Physics and Math Assignment

This assignment covers a range of topics in physics and mathematics, including trigonometry, forces, graphs, and trigonometric functions. It includes several problems that require students to apply these concepts to solve real-world situations. The document also provides solutions to each problem, making it a valuable resource for students looking to understand the material and check their work.

Task 1
(i) 7(y2)4 x 2(y3)3
solution
¿ 7 y8∗2 y9
¿ 1417
(ii)
( T 2
T 1 . 4 × V 5
V 6 × P1 . 4
P−1. 2 )3
¿( ( t2 ÷T 1.4 )∗( v5 ÷ v6 )∗( p1.4 ÷ p−1.2))3
¿ ( T 0.6∗V−1∗P2.6 ) 3
¿ T 1.8∗V −3∗P−7.2
(iii)
(6 W 3 )2×(2 W 2 )3
W 2
solution
6 W 6∗2W 6 ÷ W 2
¿ 12 w10
Task 2
a.
solution
¿(24log2-12log2+3log2)/2log2
24 log 2
2 log 2 −12 log 2
2 log 2 + 3 log 2
2 log 2
¿ 7 1
2
b. 6( x +1)=25 .158 solution
( x+1 ) log 6
log6 = log 12.158
log 6
x +1=1.39415331
x=0.39415331
4 log 82−3 log 16+ 1
2 log 64
2 log 2

Task 3
i) Plot a graph of these values of Current on the horizontal axis (x-axis) and
Voltage on the vertical axis (y-axis).
ii) Using your graph deduce the y-axis intercept of the line.
Solution
y-axis intercept is 1.5
iii) Using your graph deduce the gradient of the line.
Solution
Gradient of the line is 0.25
iv) Using your graph deduce the equation of the line.
Solution
Equation is;
Voltage=0.25 ( Current )+ 1.5

Task 4
(i) 5x + 20y
Solution
=5( x + 4 y)
(ii) 8(a + b) – 2(3a)
solution
¿ 8 a+ 8 b−6 a
¿ 2 a+8 b
¿ 2(a+4 b)
(iii) 8ax + 40cx + 3a +18c (Hint: group the common factors (i.e. a and c) together here
then factorise each group)
solution
¿ 8 ax +3 a+ 40 cx +18 c
¿ a ( 8 x +3 ) +2 c (20 x+ 9)
Task 5
Angle
measures in degrees Measures in Radians
In decimal form In pie
45 0.7875 Pie/3.98
89.76 1.5708 Pie/2
179.43 3.14 22/7
240 4.2 4pie/3
359.04 6.2831 2 pie
300 5.25 5pie/3
239.36 4.1888 4pie/3
Task 5b
solution
2450
190
¿ 12.8947 revolution per second

solution
area=108
360 ×3.142× 42
¿ 15.0816 c m2
Task 5c
solution
vertical opponent =120 sin60=103.90
horizontal component 120 cos 60=600
R= √ ( 2502+ 4802 ) N=541.2
tan A= 250
480 =27.51
¿
R=541.2 N working ∈direction ¿ ¿¿ at an angle 27.510
Task 5c
a) Calculate the horizontal and vertical components of the force.
Solution
F = √(Fx2 + Fy2)
Horizontal force is
Fy = 50 · cos 60o = 50 ·(1/2) = 25 N
Vertical force is
Fx = 50 · cos 30o = 50 · (√3)/2 = 25√3 N
b) Calculate the resultant of the two forces and the angle at which it acts to the
horizontal.
Solution
Resultant force= √4802 +2502=¿ √292900=541.2024 N ¿

Task 6
e 0 30 60 90 120 150 180 210 240 270 300 330 360
Sin ɸ 0.000 -0.988 -0.305 0.894 0.581 -0.715 -0.801 0.468 0.945 -0.176 -1.000 -0.132 0.959
Cos ɸ 1.000 0.154 -0.952 -0.448 0.814 0.699 -0.598 -0.884 0.326 0.984 -0.022 -0.991 -0.284
Tan ɸ 0.000 -6.405 0.320 -1.995 0.713 -1.022 1.339 -0.529 2.902 -0.179 45.245 0.134 -3.380
Sine graph

Task 7 a
solution
¿ AB=BC=6 cm since they are symetrical
A D2=62+52−2∗6∗5 cos 20 , AD=2.14
¿ 2.14
sin 20 = 5
sinA =53.030 , ACB=53.030
ABC=180−53.03−53.03=73.940
Task 7b
solution
180−80−70=3 00
1.4
sin 30 = a
sin 80 =2.76 m
1.4
sin 30 = b
sin 70 =2.63 m
Task 8
solution
3.142∗10∗6+ (3.142∗12∗12 ) + ( 2∗3.142∗62 )
188.52+452.448+226.224
¿ 867.192 c m2
Task 9
solution
15 p1+10 p 2=180
15 p1+12 p 2=192
¿ p 2=6
p 1=8