The range of the piece of paper is C) 1.4 m
Explanation:
The motion of the piece of paper is the motion of a projectile, which consists of two separate motions:
- A uniform motion along the horizontal direction, with constant velocity
- A uniformly accelerated motion along the vertical direction, with constant acceleration (the acceleration of gravity, 
)
From the equation of motion, it is possible to find an expression for the range (the total horizontal distance covered) of a projectile, which is given by:
where
u is the initial velocity
is the angle of projection
g is the acceleration of gravity
For the piece of paper in this problem,
u = 4.3 m/s
Substituting,
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Answer:
5.59 m/s
Explanation:
We are given;
Mass = 110 kg
Initial velocity: u = 13.41 m/s
Force = 615 N
Time(t) = 1 s
Now, the formula for force is;
Force = mass x acceleration
Thus;
615 = 110 × acceleration
\Acceleration(a) = 615/110 = 5.591 m/s²
Now, using Newton's first law of motion, we can find acceleration (a). Thus;
v = u + at
v = 13.41 + (5.591 × 1)
v ≈ 19 m/s
So,the change in velocity is;
Final velocity(v) - Initial velocity(u) = 19 - 13.41 = 5.59 m/s
Ok so it would be late and the relative velocity would be 190 m/s because 200 m/s - 10 m/s is 190 m/s. Hope this helps.
We need a and we have m and F . Now a = f÷m so therefore a = 4,9 ÷ 0,5 which is 0,98 metres per cubic second
