Given: Normal pull of gravity g = 9.8 m/s²;
g = 0.855 m/s² (at a certain distance)
Universal gravitational constant G = 6.67 x 10⁻¹¹ N.m²/Kg²
Mass of the Earth Me = 5.98 x 10²⁴ Kg
Radius r = ?
g = GMe/r²
r = √GMe/g
r = √(6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)/(0.855 m/s²)
r = 2.16 x 10⁷ m or
r = 21,610 Km
.
Explanation:
Given:
v₀ₓ = 15 m/s cos 20° = 14.10 m/s
aₓ = 0 m/s²
v₀ᵧ = 15 m/s sin 20° = 5.13 m/s
aᵧ = -9.8 m/s²
t = 1.5 s
Find: Δx and Δy
Δx = v₀ₓ t + ½ aₓ t²
Δx = (14.10 m/s) (1.5 s) + ½ (0 m/s²) (1.5 s)²
Δx = 21.1 m
Δy = v₀ᵧ t + ½ aᵧ t²
Δy = (5.13 m/s) (1.5 s) + ½ (-9.8 m/s²) (1.5 s)²
Δy = -3.33 m
Answer with Explanation:
We are given that
Y=(-4,12)
Z=(1,19)
We have to find the ordered pair which represents the vector YZ and magnitude of vector YZ.
Vector YZ=Z-Y=<1,19>-<-4,12>
Vector YZ=<5,7>
Magnitude of vector r


Using the formula

