Answer:
Image result for In covalent bonds what is being shared
A covalent bond is a chemical bond that involves the sharing of electron pairs between atoms. These electron pairs are known as shared pairs or bonding pairs, and the stable balance of attractive and repulsive forces between atoms, when they share electrons, is known as covalent bonding.
Explanation:
I am pretty sure it is A Becoming warm
Since it’s moving and causing friction which makes it warm
Hope this helps
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Answer:
Al llegar a su equilibrio térmico ambas barran tendrán una temperatura de 53 grados centígrados.
Explanation:
Dado que una barra de aluminio que está a 78 grados centígrados entra en contacto con una barra de cobre de la misma longitud y área que esta a 28 grados centígrados, y posteriormente se lleva acabo la transferencia de energía entre ambas barras llegando a su equilibrio térmico, para determinar la temperatura a la que ambas barras llegarán se debe realizar el siguiente cálculo:
(78 + 28) / 2 = X
106 / 2 = X
53 = X
Por lo tanto, al llegar a su equilibrio térmico ambas barran tendrán una temperatura de 53 grados centígrados.
Work done against gravity to climb upwards is always stored in the form of gravitational potential energy
so we can say

here h = vertical height raised
so here we know that

here we have

now from above equation


so work done will be given by above value
Answer:
(a) 
(b) 
(c) 
Explanation:
First change the units of the velocity, using these equivalents
and 

The angular acceleration
the time rate of change of the angular speed
according to:


Where
is the original velocity, in the case the velocity before starting the deceleration, and
is the final velocity, equal to zero because it has stopped.

b) To find the distance traveled in radians use the formula:


To change this result to inches, solve the angular displacement
for the distance traveled
(
is the radius).


c) The displacement is the difference between the original position and the final. But in every complete rotation of the rim, the point returns to its original position. so is needed to know how many rotations did the point in the 890.16 rad of distant traveled:

The real difference is in the 0.6667 (or 2/3) of the rotation. To find the distance between these positions imagine a triangle formed with the center of the blade (point C), the initial position (point A) and the final position (point B). The angle
is between the two sides known. Using the theorem of the cosine we can find the missing side of the the triangle(which is also the net displacement):

