Answer:
3) D: 31 m/s
4) D: 84.84 metres
Explanation:
3) Initial velocity along the x-axis is;
v_x = v_o•cos θ
Initial velocity along the y-axis is;
v_y = v_o•sin θ
Plugging in the relevant values, we have;
v_x = 31 cos 60
v_x = 31 × 0.5
v_x = 15.5 m/s
Similarly,
v_y = 31 sin 60
v_y = 31 × 0.8660
v_y = 26.85 m/s
Thus, magnitude of the initial velocity is;
v = √(15.5² + 26.85²)
v ≈ 31 m/s
4) Formula for horizontal range is;
R = (v² sin 2θ)/g
R = (31² × sin (2 × 60))/9.81
R = 84.84 m
The answers to question 4 d
momm=massxvelocity
momm=1200x2.5=120x25=600x5=3000kgm/s
Answer:
53.895 m.
Explanation:
Using the equation of motion,
v² = u² + 2as .............. Equation 1
Where v = final velocity of the swan, u = initial velocity of the swan, a = acceleration of the swan, s = distance covered by the swan.
make s the subject of the equation,
s = (v² - u²)/2a----------- Equation 2
Given: v = 6.4 m/s, u = 0 m/s ( from rest) a = 0.380 m/s².
Substitute into equation 2
s = (6.4²-0²)/(2×0.380)
s = 40.96/0.76
s = 53.895 m.
Hence the swan will travel 53.895 m before becoming airborne.
I am not completely sure, but I believe that it depends on the total mass of the Protons and Neutrons
