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Igoryamba
4 years ago
5

15. A volleyball player who weighs 650 Newtons jumps 0.500 meters vertically off the floor. What is her kinetic energy just befo

re hitting the floor?
Physics
1 answer:
FromTheMoon [43]4 years ago
5 0
<h2>Kinetic energy just before hitting the floor is 324.57 J</h2>

Explanation:

Weight of volleyball player = 650 N

That is

            Mass x Acceleration due to gravity = 650

            Mass x 9.81 = 650

            Mass = 66.26 kg

We also have equation of motion v² = u² + 2as

Initial velocity, u = 0 m/s  

Acceleration, a = 9.81 m/s²  

Final velocity, v = ?

Displacement, s = 0.5 m

Substituting  

v² = u² + 2as

v² = 0² + 2 x 9.81 x 0.5

v = 3.13 m/s

Velocity with which he lands on ground is 3.13 m/s

We have kinetic energy = 0.5 x Mass x Velocity²

Substituting

          Kinetic energy = 0.5 x 66.26 x 3.13²

           Kinetic energy = 324.57 J

Kinetic energy just before hitting the floor is 324.57 J

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Answer:

19.872 degrees

Explanation:

Mathematically;

Using Snell’s law

n1 sin A = n2 sinB

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n1 = refractive index in air = 1

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Substituting the values in the equation;

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4 years ago
compare and contrast the results of a crash while the passengers are not wearing seat belts and while they are wearing seat belt
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7. A force stretches a wire by 1 mm. a. A second wire of the same material has the same cross section and twice the length. How
Lera25 [3.4K]

Answer:

(a) The second wire will be stretched by 2 mm

(b) The third wire will be stretched by 0.25 mm

Explanation:

Tensile stress on every engineering material is given as the ratio of applied force to unit area of the material.

σ = F / A

Tensile strain on every engineering material is given as the ratio of extension of the material to the original length

δ = e / L

The ratio of tensile stress to tensile strain is known as Young's modulus of the material.

Y = \frac{FL}{Ae}

<u></u>

<u>Part A</u>

cross sectional area and applied force are the same as the original but the length is doubled

\frac{FL_1}{A_1e_1} =  \frac{FL_o}{A_oe_o} \\\\\frac{L_1}{e_1} =  \frac{L_o}{e_o}\\\\e_1 = \frac{L_1e_o}{L_o} \\\\But, L_1 =2L_o\\\\e_1 = \frac{2L_oe_o}{L_o}\\e_1 = 2e_o

The second wire will be stretched by 2 mm

<u>Part B</u>

a third wire with the same length but twice the diameter of the first

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e₁ = ¹/₄ x 1 mm = 0.25 mm

The third wire will be stretched by 0.25 mm

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