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harina [27]
3 years ago
12

How to play track and field

Physics
2 answers:
Ahat [919]3 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]3 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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Gennadij [26K]

Answer:

5 I think will be none of the above and 6 could be all of the above

8 0
2 years ago
A force in the +x-direction with magnitude ????(x) = 18.0 N − (0.530 N/m)x is applied to a 6.00 kg box that is sitting on the ho
fiasKO [112]

Answer:

v_f=8.17\frac{m}{s}

Explanation:

First, we calculate the work done by this force after the box traveled 14 m, which is given by:

W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m

Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:

W=\Delta K\\W=K_f-K_i\\W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}

The box is initially at rest, so v_i=0. Solving for v_f:

v_f=\sqrt{\frac{2W}{m}}\\v_f=\sqrt{\frac{2(200N\cdot m)}{6kg}}\\v_f=\sqrt{66.67\frac{m^2}{s^2}}\\v_f=8.17\frac{m}{s}

5 0
3 years ago
A motorist traveling with a constant speed of 15 m/s (about 34 mi/h) passes a school-crossing corner, where the speed limit is 0
nikklg [1K]

Answer:

(a) 10 s

(b) 30 m/s

(c) 150 m

Explanation:

The motorist's position at time t is:

x = 15t

The officer's position at time t is:

x = ½ (3) t² = 1.5 t²

(a) When they have the same position, the time is:

15t = 1.5 t²

t = 0 or 10 s

(b) The officer's speed is:

v = 3t

v = 30 m/s

(c) The position is:

x = 15t = 150 m

6 0
3 years ago
The climate of an area is based on A) Rivers and lakes B) natural resources. C) mountain and valleys. D) Precipitation and tempe
LenaWriter [7]
It is D as it is the only answer referring to weather and climate.
3 0
3 years ago
What is the net force on this object?​
Mariulka [41]
Answer=8N

Explantion:
20-20=0
22-14=8N
5 0
3 years ago
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