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3 years ago

Who is???????????????????

Engineering

1 answer:

aleksley [76]3 years ago

8 0

Woah that’s pretty cool! Did you draw/make that?!

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What is 2+2 equal to

Anit [1.1K]

The answer is 4, lol.

4 0

3 years ago

Read 2 more answers

In class we learned that in order to uniquely identify one of N equally likely symbols, ceiling (〖log〗_2 N) bits of information

AleksAgata [21]

Answer:

a. 10bits

b. 7 bits

c. 6 bits

Explanation:

a. for 0 to 512

# of numbers = 512 - 0 + 1 =513

[log ₂513] = 9 bits

we actually need 10 bits

b. for 0 to 75

# of numbers = 75 - 0 + 1 =76

[log ₂76] = 6 bits

we actually need 7 bits

c. for -20 to 13

# of numbers = 13 - (-20) + 1 =34

[log ₂34] = 5 bits

we actually need 6 bits

6 0

3 years ago

Two transmission belts pass over a double-sheaved pulley that is attached to an axle supported by bearings at A and D. The radiu

Marat540 [252]

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

4 0

3 years ago

The cross section of a heat exchanger consists of three circular pipes inside a larger pipe. The internal diameter of the three

omeli [17]

Answer:

0.0432 m^3/s

Explanation:

Internal diameter of smaller pipes = 2.5 cm = 0.025 m

pipa wall thickness = 3 mm = 0.003 m

internal diameter of larger pipes = 8 cm = 0.8 m

velocity of region between smaller and larger pipe = 10 m/s

Calculate discharge in m^3/s

First we calculate the area of the smaller pipe

A = $\pi Dt$ = $\pi ( 0.025 ) ( 0.003 )$ = 0.00023571 m^2

next we calculate area of fluid between the smaller pipes and larger pipe

A = $[\frac{\pi }{4} D^{2} _{L} ] - 3(A_{s})$

= $[ \frac{\pi }{4} (0.08 )^2 - 3 ( 0.00023571 )]$

= [ 0.00502857 - 0.00070713 ]

= 0.00432144 m^2

hence the discharge in m^3/s

Q = AV

= 0.00432144 * 10

= 0.0432 m^3/s

3 0

2 years ago

A peripheral milling operation is performed on the top surface of a rectangular workpart which is 400 mm long and 60 mm wide. Th

oee [108]

Answer:

N = 278.5 rpm

F = 348.15 mm/min

machine time = 10.34 seconds

material removal rate = 33420000 mm³/min

Explanation:

given data

length L = 400 mm

width W = 60 mm

diameter D = 80 mm

no of cutting n = 5

velocity V = 70 m/min = 70000 mm/min

chip load P = 0.25 mm/tooth

depth = 5 mm

solution

we know velocity that is express as

velocity V = $\pi$ D N ........1

N = $\frac{70000}{\pi \times 80}$

N = 278.5 rpm

and

now we get feed rate in milling operation is

F n P N

F = 5 × 0.25 × 278.5

F = 348.15 mm/min

and

now we get actual machine time to make 1 pass across surface of worl is

machine time = $\frac{W}{F}$

machine time = $\frac{60}{348.15}$

machine time = 0.1723 min = 10.34 seconds

and

max material removal rate during these cut is

material removal rate = A × d × N

material removal rate = 400 × 60 × 5 × 278.5

material removal rate = 33420000 mm³/min

7 0

3 years ago