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inessss [21]
2 years ago
14

During a race, a runner runs with an average velocity of 4.26 m/s toward city hall. What is the runner’s displacement after 167s

?
Physics
1 answer:
Artyom0805 [142]2 years ago
3 0

Answer:

d = 711.42 m

Explanation:

Given that,

The average velocity of a runner, v = 4.26 m/s

Time, t = 167 s

We need to find the runner's displacement.

The average velocity of an object is equal to the displacement per unit time. It can be given by :

v=\dfrac{d}{t}\\\\d=vt\\\\d=4.26\ m/s\times 167\ s\\\\d=711.42\ m

So, the runner's displacement is 711.42 m.

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A 4.0-kg object is supported by an aluminum wire of length 2.0 m and diameter 2.0 mm. How much will the wire stretch?
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Answer:

The extension of the wire is 0.362 mm.

Explanation:

Given;

mass of the object, m = 4.0 kg

length of the aluminum wire, L = 2.0 m

diameter of the wire, d = 2.0 mm

radius of the wire, r = d/2 = 1.0 mm = 0.001 m

The area of the wire is given by;

A = πr²

A = π(0.001)² = 3.142 x 10⁻⁶ m²

The downward force of the object on the wire is given by;

F = mg

F = 4 x 9.8 = 39.2 N

The Young's modulus of aluminum is given by;

Y = \frac{stress}{strain}\\\\Y = \frac{F/A}{e/L}\\\\Y = \frac{FL}{Ae} \\\\e = \frac{FL}{AY}

Where;

Young's modulus of elasticity of aluminum = 69 x 10⁹ N/m²

e = \frac{FL}{AY} \\\\e = \frac{(39.2)(2)}{(3.142*10^{-6})(69*10^9)} \\\\e = 0.000362 \ m\\\\e = 0.362 \ mm

Therefore, the extension of the wire is 0.362 mm.

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