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jekas [21]
2 years ago
6

A 90-kg skydiver jumps from a height of 6000 m above the ground, falling head-first (pike position). The area of the diver is 0.

14 m^2. The value for C is 1.0, and the density of air is 1.21 kg/m^3. Determine the terminal velocity.
a. 49 m/s
b. 68 m/s
c. 83 m/s
d. 102 m/s
Physics
1 answer:
svp [43]2 years ago
4 0

Answer:

The correct option is d: 102 m/s.

Explanation:

The terminal velocity is given by:

v = \sqrt{\frac{2mg}{\rho AC}}   (1)

Where:

m: is the mass = 90 kg

g: is the gravity = 9.81 m/s²

A: is the area = 0.14 m²

C: is the drag coefficient = 1.0  

ρ: is the density of the air = 1.21 kg/m³  

Now, by entering the above values into equation (1) we have:

v = \sqrt{\frac{2mg}{\rho AC}} = \sqrt{\frac{2*90 kg*9.81 m/s^{2}}{1.21 kg/m^{3}*0.14 m^{2}*1.0}} = 102 m/s

Therefore, the correct option is d: 102 m/s.

I hope it helps you!  

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julia-pushkina [17]

30,869.2 J

Explanation:

Given parameters:

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Initial temperature = -30°C

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Specific heat capacity of water = 4.2J/g°C

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Unknown:

Quantity of heat required = ?

Temperature-energy graph = ?

Solution:

The temperature energy profile is attached to this solution.

q = mc∅ₓ + mlₓ + mc∅ₙ + mlₙ + mc∅ₐ

qₓ = mc∅ₓ in converting ice from -30°C to ice at 0°C

qₓ is the amount of heat supplied to the ice that changes its temperature from -30°C to that at freezing point:

qₓ  = 10 x 4.2 x (0-(-30)) = 10 x 2.1 x 30 = 630J

qₓ = mlₓ in converting ice to water

qₓ here is the latent heat used to break the ice bonds without a change in temperature:

qₓ = ml = 10 x 334 = 3340J

qₙ = mc∅ₙ is the heat from water at 0°C to boiling point.

This is the heat required to take water from freezing temperature to its boiling point

qₙ = mc∅ₙ  = 10 x 4.2 x (100 - 0) = 4200J

qₙ  = mlₙ is the heat in vaporizing water

In vaporizing water, there is no temperature change when the hydrogen bonds are broken. The heat supplied is not used to raise temperature. We use the latent heat of vaporization:

qₙ   = 10 x 2230 = 22300J

qₐ = mc∅ₐ from vapor at 100°C to 120°C

This is the heat used to raise the temperature of vapor:

qₐ = 10 x 1.996 x (120-100) = 399.2J

The total heat:

  q = qₓ + qₙ + qₐ = 630J + 3340J + 4200J + 22300J + 399.2J =30,869.2 J

Learn more:

Specific heat brainly.com/question/7210400

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