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3 years ago

When is an object that has been dropped from a great height through the air closest to vein in a free fall?

2 answers:

7 0

-- "Free fall" means no forces acting on the object except for gravity.

-- An object dropped through air has two forces on it -- gravity and air resistance.

-- So we're looking for the part of the drop where air resistance is smallest.

-- The force of air resistance depends on the speed through the air, so the
force of air resistance is smallest when <em>velocity downward is smallest</em>.

Degger [83]3 years ago

6 0

If there were no air, the object would be in free fall the entire time, and there would be no "terminal velocity". So, the object is closest to being in free fall when the effects of the air are minimized--and that is when the object has the smallest speed relative to the air.

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REY [17]

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3 years ago

Read 2 more answers

Why are the temperatures on each of the other terrestrial planets more extreme than earth?

netineya [11]

Other terrestrial planets have more extreme temperatures mainly because of their atmospheres
Explanation:
for example the atmosphere of Venus is composed mainly of carbon dioxide, this carbon dioxide traps the heat or energy from the sun and makes the planet have higher temperatures. where on mars the atmosphere is very thin so it takes in lots of heat and doesn't keep it in very well so it gets very hot and very cold

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3 years ago

How many types of electrical charges exists

Dafna11 [192]

There are two main types of electrical charges.

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3 years ago

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The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume e =

Svetllana [295]

Answer:

Explanation:

(a) Rigel, 2.7x10^32W, T = 11,000K

But L = 4pR²sT⁴

L = 2.7x10^32W, T = 11,000K, s= 5.67 x 10^-8, R= radius in meters

Rigel parallax, p = 0.00378 arc sec

Substituting the various values and making R the subject of the formula

R² = L/(4psT⁴)

R² = 2.7x10^32/(4 x 0.003878 x 5.67x10^-8 x (11,000)⁴)

R² = 2.7x10^32/1.2877x10^7

R² = 2.096761668 x 10^25

R = 4.579041021 x 10^12meters

(b)

Procyon B, 2.1x10^23W, T = 10,000K

But L = 4pR²sT⁴

L = 2.1x10^23W, T = 10,000K, s= 5.67 x 10^-8, R= radius in meters

Procyon B parallax, p = 0.00284 arc sec

R² = 2.1x10^23/(4 x 0.00284 x 5.67x10^-8 x (10,000)⁴)

R² = 2.1x10^23/(6.441 x 10^6)

R² = 3.26036 x 10^16

R = 1.80565 x 10^8 meters

(c) The radius of Rigel is given as 4.579041021 x 10^12meters and the radius of Procyon B is given by 1.80565 x 10^8 meters shows the remarkable difference between a super-giant star(Rigel) and a white dwarf star (Procyon B)

The radius of the sun a red star is 6.96 x 10^8meters which shows a certain level of resemblance with the size of a dwarf white star Procyon B.

The sun is larger than Procyon B as estimated above.

7 0

3 years ago

A merry-go-round with a a radius of R = 1.63 m and moment of inertia I = 196 kg-m2 is spinning with an initial angular speed of

Ulleksa [173]

Answer:

299.88 kgm²/s

499.758 kgm²/s

Explanation:

R = Radius of merry-go-round = 1.63 m

I = Moment of inertia = 196 kgm²

$\omega_i$ = Initial angular velocity = 1.53 rad/s

m = Mass of person = 73 kg

v = Velocity = 4.2 m/s

Initial angular momentum is given by

$L=I\omega_i\\\Rightarrow L=196\times 1.53\\\Rightarrow L=299.88\ kgm^2/s$

The initial angular momentum of the merry-go-round is 299.88 kgm²/s

Angular momentum is given by

$L=mvR\\\Rightarrow L=73\times 4.2\times 1.63\\\Rightarrow L=499.758\ kgm^2/s$

The angular momentum of the person 2 meters before she jumps on the merry-go-round is 499.758 kgm²/s

5 0

3 years ago