Physics, 16.07.2020 02:01 bernardhenderso4459

1. A sphere with a mass of 10 kg and radius of 0.5 m moves in free fall at sea level (where the air density is 1.22 kg/m3). If the object has a drag coefficient of 0.8, what is the object’s terminal velocity? What is the terminal velocity at an altitude of 5,000 m, where the air density is 0.736 kg/m3? Show all calculations in your answer.

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Answer from: zee786

The value of terminal velocity of object is 7.99 m/s and the value of terminal velocity at an altitude of 5000 m is 10.30 m/s.

Given data:

The mass of sphere is, m = 10 kg.

The radius of sphere is, r = 0.5 m.

The density of air is, $\rho = 1.22 \;\rm kg/m^{3}$ .

The drag coefficient of object is, $C_{d}=0.8$ .

The altitude is, h = 5000 m.

The density of air at altitude is, $\rho' =0.736 \;\rm kg/m^{3}$ .

The mathematical expression for the terminal velocity of an object is,

$v_{t}=\sqrt\dfrac{2mg}{\rho \times A \times C_{d}}$

here,

g is the gravitational acceleration.

A is the area of sphere.

Solving as,

$v_{t}=\sqrt{\dfrac{2 \times 10 \times 9.8}{1.22 \times (4 \pi r^{2}) \times C_{d}}}\\\\\\v_{t}=\sqrt{\dfrac{2 \times 10 \times 9.8}{1.22 \times (4 \pi \times 0.5^{2}) \times 0.8}}\\\\\\\v_{t}=7.99 \;\rm m/s$

Now, the terminal velocity at the altitude of 5000 m is given as,

$v_{t}'=\sqrt\dfrac{2mg}{\rho' \times A \times C_{d}}$

Solving as,

$v_{t}'=\sqrt{\dfrac{2 \times 10 \times 9.8}{0.736 \times (4 \pi \times 0.5^{2}) \times 0.8}}\\\\\\v_{t}'=10.30 \;\rm m/s$

Thus, we can conclude that the value of terminal velocity of object is 7.99 m/s and the value of terminal velocity at an altitude of 5000 m is 10.30 m/s.

Learn more about the terminal velocity here:

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Answer from: Rogeartest4

The terminal velocity at sea level is 7.99 m/s

The terminal velocity at an altitude of 5000 m is 10.298 m/s

Explanation:

mass of sphere m = 10 kg

radius of sphere r = 0.5 m

air density at sea level p = 1.22 kg/m^3

drag coefficient Cd = 0.8

terminal velocity = ?

Area of the sphere A = $4\pi r^{2}$ = 4 x 3.142 x $0.5^{2}$ = 3.142 m^2

terminal velocity is gotten from the relationship

$Vt = \sqrt{\frac{2mg}{pACd} }$

where g = acceleration due to gravity = 9.81 m/s^2

imputing values into the equation

$Vt = \sqrt{\frac{2*10*9.81}{1.22*3.142*0.8} }$ = 7.99 m/s

If at an altitude of 5000 m where air density = 0.736 kg/m^3, then we replace value of air density in the relationship as 0.736 kg/m^3

$Vt = \sqrt{\frac{2mg}{pACd} }$

$Vt = \sqrt{\frac{2*10*9.81}{0.736*3.142*0.8} }$ = 10.298 m/s

Answer from: Quest

the grams of oxygen gas that can be produced is 20.6 grams of o₂

calculation

2liclo₃ → 2 licl + 3 o₂

step 1: find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5 +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281 moles

step 2: use the mole ratio to calculate the moles of o2

from equation above, liclo₃: o₂ is 2: 3

therefore the moles of o₂=0.4281 moles x3/2 =0.6422 moles

step 3: find mass of o₂

mass= moles x molar mass

from periodic table the molar mass of o₂ = 16 x2= 32 g/mol

mass=0.6422 moles x 32 g/mol = 20.6 grams of o₂

Answer from: Quest

come closer and become far apart and become steam/gas

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1. A sphere with a mass of 10 kg and radius of 0.5 m moves in free fall at sea level (where the air...

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