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irga5000 [103]
3 years ago
12

How is a game of Flag Rugby started?

Physics
2 answers:
Snowcat [4.5K]3 years ago
5 0
The game is started by a PUNT or DROP KICK to the opposing team from the MID FIELD line. The object of the game is to score by placing the ball on or behind the opponents' goal goal line.
viktelen [127]3 years ago
5 0
Punt it or drop it by throwing the ball
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A woman carries a 10kg box up a set of 5m high stairs and then down a 12m long hallway. How much work does she do on the box?
nika2105 [10]

Answer: 1666J

Explanation:

Given that,

Mass of box (m) = 10kg

Total distance covered by box (h)

= (5m + 12m)

= 17m

work done on the box = ?

Work is done when force is applied on an object over a distance. Hence, the magnitude of work done on the box depends on its mass (m), distance covered (h), and acceleration due to gravity (g)

(g has a value of 9.8m/s²

i.e Work = mgh

Work = 10kg x 9.8m/s² x 17m

Work = 1666J

Thus, 1666 joules of work was done by the woman on the box.

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3 years ago
What is the effect on the force of gravity between two objects if the mass of one object doubles?
Leni [432]
Then the force will also be doubled
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An engineer weighs a sample of mercury (ρ = 13.6 × 103 kg/m3 ) and finds that the weight of the sample is 6.0 n. what is the sam
Amanda [17]
Given:
ρ = 13.6 x 10³ kg/m³, density of mercury
W = 6.0 N, weight of the mercury sample
g = 9.81 m/s², acceleration due to gravity.

Let V =  the volume of the sample.
Then
W = ρVg
or
V =  W/(ρg)
   = (6.0 N)/[(13.6 x 10³ kg/m³)*(9.81 m/s²)]
   = 4.4972 x 10⁻⁵ m³

Answer: The volume is 44.972 x 10⁻⁶ m³
5 0
3 years ago
An arrow is shot at a target 20 m away. The arrow is shot with a horizontal velocity of 80 m/s.
Verdich [7]

(1) The time of motion of the arrow is 0.25 s.

(2) The vertical height dropped by the arrow as it approaches the target is 0.31 m.

The given parameters:

  • <em>Horizontal distance of the arrow, X = 20 m</em>
  • <em>Horizontal speed of the arrow, v = 80 m/s</em>

<em />

The time of motion of the arrow is calculated as follows;

t = \frac{X}{v} \\\\t = \frac{20 }{80} \\\\t  = 0.25 \ s

The vertical height dropped by the arrow as it approaches the target is calculated as follows;

h = v_0_y t + \frac{1}{2} gt^2\\\\h = 0 \ + \ \frac{1}{2} \times 9.8 \times 0.25^2\\\\h =0.31 \ m

Learn more about time of motion of projectile here:  brainly.com/question/1912408

4 0
2 years ago
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