Comprehensive Physics Assignment: Kinematics, Motion, Speed, Distance

This assignment provides detailed solutions to a set of physics problems. It covers concepts such as calculating distance using the speed of light, determining travel time between the Earth and the Moon, simplifying expressions with scientific notation, calculating the surface area of a sphere, comparing volumes, and analyzing average speed and velocity. Additionally, it addresses kinematics problems, including determining the time it takes for an object to stop under constant deceleration and analyzing the motion of objects thrown upward under gravitational acceleration. The assignment uses relevant formulas and provides step-by-step solutions to each problem, offering a comprehensive understanding of the physics principles involved. Desklib offers more solved assignments and past papers for students.

1.
 Distance in meters is given by:Light travels 3.0 ×1 08 meters∈1 second .
1 year has ( 365 ×24 × 3600 ) seconds.
Now Distance=Speed ×Time
⟹ distance=3.0 ×1 08 × ( 365 ×24 × 3600 )=9.4608× 1015 meters
 Light travels at 3.0 × 108 meters∈1 second .
T he distance between the earth∧moon=3.84 ×1 08 m.
Time= Distance
Speed =3.84 × 108
3.0 ×1 08 =1.28 seconds
2. 1.783 ×1 02 × 4.4 × 10−3=0.78452
3. S . A of a sphere=4 π r2 . r= D
2 =2.4 × 1 02
2 =1.2× 102
S . A=4 × 22
7 × ( 1.2 ×1 02 )2
=181,028.5714 c m2=18.10286 m2
4. 1 L=1000 c m3 ⟹ 4000 L×1000 c m3=4.0 × 106 c m3 .
N ow that thetwo measurements have the sameunits , we can compare them directly i. e
4 ×1 05 c m3 vs 4.0 ×1 06 c m3 .Clearly 4.0 ×1 06 c m3 islarger .
5.
Average Speed= Distance
Time = ( 120+(0.75 ×120) )
20+10 =7 m/s
Average Velocity= displacement
time =120−(0.75× 120)
20+10 =1 m/s
6.
U =150
4 =37.5 m/s
V =0 m/s
a=−6.0 m s−2
t=?
we use V =U + at
¿ substitute the values above ¿ get the unknownt i . e
0=37.5+ (−6.0 ×t )

⟹ 37.5=6 t ⟹ t=6.25 seconds
7. When a stone is thrown upward with velocity u m/s (u is positive). it moves up to a
maximum height with its velocity decreasing to zero. At maximum height, the stone
changes direction and starts falling back. At this time, the stone is acted upon by
gravitational acceleration. note the velocity is negative and increases up to a magnitude
value equal to u.
8. initial velocity , U=0 , t=1.25 s∧g=9.8 m/s2
we use h=ut + 1
2 g t2 ¿ find the value of h when stone 2 was released
i .e h=0 ( 1.25 )+ 1
2 ( 9.8× 1.252 ) =7.65625 m.
Now we need ¿ define the motion of the first stone after 1.25 s .
i .e the velocity v =u+at=0+ ( 9.8× 1.25 )=12.25 m s−1 .
Now thetwo stones will be 25 m apart after stone1 has moved
25+7.65625=32.65625 m. ¿ get t , we use h=ut+ 1
2 g t2
i .e 32.65625=¿
using quadratic formula, t=1.63 seconds 5 ¿ aeasedu . ne is acted upon by gravitational acceleration. note th
References
Gerd Baumann (2005). Mathematica for Theoretical Physics: Classical Mechanics and
Nonlinear Dynamics 2nd ed. Springer
Atam p. arya(1998). Introduction to classical mechanics 2nd ed. Prentice hall.