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harina [27]
4 years ago
12

How to play track and field

Physics
2 answers:
Ahat [919]4 years ago
3 0

Explanation:

Track and Field is a sport, which is includes disciplines of running, jumping, and throwing events. The sport traces back to Ancient Greece. The first recorded examples of this sport were at the Ancient Greek Olympics. In Ancient Greece, only one event was contested, the stadion footrace. Later on, the game expanded to more events.Events of track and field are divided into three: track events, field events, and combined events. Track events consist of Sprints, middle-distance, long distance, hurdles and relays; Field events consist of jumps and throws; while combined events consist of pentathlon, heptathlon, and decathlon. Track and field is usually played outdoors in stadiums. The usual features of a track and field stadium are the outer running track, and the field within the track

KATRIN_1 [288]4 years ago
3 0
When Konami ported their arcade hit Track & Field to the Famicom (as Hyper Olympic), they only included four out of six events. Afterwards, they converted Hyper Sports to the Famicom as well, this time including three of the Hyper Sports events and one more event from Track & Field. By the time the NES gained popularity in the United States, Konami retooled the game for release in America by including all eight events from both games in one cartridge. Of the original six events from Track & Field, only the hammer throw is missing; in its place, however, are skeet shooting, archery, and triple jump.
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Read 2 more answers
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kakasveta [241]

(a) The value of the gravitational acceleration is given by:

g=\frac{GM}{r^2}

where

G=6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2} is the gravitational constant

M is the Earth's mass

r=6371 km=6.371 \cdot 10^6 m is the Earth's radius at the pole

If we use the value for g given by the problem, g=9.830 m/s^2, and we rearrange the equation above, we find the value of the Earth's mass:

M=\frac{gr^2}{G}=\frac{(9.830 m/s^2)(6.371 \cdot 10^6 m)^2}{6.67 \cdot 10^{-11} m^3 kg^{-1}s^{-2}}=5.982 \cdot 10^{24} kg

(b) The value we found for the Earth's mass is 5.982 \cdot 10^{24} kg, and we see that this value is slightly larger than the accepted value of 5.979 \cdot 10^{24} kg.

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