Please answer the questions !

You are working in a lab where RC circuits are used to delay the initiation of a process. One particular experiment involves an

Ymorist [56]

Answer:

$t'_{1\2} = 6.6 sec$

Explanation:

the half life of the given circuit is given by

$t_{1\2} =\tau ln2$

where [/tex]\tau = RC[/tex]

$t_{1\2} = RCln2$

Given $t_{1\2} = 3 sec$

resistance in the circuit is 40 ohm and to extend the half cycle we added new resister of 48 ohm. the net resitance is 40+48 = 88 ohms

now the new half life is

$t'_{1\2} =R'Cln2$

Divide equation 2 by 1

$\frac{t'_{1\2}}{t_{1\2}} = \frac{R'Cln2}{RCln2} = \frac{R'}{R}$

$t'_{1\2} = t'_{1\2}\frac{R'}{R}$

putting all value we get new half life

$t'_{1\2} = 3 * \frac{88}{40} = 6.6 sec$

$t'_{1\2} = 6.6 sec$

7 0

3 years ago

Three-dimensional measuring references all of these EXCEPT:

cricket20 [7]

Except the Table of Contents

4 0

2 years ago

A sum of $500,000 will be invested by a firm two years from now. If money is worth 12%, what will be the worth of this investmen

notka56 [123]

Answer:

investment 10 years from now is $1,238,000 .

Explanation:

given data

sum = $500,000

rate = 12% =0.12

total time = 10 year

solution

as present value After 2 years from now is $500,000

so time period is now = 8 year ( 10 - 2 )

so we apply future value formula that is

Future value = present value × $(1+r)^{t}$ ............1

put here value we get

Future value = $500,000 × $(1+0.12)^{8}$

Future value = $500,000 × 2.476

Future value = $1,238,000

so investment 10 years from now is $1,238,000 .

8 0

3 years ago

What is the pressure at the bottom of a 25 ft volume of hydraulic fluid with a weight density of 55 lb/ft3 a. 114.6 psi b. 1375p

Assoli18 [71]

Answer:

d) 9.55 psi

Explanation:

pressure at the bottom is =ρgh

weight density is ρg=55 lb/ft³

h=25ft

pressure at the bottom is = $55\times 25$

=1375psf

1 ft = 12 inch

pressure at bottom = $\frac{1375}{12^2}$

= 9.55 psi

so, answer will be option (d) which is 9.55 psi

3 0

2 years ago

A car is traveling at sea level at 78 mi/h on a 4% upgrade before the driver sees a fallen tree in the roadway 150 feet away. Th

Dmitrij [34]

Answer: V = 47.7 mi/hr

Explanation:

first we calculate elements of aero-dynamic resistance

Ka = p/2 * CD * A.f

p is the density of air(0.002378 slugs/ft^3) for zero altitude, CD is the drag coefficient(0.35) and A.f is the front region of the vehicle

so we substitute

Ka = 0.002378/2 * 0.35 * 18

Ka = 0.00749

Now we calculate the final speed of the vehicle (V2) using the relation;

S = (YbW/2gKa)In[ (UW + KaV1^2 + FriW ± Wsinθg) / (UW + KaV2^2 + FriW ± Wsinθg)

so

WE SUBSTITUTE

150 = (1.04 * 2700 / 2 * 32.2 * 0.0075) In [(0.8 * 2700 + 0.0075 *(78mil/hr * 5280ft/1min * 1hr/3600s)^2 + 0.017 * 2700 ± 2700 * 0.04) / (0.8 * 2700 + 0.0075 * V2^2 + 0.017 * 2700 ± 2700 * 0.04)]

150 = (2808/0.483) In [(2160 + 98.16 + 153.9) / ( 2160 + 0.0075V2^2 + 153.9)]

150 = 5813.66 In [ (2160 + 98.16 + 153.9) / ( 2160 + 0.0075V2^2 + 153.9)]

divide both sides by 5813.66

0.0258 = In [ (2412.06) / ( 0.0075V2^2 + 2313.9)]

take the e^ of both side

e^0.0258 = (2412.06) / ( 0.0075V2^2 + 2313.9)

1.0261 = (2412.06) / ( 0.0075V2^2 + 2313.9)]

(0.0075V2^2 + 2313.9) = 2412.06 / 1.0261

(0.0075V2^2 + 2313.9) = 2350.7

0.0075V2^2 = 2350.7 - 2313.9

0.0075V2^2 = 36.8

V2^2 = 36.8 / 0.0075

V2^2 = 4906.6666

V2 = √4906.6666

V2 = 70.0476 ft/s

converting to miles per hour

V2 = 70.0476 ft/s * 1 mil / 5280 ft * 3600s / 1hr

V = 47.7 mi/hr

8 0

3 years ago