Rounding What is 13.428 to the nearest tenths positions
1 answer:
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Answer:
hey sandra! i hope this will help you!
Explanation:
The geographic coordinate system is a reference system that uses the two angular latitude (north or south) and longitude (east or west) coordinates to determine the positions of the land surface points. These two angular coordinates measured from the center of the Earth are from a spherical coordinate system that is aligned with its axis of rotation. These coordinates are usually expressed in sexagesimal degrees:
Latitude measures the angle between any point and the equator. Latitude lines are called parallel and are circles parallel to the equator on the Earth's surface. Latitude is the distance between any point and Ecuador, measured on the meridian that passes through that point.
• All points located on the same parallel have the same latitude.
• Those who are in the north of Ecuador receive the denomination North (N).
• Those who are in the south of Ecuador receive the denomination South (S).
• It is measured from 0º to 90º.
• The latitude of 0º corresponds to Ecuador.
• The North and South poles have latitude 90º N and 90º S respectively.
The length measures the angle along the equator from any point on Earth. It is accepted that Greenwich in London is length 0 in most modern societies. The lines of length are maximum circles that pass through the poles and are called meridians.
• All points located on the same meridian have the same length.
• Those who are east of Meridian Zero receive the designation East (E).
• Those who are west of Meridian Zero receive the denomination West (O).
• It is measured from 0º to 180º.
• The Greenwich meridian corresponds to the length 0º.
By combining these two angles, the position of any point on the Earth's surface can be expressed.
Answer:
a) = 928 J
, b)U = -62.7 J
, c) K = 0
, d) Y = 11.0367 m, e) v = 15.23 m / s
Explanation:
To solve this exercise we will use the concepts of mechanical energy.
a) The elastic potential energy is
= ½ k x²
= ½ 2900 0.80²
= 928 J
b) place the origin at the point of the uncompressed spring, the spider's potential energy
U = m h and
U = 8 9.8 (-0.80)
U = -62.7 J
c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also
K = ½ m v²
K = 0
d) write the energy at two points, maximum compression and maximum height
Em₀ = ke = ½ m x²
= mg y
Emo =
½ k x² = m g y
y = ½ k x² / m g
y = ½ 2900 0.8² / (8 9.8)
y = 11.8367 m
As zero was placed for the spring without stretching the height from that reference is
Y = y- 0.80
Y = 11.8367 -0.80
Y = 11.0367 m
Bonus
Energy for maximum compression and uncompressed spring
Emo = ½ k x² = 928 J
= ½ m v²
Emo =
Emo = ½ m v²
v =√ 2Emo / m
v = √ (2 928/8)
v = 15.23 m / s