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Physics Assignment 1st smester BSCS, Assignments of Physics

Virtual University of Pakistan (VUP)Physics

This is physics asssignment for university students.

Typology: Assignments

2021/2022

Available from 01/15/2022

moazzam-ali-7
moazzam-ali-7 🇵🇰

2 documents

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PHY101 Assignment 1
Question No 1
Given Data:
Mass of car = 2.0 × 103 kg
Initial velocity of car = 45 ms-1
Final velocity of car = 0
Frictional Force = 7.5 × 103 N
Solution:
Part a:
As we know,
Friction force = normal force in opposition direction
Fn = - Ff
ma = - Ff
a = - Ff / m
a = - 7.5 × 103 N / 2.0 × 103 kg
a = - 3.75 ms-2
By Using Third Equation of motion
2aS = vf2 – vi2
S = (vf2 – vi2) / 2a
S = (0-(45ms-1)2) / 2(-3.75 ms-2)
S = - 2025 m2s-2 / -7.5 ms-2
S = 270m
Part (b):
Distance = S = 40.0 m?
pf3

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 PHY101 Assignment 1

Question No 1 Given Data: Mass of car = 2.0 × 10^3 kg Initial velocity of car = 45 ms- Final velocity of car = 0 Frictional Force = 7.5 × 10^3 N Solution: Part a: As we know, Friction force = normal force in opposition direction Fn = - Ff ma = - Ff a = - Ff / m a = - 7.5 × 10^3 N / 2.0 × 10^3 kg a = - 3.75 ms- By Using Third Equation of motion 2aS = vf^2 – vi^2 S = (vf^2 – vi^2 ) / 2a S = (0-(45ms-1)^2 ) / 2(-3.75 ms-2) S = - 2025 m^2 s-2^ / -7.5 ms- S = 270m Part (b): Distance = S = 40.0 m?

Using Third Equation of motion 2aS = vf^2 – vi^2 2(-3.75 ms-2) (40m) = vf^2 - (45 ms-1)^2 -300 m^2 s-2^ = vf^2 - 2025 m^2 s- vf^2 = 1725 m^2 s- By Applying Square root on both sides, we have vf = 41.5 ms- Part (c) Physics plays a vital role in our daily life. We use the principle of physics in our everyday life activities such as walking, cooking, cutting, watching, and opening and closing objects.Now we will show some examples newton’s first law of motion examples in everyday life:  The electric fan continues to move for a period after the electricity is turned off.  The motion of a ball falling through the atmosphere or a model rocket launched into the atmosphere are excellent examples of Newton’s 1st law of motion. Question No. 2 Given Data: Initial Velocity = v mph = (5280v/3600) = 1.467fts-1^ v Final Velocity = 0 Skid Distance = d feet Solution: Using Third Equation of motion 2aS = vf^2 – vi^2 a = (vf^2 – vi^2 ) / 2S a = (0 – 2.151 v^2 ) / 2d a = - 1.0760fts-2^ v^2 d-