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swat32
3 years ago
5

BRO FNAF ANIMATRONICS ALL OF THEM INCLUDING THE BOOKS AND THE OTHER ONES IN FNAF SPECIAL DELIVERY!

Engineering
3 answers:
Nezavi [6.7K]3 years ago
9 0
Nice nice. . . . . . .
MrRissso [65]3 years ago
8 0

uhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

YNwastaken2 years ago
0 0

heh heh heh what.

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1) A cylinder has a volume of 20 cubic feet. What is that volume in cubic inches? (1 ft = 12 in) ​
hram777 [196]
Before calculating the volume of this cylinder, we must either convert the diameter to feet or the height to inches. I'll do the latter, to avoid getting a very small number.
Recall that there are
12
inches in
1
foot. Thus, we can state:
12
in.
1
ft.
=
x
4
ft.
x
=
12
in.
×
4
ft.
1
(ft.)
x
=
48
in.
Hence, the cylinder has a height of
48
inches.
Now, we can apply the formula for volume of a cylinder to effectuate our calculation. The formula in question is
V
=
a
base
×
h
, or
V
=
r
2
π
×
h
.
However, we know our diameter but we don't know our radius. As you probably know, the diameter is linked to the radius b the formula
d
=
2
r
. Solving for
r
and substituting:
r
=
d
2
r
=
12
2
r
=
6
∴
The radius of the cylinder measures
6
inches.
V
=
r
2
π
×
h
V
=
6
2
π
×
48
V
=
(
36
×
48
)
π
V
=
1728
π
in
2
Note that this answer is in exact value. Rounded to two decimal places, the volume is
5428.67
in
7 0
2 years ago
Order of Design Process steps ?
gregori [183]

Answer:

The five stages of Design Thinking, according to d.school, are as follows: Empathise, Define , Ideate, Prototype, and Test. Let's take a closer look at the five different stages of Design Thinking

Explanation:

5 0
2 years ago
A car generator turns at 400 rpm (revolutions per minute) when the engine is idling. It has a rectangular coil with 300 turns of
larisa86 [58]

Answer:

The field strength needed is 0.625 T

Explanation:

Given;

angular frequency, ω = 400 rpm = (2π /60) x (400) = 41.893 rad/s

area of the rectangular coil, A =  L x B = 0.0611 x 0.05 = 0.003055 m²

number of tuns of the coil, N = 300 turns

peak emf = 24 V

The peak emf is given by;

emf₀ = NABω

B = (emf₀ ) / (NA ω)

B = (24) / (300 x 0.003055 x 41.893)

B = 0.625 T

Therefore, the field strength needed is 0.625 T

4 0
3 years ago
An 80-percent-efficient pump with a power input of 20 hp is pumping water from a lake to a nearby pool at a rate of 1.5 ft3/s th
photoshop1234 [79]

Answer:

mechanical power used to overcome frictional effects in piping is 2.37 hp

Explanation:

given data

efficient pump = 80%

power input = 20 hp

rate = 1.5 ft³/s

free surface = 80 ft

solution

we use mechanical pumping power delivered to water is

{W_{u}}= \eta  {W_{pump}}  .............1

put here value

{W_{u}}  = (0.80)(20)

{W_{u}} = 16 hp

and

now we get change in the total mechanical energy of water is equal to the change in its potential energy

\Delta{E_{mech}} = {m} \Delta pe   ..............2

\Delta {E_{mech}} = {m} g \Delta z  

and that can be express as

\Delta {E_{mech}} = \rho Q g \Delta z     ..................3

so

\Delta {E_{mech}} = (62.4lbm/ft^3)(1.5ft^3/s)(32.2ft/s^2)(80ft)[\frac{1lbf}{32.2lbm\cdot ft/s^2}][\frac{1hp}{550lbf \cdot ft/s}]      ......4

solve it we get

\Delta {E_{mech}} = 13.614 hp

so here

due to frictional effects, mechanical power lost in piping

we get here

{W_{frict}} = {W_{u}}-\Delta {E_{mech}}  

put here value

{W_{frict}} = 16 -13.614

{W_{frict}} = 2.37  hp

so mechanical power used to overcome frictional effects in piping is 2.37 hp

4 0
3 years ago
A manufacturer makes two types of drinking straws: one with a square cross-sectional shape, and the other type the typical round
Harlamova29_29 [7]

Answer:

\frac{Q_{square}}{Q_{circle}} =  0.785  

Explanation:

given data

types of drinking straws

  1. square cross-sectional shape
  2. round shape

solution

we know that both perimeter of the cross section are equal

so we can say that

perimeter of square  = perimeter of circle  

4 × S = π × D

here S is length and D is diameter

S = \frac{\pi D}{4}        ....................1

and

ratio of  flow rate through the square and circle is here

\frac{Q_{square}}{Q_{circle}} = \frac{AV^2}{AV^2}  

\frac{Q_{square}}{Q_{circle}} = \frac{S^2}{\frac{\pi D^2}{4}}  

\frac{Q_{square}}{Q_{circle}} = \frac{(\frac{\pi D}{4})^2}{\frac{\pi D^2}{4}}  

\frac{Q_{square}}{Q_{circle}} = \frac{\pi }{4}  

\frac{Q_{square}}{Q_{circle}} =  0.785  

4 0
3 years ago
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