Who invented a control unit for an artificial heart?<br> ements<br> ante

Nonshielded cable with a 1.5-inch diameter should have a minimum bending radius of

Flura [38]

Answer:

c is the answer because we have to double the number

6 0

1 year ago

Find the percent change in cutting speed required to give an 80% reduction in tool life when the value of n is 0.12.

vaieri [72.5K]

Answer:21.3%

Explanation:

Given

80 % reduction in tool life

According to Taylor's tool life

$VT^n$ =c

where V is cutting velocity

T=tool life of tool

80 % tool life reduction i.e. New tool Life is 0.2T

Thus

$VT^{0.12}=V'\left ( 0.2T\right )^{0.12}$

$V'=\frac{V}{0.2^{0.12}}$

$V'=\frac{V}{0.824}$ =1.213V

Thus a change of 21.3 %(increment) is required to reduce tool life by 80%

6 0

2 years ago

A compressible clay layer has a thickness of 3.8 m. After 1.5 yr, when the clay is 50% consolidated, 7.3 cm of settlement has oc

grigory [225]

The amount of settlement that would occur at the end of 1.5 year and 5 year are 7.3 cm and 13.14 cm respectively.

<h3>How to determine the amount of settlement?</h3>

For a layer of 3.8 m thickness, we were given the following parameters:

U = 50% = 0.5.

Sc = 7.3 cm.

For Sf, we have:

Sf = Sc/U

Sf = 7.3/0.5

Sf = 14.6

Therefore, Sf for a layer of 38 m thickness is given by:

Sf = 14.6 × 38/3.8

Sf = 146 cm.

At 50%, the time for a layer of 3.8 m thickness is: $t_{50}$ = 1.5 year.

At 50%, the time for a layer of 38 m thickness is:

$t_{50}$ = 1.5 × (38/3.8)²

$t_{50}$ = 150 years.

For the thickness of 38 m, U₂ is given by:

$\frac{U_1^2}{U_2^2} =\frac{(T_v)_1}{(T_v)_2} = \frac{t_1}{t_2} \\\\U_2^2 = U_1^2 \times [\frac{t_2}{t_1} ]\\\\U_2^2 = 0.5^2 \times [\frac{1.5}{150} ]\\\\U_2^2 = 0.25 \times 0.01\\\\U_2=\sqrt{0.0025} \\\\U_2=0.05$

The new settlement after 1.5 year is:

Sc = U₂Sf

Sc = 0.05 × 146

Sc = 7.3 cm.

For time, t₂ = 5 year:

$U_2^2 = U_1^2 \times [\frac{t_2}{t_1} ]\\\\U_2^2 = 0.5^2 \times [\frac{5}{150} ]\\\\U_2^2 = 0.25 \times 0.03\\\\U_2=\sqrt{0.0075} \\\\U_2=0.09$

The new settlement after 5 year is:

Sc = U₂Sf

Sc = 0.09 × 146

Sc = 13.14 cm.

Read more on clay layer here: brainly.com/question/22238205

8 0

2 years ago

A source outputs data at the rate of 1,000,000 bits/sec. The transmitter uses binary PAM with raised cosine pulse shaping. If th

musickatia [10]

Answer:

maximum roll-off factor is 0.7

Explanation:

given data

rate of output data = 1,000,000 bits/sec

cutoff frequency fc = 850 kHz

solution

we know bandwidth frequency is same as that of bandwidth

we apply here fc formula that is

fc = ω ( 1 + α ) ..........1

here

ω = data rate ÷ 2 .............2

ω = 1000 ÷ 2

ω = 500000 Hz

so put now value in eq 1

fc = ω ( 1 + α )

850000= 500000 ( 1 + α )

solve it we get

α = 0.7

so, maximum roll-off factor is 0.7

7 0

2 years ago

Designating a standard part is an example of a use for _______ notes.

Y_Kistochka [10]

Answer:

standard references

Explanation:

hope this helps :D

8 0

2 years ago

Who invented a control unit for an artificial heart? ements ante