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g100num [7]
2 years ago
9

An automobile with an initial speed of 4.92 m/s accelerates uniformly at the rate of 3.2 m/s2 . Find the final speed of the car

after 4.5 s. Answer in units of m/s. 008 (part 2 of 2) 10.0 points Find the displacement of the car after 4.5 s.
Physics
1 answer:
Rudik [331]2 years ago
4 0

Answer:19.32 m/s

Explanation:

Given

initial speed of car(u)=4.92 m/s

acceleration(a)=3.2 m/s^2

Speed of car after 4.5 s

using equation of motion

v=u+at

v=4.92+3.2\times 4.5=4.92+14.4

v=19.32 m/s

Displacement of the car after 4.5 s

v^2-u^2=2as

19.32^2-4.92^2=2\times 3.2\times s

349.05=2\times 3.2\times s

s=54.54 m

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How large a force is necessary to stretch a 4.0-mm-diameter steel wire from its original length by 1.0%?
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The force needed to stretch the steel wire by 1% is 25,140 N.

The given parameters include;

  • diameter of the steel, d = 4 mm
  • the radius of the wire, r = 2mm = 0.002 m
  • original length of the wire, L₁
  • final length of the wire, L₂ = 1.01 x L₁ (increase of 1% = 101%)
  • extension of the wire e = L₂ - L₁ = 1.01L₁ - L₁ = 0.01L₁
  • the Youngs modulus of steel, E = 200 Gpa

The area of the steel wire is calculated as follows;

A = \pi r^2\\\\ A= 3.142 \times (0.002)^2\\\\ A= 1.257 \times 10^{-5} \ m^2

The force needed to stretch the wire is calculated from Youngs modulus of elasticity given as;

E = \frac{stress}{strain} = \frac{F/A}{e/L} = \frac{FL}{Ae} \\\\F = \frac{EAe}{L}

F = \frac{200 \times 10^9\  \times\  1.257\times 10^{-5}\  \times \ 0.01l_1}{l_1} \\\\F = 25,140\ N

Thus, the force needed to stretch the steel wire by 1% is 25,140 N.

Learn more here: brainly.com/question/21413915

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When an electron moves 2.5 m in the direction of an electric field, the change in electrical potential energy of the electron is
anygoal [31]
In the field of electromagnetism, when two charged plates that are situated opposite to each other by a certain distance, it forms an energy called the electric field. This energy is due to the difference in potential energy with respect to distance. Thus,

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