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Arada [10]
3 years ago
13

Defination of rolling

Engineering
1 answer:
Bingel [31]3 years ago
3 0

Answer:

moving by turning over and over on an axis

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A 0.25" diameter A36 steel rivet connects two 1" wide by .25" thick 6061-T6 Al strips in a single lap shear joint. The shear str
just olya [345]

Answer:

Option B

1025 psi

Explanation:

In a single shear, the shear area is \frac {\pi d^{2}}{4}=\frac {\pi 0.25^{2}}{4}

The shear strength=0.58\sigma_y and in this case \sigma_y=36 000 psi

Shear strength=\frac {Load}{Shear area} hence making load the subject then

Load=Shear area X Shear strength

Load=\frac {\pi 0.25^{2}}{4} \times 0.58\times 36000\approx 1025 psi

3 0
3 years ago
A step-up transformer has 20 primary turns and 400 secondary turns. If the primary current is 30 A, what is the secondary curren
-BARSIC- [3]
150
A
Explanation:
V
s
V
p
=
N
s
N
p

(
1
)
N
refers to the number of turns
V
is voltage
s
and
p
refer to the secondary and primary coil.
From the conservation of energy we get:
V
p
I
p
=
V
s
I
s

(
2
)
From
(
1
)
:
V
s
V
p
=
900
00
3
00
=
300
∴
V
s
=
300
V
p
Substituting for
V
s
into
(
2
)
⇒
V
p
I
p
=
300
V
p
×
0.5
∴
I
p
=
150
A
Seems a big current.
3 0
3 years ago
Volume of sale (i.e., the number of parts sold) is a factorwhen determining which
nadya68 [22]

Answer: N has to be lesser than or equal to 1666.

Explanation:

Cost of parts N in FPGA = $15N

Cost of parts N in gate array = $3N + $20000

Cost of parts N in standard cell = $1N + $100000

So,

15N < 3N + 20000 lets say this is equation 1

(cost of FPGA lesser than that of gate array)

 Also. 15N < 1N + 100000  lets say this is equation 2  

(cost of FPGA lesser than that of standardcell)

Now

From equation 1

12N < 20000

N < 1666.67

From equation 2

14N < 100000

N < 7142.85

AT the same time, Both conditions must hold true

So N <= 1666 (Since N has to be an integer)

N has to be lesser than or equal to 1666.

3 0
3 years ago
During the pre-drive check, you'll want to observe the car from the _______.
DENIUS [597]

Answer:

Passenger seat

Explanation:

If im wrong correct me

5 0
3 years ago
A large particle composite consisting of tungsten particles within a copper matrix is to be prepared. If the volume fractions of
OverLord2011 [107]

Answer:

Upper bounds 22.07 GPa

Lower bounds 17.59 GPa

Explanation:

Calculation to estimate the upper and lower bounds of the modulus of this composite.

First step is to calculate the maximum modulus for the combined material using this formula

Modulus of Elasticity for mixture

E= EcuVcu+EwVw

Let pug in the formula

E =( 110 x 0.40)+ (407 x 0.60)

E=44+244.2 GPa

E=288.2GPa

Second step is to calculate the combined specific gravity using this formula

p= pcuVcu+pwTw

Let plug in the formula

p = (19.3 x 0.40) + (8.9 x 0.60)

p=7.72+5.34

p=13.06

Now let calculate the UPPER BOUNDS and the LOWER BOUNDS of the Specific stiffness

UPPER BOUNDS

Using this formula

Upper bounds=E/p

Let plug in the formula

Upper bounds=288.2/13.06

Upper bounds=22.07 GPa

LOWER BOUNDS

Using this formula

Lower bounds=EcuVcu/pcu+EwVw/pw

Let plug in the formula

Lower bounds =( 110 x 0.40)/8.9+ (407 x 0.60)/19.3

Lower bounds=(44/8.9)+(244.2/19.3)

Lower bounds=4.94+12.65

Lower bounds=17.59 GPa

Therefore the Estimated upper and lower bounds of the modulus of this composite will be:

Upper bounds 22.07 GPa

Lower bounds 17.59 GPa

7 0
3 years ago
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