I need my math homework for my statics and strengths of materials class. Answers and questions are provided in the file I attached. I need to show work on each problem to recieve credit

Supplementary problems
1.30:
1 kips – 4418.2216 N
15000 kipq = 15000*4418.2216 = 6.7x10^7 newtons
1.31
1 Newton metre =0.7376 foot pounds
680 N =680*0.7376 = 5.02x10^2 foot pounds
1.32
1 inch = 2.54 cm =2.54x10-2m
1 inch^3 =1 .63871x10-5m^3
1250 inch^3=1250*1.63871x10-5m^3=2.05x10-2m^3
1.33
1m^3 =264.172 galons
750 m^3 = 750x264.172=198129 galons=1.98x105galons

Foundations
1.34
1 HP = 745.7 watts
2500 HP =2500x745.7 Watts=1864250=1.86x106watts
1.35
1 radian – 57.29 degrees
0.261 radians = 14.96 degrees
1.36
1 ft^2=9.2903x10-8sq km
950000 ft^2 = 8825785x10-8sq km=8.83x10-2sq km
ALGEBRA
1.37
Substitute the values in
L
1.38
1.39
1.40
Substitute to get
=92N
1.41
1.42
Rounded off it equals = 60 m
(answer differs from what is given slightly)
1.43
165.6 kip
Trignometry:
1.44:
Use Pythagorean theorem to find MO = 234 cm.
MO = 234 cm.
Tan O = 143/185 = 0.7730
Angle O=37.7 degrees
Angle M = 90-37.7 = 52.3 degrees.
1.45
Using Pythagoren theorem, ST = 23481.09
=2.35x104Newtons
Tan T = 7650/22200
Hence angle T = 19 degrees and angle S = 71 degrees
1.46
Round off b = 180 and c = 218
1.47
Hence we have Beta = 34 degrees gamma = 114 degrees and c = 28.4 mm
(c correct answer is 28.4 only not 28.2 mm)
1.48
We are given three sides of the triangle as a=2320 : b=2490:c=1330 kps. To find angles.
After rounding off we get angles as
A=67 deg. B= 81,1 deg and C=31.9 degrees
1.49
We are given three sides of the triangle as a=26.9 MN : b=38.4 MN :c=15.7MN. To find angles.
After rounding off we get
A=0.596 radian: B = 2.21 radian and C=0.334 radian
1.50
Area of the given figure = Area of semi ellipse (on left) + Area of rectangle (middle)+area of semi ellipse (right)
=Area of rectangle +area of full ellipse
= lw+pi ab
L=w=a =1.51 m, a= semi minor axis =a/2 = 0.755 m and b = semi minor axis =0.623 m
Area = 1.51^2+ 3.14 (0.755)(0.623)
= 3.76 m^2
1.51
Here the figure comprises of area of a semi circle with radius 1.7 mm, a rectangle with length 29 mm and width 1.7 mm and a fillet with radius 1.7 mm
Area of the fillet = area of square with side R – area of quarter circle
Hence area of figure = area of rectangle + area of semi circle + area of square with side R –area of quarter circle
=
(Given answer is wrong)
1.52
Area of cross sectional area = Area of big circle with diameter 10.75 – area of small circle with diameter 9.75
=
1.53
Lateral area = perimeter x height
Height = e sin 75
E= 0.0871
H=0.08421
Hence LSA = 0.038(0.08421) = 0.0032 m^2
1.54
Mass of the cylinder = Volume x density
Volume of cylinder =
Hence volume of pipe = where 2R = 6.625 in and 2r=4.897 in.
=
Mass = 820.6233 (0.2840)=23.305
=23.31 lb ft.
2.24
We have resolved vectors as F1 = F cos t and F2 = F sin t
Where t is the angle between the force and the relevant axis.
2.25
Here for S angle is 32 and for y it is negative angle 32.
Hence we have
2.26
Here for S angle formed is 90-21 and for Perpendicular angle = 21
Hence
2.27
Here F1 acts along x axis and F2 along y axis i.e. angle between is 90 degrees.
Hence resultant =
Angle formed = tan inverse of F2/F1 = 51.34 degree = 51 degrees
2.28
Let us resolve F! and F2 together along x and y axis
Along X axis = 34 cos 72 +41 cos (109) = 10.506-38.766=-28.26N
Along Y axis = 34 cos 18+41 cos 109 = 32.336-13.348=18.988 N
R =34.04
Angle t = arctan 18.988/-28.26 = arctan (-0.6719) =123 degrees
(Here answer differs)
2.29
If R is the resultant,
2.30
Let us solve graphically
F1 = 140 sin 36 I + 140cos 36 j=82.28i+113.26j
F2=-170j
F3 =165cos39i-165 sin39j = 128.23i+-102.579j
R = 211.63i-159.319j
|R}=265
T = tan inverse of y/x =323 (since 4th quadrant)
2.31
    2.31
    Ozf
    
    
    X
    Y
    
    
    
    force
    Angle X
    Angle Y
    Cos x
    Cos...