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seropon [69]
3 years ago
14

What if the earth was flat

Physics
2 answers:
Zinaida [17]3 years ago
8 0
If the Earth were flat, then ...

-- the sun would never rise;

-- the sun would never set;

-- the sun would look larger and smaller at different times of day;

-- you would be able to see Mt. McKinley, Mt. Everest, Mt. Pinatubo,
and Mt. Kilimanjaro, all from your front yard or your bedroom window;

-- in fact, when the air is clear, you would be able to see the Sears Tower
(here in Chicago) from your front yard or your bedroom window;

-- the shadow of the Earth, falling across the moon, would be some times
circular, some times elliptical, and some times straight, instead of always
circular every time;

-- when you're on the beach, watching a ship sail away from you,
it would just get smaller and smaller and smaller as it got farther
and farther and farther from you; you would never see any of it
disappear, instead of seeing it disappear from the bottom up
the way it actually does.

Lots and lots of other things would also look different and happen
different from the way they actually do look and happen now.
mina [271]3 years ago
5 0
Then everyone would fall off the surface
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A 160 g basketball has a 32.7 cm diameter and may be approximated as a thin spherical shell. Starting from rest, how long will i
mafiozo [28]

Answer:

   t = 0.24 s

Explanation:

As seen in the attached diagram, we are going to use dynamics to resolve the problem, so we will be using the equations for the translation and the rotation dyamics:

Translation:  ΣF = ma

Rotation:      ΣM = Iα ; where α = angular acceleration

Because the angular acceleration is equal to the linear acceleration divided by the radius, the rotation equation also can be represented like:

                    ΣM = I(a/R)

Now we are going to resolve and combine these equations.

For translation:     Fx - Ffr = ma

We know that Fx = mgSin27°, so we substitute:

         (1)                 mgSin27° - Ffr = ma  

For rotation:         (Ffr)(R) = (2/3mR²)(a/R)

The radius cancel each other:

        (2)                Ffr = 2/3 ma

We substitute equation (2) in equation (1):

                            mgSin27° - 2/3 ma = ma

                            mgSin27° = ma + 2/3 ma

The mass gets cancelled:

                            gSin27° = 5/3 a

                            a = (3/5)(gSin27°)

                            a = (3/5)(9.8 m/s²(Sin27°))

                            a = 2.67 m/s²

If we assume that the acceleration is a constant we can use the next equation to find the velocity:

                           V = √2ad; where  d = 0.327m

                           V = √2(2.67 m/s²)(0.327m)

                            V = 1.32 m/s

Because V = d/t

                             t = d/V

                             t = 0.327m/1.32 m/s

                             t = 0.24 s

7 0
3 years ago
places with continental climates typically have _______ summers and ________ winters warm; cool hot; cool warm; cold hot; cold
ipn [44]
Places with continental climates typically have hot summers and cold winters. As compared to places with mild climates, places with continental climates tend to experience the two extremes when it comes to the seasons. Summers can get very hot, and winters can get very cold. 
8 0
3 years ago
An echo is a reflection of sound against another surface. Suppose you are standing on one side of a canyon and you
Morgarella [4.7K]

Answer:

10-1 Temperature and Expansion. 135 ... text, the laboratory work that you do, or your physics teacher. ... Assume that the speed of sound in air is 340 m/s. How.

Explanation:

hope that helps

4 0
3 years ago
Hola! necesito las respuestas de "fuerzas"
Alexus [3.1K]
I need help on this to I have to get the answer fast
5 0
3 years ago
N2 + O2 → 2NO N-N triple bond: 941 kJ/mol O-O double bond: 495 kJ/mol N-O bond: 201 kJ/mol
kiruha [24]

Answer:

\large \boxed{\text{761 kJ}}

Explanation:

You calculate the energy required to break all the bonds in the reactants.

Then you subtract the energy needed to break all the bonds in the products.

                        N₂  +   O₂   ⟶         2NO

                     N≡N  + O=O ⟶       2O-N=O

Bonds:         2N≡N   1O=O        2N-O + 2N=O

D/kJ·mol⁻¹:     941      495           201      607

\begin{array}{rcl}\Delta H & = & \sum{D_{\text{reactants}}} - \sum{D_{\text{products}}}\\& = & 2 \times 941 +1 \times 495 - (2 \times 201 + 2\times 607)\\&=& 2377 - 1616\\&=&\textbf{761 kJ}\\\end{array}\\\text{The enthalpy of reaction is $\large \boxed{\textbf{761 kJ}}$}.

3 0
2 years ago
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