-0 m/s
- average velocity=displacement/time
- the runners displacement is zero so her average velocity must be zero
The height at time t is given by
h(t) = -4.91t² + 34.3t + 1
When the ball reaches maximum height, its derivative, h'(t) = 0.
That is,
-2(4.91)t+34.3 = 0
-9.82t + 34.3 = 0
t = 3.4929 s
Note that h''(t) = -9.82 (negative) which confirms that h will be maximum.
The maximum height is
hmax = -4.91(3.4929)² + 34.3(3.4929) + 1
= 60.903 m
Answer:
The ball attains maximum height in 3.5 s (nearest tenth).
The ball attains a maximum height of 60.9 m (nearest tenth)
See this suggested solution.
1. Let a force F' is the vector sum of the forces P and Q, then it is shown on the attached picture and marked with red color.
2. according to the condition the force F holds the object, then F should have the same length as the force F' and the opposite direction.
3. using the conditions described in 2. the answer is C.
Answer:
1 × 10⁶ N/C
Explanation:
The magnitude of the electric field between the membrane = surface density / permittivity of free space = 10 ⁻⁵C/ m² / (8.85 × 10⁻¹²N⁻¹m⁻²C²) = 1.13 × 10⁶ N/C approx 1 × 10⁶ N/C