answer:

the proper diameter at its base so as to maintain the same volume flow rate is 1.6034 cm

Explanation:

Given the data in the question ;

flowrate Q = 400 cm³/s

cross section of the sprue is round

Diameter of sprue at the top d_top = 3.4 cm

Height of sprue = 20 cm = 0.2 m³

the proper diameter at its base so as to maintain the same volume flow rate = ?

first we determine the velocity at the sprue base

V_base = √2gh = √( 2×9.81×0.2) = √3.924 = 1.980908 m = 198.0908 cm

so, diameter of the sprue at the bottom  will be

Q = AV = [ (( πd²_bottom)/4) × V_bottom ]

d_bottom =  √(4Q/πV_bottom)

we substitute

d_bottom =  √((4×400)/(π×198.0908 ))

d_bottom =  √( 1600/622.3206)

d_bottom =  √2.571022

d_bottom =  1.6034 cm

Therefore, the proper diameter at its base so as to maintain the same volume flow rate is 1.6034 cm

8 0
3 years ago
If the specific surface energy for magnesium oxide is 1.0 J/m2 and its modulus of elasticity is (225 GPa), compute the critical
lubasha [3.4K]

Answer:critical stress= 20.23 MPa

Explanation:

Since there was an internal crack, we will divide the length of the internal crack by 2

Length of internal crack, a = 0.7mm,

Half length = 0.7mm/2= 0.35mm  changing to meters becomes

0.35/ 1000= 0.35 x 10 ^-3m

The formulae for critical stress is calculated using

σC = (2Eγs /πa) ¹/₂

σC = critical stress=?

Given

E= Modulus of Elasticity= 225GPa =225 x 10 ^ 9 N/m²

γs= Specific surface energy = 1.0 J/m2 = 1.0 N/m

a= Half Length of crack=0.35 x 10 ^-3m

σC= (2 x 225 x 10 ^ 9 N/m² x 1.0 N/m /π x 0.35 x 10 ^-3m)¹/₂

=(4.5 x 10^11/π x 0.35 x 10 ^-3)¹/₂

=(4.0920 x10 ^14)¹/₂

σC=20.23 x10^6 N/m² = 20.23 MPa

​  

​

8 0
3 years ago
1. When jump starting a car, make sure it is in __________.
____ [38]

Answer:

Make sure both cars are in park or neutral with the parking brake engaged. This makes sure that all electrical functions (headlights, radios, etc.) are turned off.

8 0
3 years ago
An object is supported by a crane through a steel cable of 0.02m diameter. If the natural swinging of the equivalent pendulum is
devlian [24]

Answer:

22.90 × 10⁸ kg

Explanation:

Given:

Diameter, d = 0.02 m

ωₙ = 0.95 rad/sec

Time period, T = 0.35 sec

Now, we know

T= 2\pi\sqrt{\frac{L}{g}}

where, L is the length of the steel cable

g is the acceleration due to gravity

0.35= 2\pi\sqrt{\frac{L}{9.81}}

or

L = 0.0304 m

Now,

The stiffness, K is given as:

K = \frac{\textup{AE}}{\textup{L}}

Where, A is the area

E is the elastic modulus of the steel = 2 × 10¹¹ N/m²

or

K = \frac{\frac{\pi}{4}d^2\times2\times10^11}{0.0304}

or

K = 20.66 × 10⁸ N

Also,

Natural frequency, ωₙ = \sqrt{\frac{K}{m}}

or

mass, m = \sqrt{\frac{K}{\omega_n^2}}

or

mass, m = \sqrt{\frac{20.66\times10^8}{0.95^2}}

mass, m = 22.90 × 10⁸ kg

4 0
3 years ago
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