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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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In an experiment, a ringing bell is placed in a vacuum jar that does not have any air in it. What best describes why the bell is
madam [21]

Answer:

Light does not need a medium to travel travel through, but since waves must have a medium to vibrate, sound is not created where no air is present.

Explanation:

3 0
3 years ago
the amount of time for one particle of the medium to make one comlpete vibration cycle is a good description of
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That's the definition of the PERIOD of the vibration.
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The average force of a baseball is 18.9 N . It’s mass is 0.145kg fine the acceleration in m/s^2
Artyom0805 [142]

Answer:

the acceleration is 130.3m/s²

Explanation:

Given data

Force F= 18.9N

Mass of ball m= 0.145kg

Acceleration a=?

Applying the Newton's second law of motion

"The rate of change of momentum of a body is proportional to the external force".

F=ma

a= F/m

a= 18.9/0.142

a= 130.3m/s²

3 0
3 years ago
A projectile is launched at an angle of 36.7 degrees above the horizontal with an initial speed of 175 m/s and lands at the same
Softa [21]

Answer:

a) The maximum height reached by the projectile is 558 m.

b) The projectile was 21.3 s in the air.

Explanation:

The position and velocity of the projectile at any time "t" is given by the following vectors:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v = (v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time "t"

x0 = initial horizontal position

v0 = initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration due to gravity (-9.80 m/s² considering the upward direction as positive).

v = velocity vector at time t

a) Notice in the figure that at maximum height the velocity vector is horizontal. That means that the y-component of the velocity (vy) at that time is 0. Using this, we can find the time at which the projectile is at maximum height:

vy = v0 · sin α + g · t

0 = 175 m/s · sin 36.7° - 9.80 m/s² · t

-  175 m/s · sin 36.7° /  - 9.80 m/s² = t

t = 10.7 s

Now, we have to find the magnitude of the y-component of the vector position at that time to obtain the maximum height (In the figure, the vector position at t = 10.7 s is r1 and its y-component is r1y).

Notice in the figure that the frame of reference is located at the launching point, so that y0 = 0.

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 175 m/s · 10.7 s · sin 36.7° - 1/2 · 9.8 m/s² · (10.7 s)²

y = 558 m

The maximum height reached by the projectile is 558 m

b) Since the motion of the projectile is parabolic and the acceleration is the same during all the trajectory, the time of flight will be twice the time it takes the projectile to reach the maximum height. Then, the time of flight of the projectile will be (2 · 10.7 s) 21.4 s. However, let´s calculate it using the equation for the position of the projectile.

We know that at final time the y-component of the vector position (r final in the figure) is 0 (because the vector is horizontal, see figure). Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

0 = 175 m/s · t · sin 36.7° - 1/2 · 9.8 m/s² · t²

0 = t (175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t)

0 = 175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t

-  175 m/s ·  sin 36.7 / -(1/2 · 9.8 m/s²) = t

t = 21.3 s

The projectile was 21.3 s in the air.

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An external force that is being applied in the direction of the displacement
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