Answer:
-6.44 m/s²
Explanation:
Given:
Δx = 60 m
v₀ = 27.8 m/s
v = 0 m/s
Find: a
v² = v₀² + 2aΔx
(0 m/s)² = (27.8 m/s)² + 2a (60 m)
a = -6.44 m/s²
Answer:
q = 3.87 x 10⁵ C
Explanation:
given,
Electric field, E = 8.60 x 10¹ = 86 N/C
radius of earth, R = 6371 Km = 6.371 x 10⁶ m
Coulomb constant, K = 9 x 10⁹ N · m²/C²
Charge on the earth = ?
the electric field at the point


inserting all the values

q = 3.87 x 10⁵ C
The electric charge on the earth is equal to 3.87 x 10⁵ C
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