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ch4aika [34]
3 years ago
7

Four samples, each having a void ratio of 0.76 and a specific gravity of 2.74, have degrees of saturation (S) of 85, 90, 95, and

100%. Determine the total (wet/moist) unit weight (γt) in pcf for each of the four samples.
Physics
1 answer:
Bas_tet [7]3 years ago
6 0

Answer:

a) γt = 120.107 lbf/ft³

b) γt = 121.455 lbf/ft³

c) γt = 122.803 lbf/ft³

d) γt = 124.150 lbf/ft³

Explanation:

Given:

e = 0.76

Gs = 2.74

γw = 62.43 lbf/ft³

a) S = 85%

b) S = 90%

c) S = 95%

d) S = 100%

We can apply the formula

γt = γw*(Gs+S*e)/(1+e)

a) γt = (62.43 lbf/ft³)*(2.74+0.85*0.76)/(1+0.76)

γt = 120.107 lbf/ft³

b) γt = (62.43 lbf/ft³)*(2.74+0.90*0.76)/(1+0.76)

γt = 121.455 lbf/ft³

c) γt = (62.43 lbf/ft³)*(2.74+0.95*0.76)/(1+0.76)

γt = 122.803 lbf/ft³

d) γt = (62.43 lbf/ft³)*(2.74+1*0.76)/(1+0.76)

γt = 124.150 lbf/ft³

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a tank circuit contains a capacitor and an inductor that produce 30 of reactance at the resonant frequency. the inductor has a q
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The total circuit current at the resonant frequency is 0.61 amps

What is a LC Circuit?

  • A capacitor and an inductor, denoted by the letters "C" and "L," respectively, make up an LC circuit, also referred to as a tank circuit, a tuned circuit, or a resonant circuit.
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Q =15 = (wL)/R

wL = 30 ohms = Xl

R = 2 ohms

Zs = R + jXl = 2 +j30 ohms where Zs is the series LR impedance

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The impedance of the LC circuit is found from:

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Answer:

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(b)Kinetic Energy , K=958546.875 Joule

(c)Distance, s=268.8m

(d)Acceleration, a= - 2.38 m/s^2

<u>Explanation</u>:

<u>Given</u>:

Distance travelled = 40 miles

Time taken = 30 minutes.

(A) The average velocity in kilometres/hour

Converting 40 miles into km ,

we know that,

1 mile = 1.60934

40 miles =  40 x 1.60934

so 40 miles  =  64.3738 Km

similarly converting 30 minutes into hours

1 minute = \frac{1}{60}hours

30 minute = \frac{30}{60}hours

30 minute = \frac{1}{2}hours

Now

Average velocity = \frac{Speed}{time}

Substituting Values,

Average velocity = \frac{64.3738}{\frac[1}{2}}

Average velocity = 64.3738 \times 2

Average velocity =128.74 Km/hr

(B) If the car weighs 1.5 tons, what is its If the car weighs 1.5 tons, what is its kinetic energy in joules (Note: you will need to convert your velocity to m/s)? in joules (Note: you will need to convert your velocity to m/s)?

Converting 1.5 tons into kg we get

1 ton = 1000 kg

so 1.5 ton =1500 kg

converting  velocity to m/s

128.74  \times \frac{5}{18}

=>35.75 m/s----------------------------------------------------------(1)

kinetic energy  K= \frac{1}{2}mv^2

Substituting the values,

K= \frac{1}{2}1500(35.75)^2

K= \frac{1}{2}1500(1278.06)

K= \frac{1500 \times (1278.06)}{2}

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K=958546.875 Joule---------------------------------------------(2)

(c)When the driver applies the brake, it takes 15 seconds to stop. How far does the car travel (in meters) while stopping

Lets use Distance formula,

S= ut+\frac{1}{2}at^2

Substituting the known values,

s= ut+\frac{1}{2}at^2

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s=566.25+\frac{(225a)}{2}-------------------------------------(3)

(D) What is the average acceleration of the car (in m/s2) during braking?

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v=u +at

re arranging the formula we get,

a = \frac{v - u}{t}

Substituting the values

a = \frac{0 - 35.75}{15}

a = \frac{- 35.75}{15}

a= - 2.38 m/s^2----------------------------------------(4)

Now substituting 4 in 3 we get

s=566.25+\frac{(225( - 2.38)}{2}

s=566.25+\frac{-535.5}{2}

s=536.25-267.75

s=268.8m--------------------------------------------------------------(5)

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