Answer:
10.4 m/s
Explanation:
First, find the time it takes for the projectile to fall 6 m.
Given:
y₀ = 6 m
y = 0 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: t
y = y₀ + v₀ t + ½ at²
(0 m) = (6 m) + (0 m/s) t + ½ (-9.8 m/s²) t²
t = 1.11 s
Now find the horizontal position of the target after that time:
Given:
x₀ = 6 m
v₀ = 5 m/s
a = 0 m/s²
t = 1.11 s
Find: x
x = x₀ + v₀ t + ½ at²
x = (6 m) + (5 m/s) (1.11 s) + ½ (0 m/s²) (1.11 s)²
x = 11.5 m
Finally, find the launch velocity needed to travel that distance in that time.
Given:
x₀ = 0 m
x = 11.5 m
t = 1.11 s
a = 0 m/s²
Find: v₀
(11.5 m) = (0 m) + v₀ (1.11 s) + ½ (0 m/s²) (1.11 s)²
v₀ = 10.4 m/s
Speed = Distance ÷ Time so divide .5 km by .1h. .5 km÷.1h=5 km/h, so the answer is B. 5km/h.
The displacement vector (SI units) is
![\vec{r} =At\hat{i}+A[t^{3}-6t^{2}]\hat{j}](https://tex.z-dn.net/?f=%5Cvec%7Br%7D%20%3DAt%5Chat%7Bi%7D%2BA%5Bt%5E%7B3%7D-6t%5E%7B2%7D%5D%5Chat%7Bj%7D)
The speed is a scalar quantity. Its magnitude is
Answer: At√(t⁴ - 12t³ + 36t² + 1)
Answer:
288.0 units; that is the electrostatic force of attraction become quadruple of its initial value.
Explanation:
If all other parameters are constant,
Electrostatic Force of attraction ∝ (1/r²)
F = (k/r²) = 72.0
If r₁ = r/2, what happens to F₁
F₁ = (k/r₁²) = k/(r/2)² = (4k/r²) = 4F = 4 × 72 = 288.0 units
