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Vikki [24]
2 years ago
9

A 75 kg skydiver can be modeled as a rectangular "box" with dimensions 20 cm * 40 cm * 180 cm. what is his terminal speed if he

falls feet first? use 0.8 for the drag coefficient.
Physics
1 answer:
Zina [86]2 years ago
4 0

Terminal speed=136.7 m/s

the formula for the terminal speed is given by :

V= \sqrt{\frac{2 mg}{\rho a C}}

m= mass=75 kg

a= area=0.2 x 0.4=0.08 m²

ρ= density of air=1.23 kg/m³

C= drag constant =0.8

so V= \sqrt{\frac{2 *75*9.8}{1.23*0.08*0.8}}

v= 136.7 m/s

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If this energy were used to vaporize water at 100.0 ∘C, how much water (in liters) could be vaporized? The enthalpy of vaporizat
Zanzabum

Answer:

0.429 L of water

Explanation:

First to all, you are not putting the value of the energy given to vaporize water, so, to explain better this problem, I will assume a value of energy that I took in a similar exercise before, which is 970 kJ.

Now, assuming that the water density is 1 g/mL, this is the same as saying that 1 g of water = 1 mL of water

If this is true, then, we can assume that 1 kg of water = 1 L of water.

Knowing this, we have to use the expression to get energy which is:

Q = m * ΔH

Solving for m:

m = Q / ΔH

Now "m" is the mass, but in this case, the mass of water is the same as the volume, so it's not neccesary to do a unit conversion.

Before we begin with the calculation, we need to put the enthalpy of vaporization in the correct units, which would be in grams. To do that, we need the molar mass of water:

MM = 18 g/mol

The enthalpy in mass:

ΔH = 40.7 kJ/mol / 18 g/mol = 2.261 kJ/g

Finally, solving for m:

m = 970 / 2.261 = 429 g

Converting this into volume:

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3 0
2 years ago
An 80.0-kg object is falling and experiences a drag force due to air resistance. The magnitude of this drag force depends on its
FromTheMoon [43]

Answer:

The terminal speed of this object is 12.6 m/s

Explanation:

It is given that,

Mass of the object, m = 80 kg

The magnitude of drag force is,

F_{drag}=12v+4v^2

The terminal speed of an object is attained when the gravitational force is balanced by the gravitational force.

F_{drag}=mg

12v+4v^2=80\times 9.8

4v^2+12v=784

On solving the above quadratic equation, we get two values of v as :

v = 12.58 m/s

v = -15.58 m/s (not possible)

So, the terminal speed of this object is 12.6 m/s. Hence, this is the required solution.

6 0
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The H line in Calcium is normally at 396.9 nm. However, in a star's spectrum, it is measured at 398.1nm. How fast is the star mo
agasfer [191]

As we know by Doppler's effect of light we have

\frac{\Delta \lambda}{\lambda} = \frac{v}{c}

here we will have

[tex}\frac{398.1 nm - 396.9 nm}{398.1 nm} = \frac{v}{c}[/tex]

here by solving above we have

3.01 \times 10^{-3} = \frac{v}{c}

here we have

v = 904.3 km/s

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2 years ago
Which statement correctly compares and contrasts the information represented by the chemical formula and model of a compound?
Alina [70]
Models show how the atoms in a compound are connected.
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An early submersible craft for deep-sea exploration was raised and lowered by a cable from a ship. When the craft was stationary
Assoli18 [71]

Answer:

The tension in the cable when the craft was being lowered to the seafloor is 4700 N.

Explanation:

Given that,

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The drag force of 1800 N will act in the upward direction. As it was lowered or raised at a steady rate, so its acceleration is 0. As a result, net force is 0. So,

T + F = W

Here, T is tension

F = 1800 N

W = 6500 N

Tension becomes :

T=W-F\\\\T=6500-1800\\\\T=4700\ N

So, the tension in the cable when the craft was being lowered to the seafloor is 4700 N.

7 0
2 years ago
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