Answer:
B
Explanation:
From Newton's law of motion, we have:
V^2 = U^2 + 2gH
Where V and U are final and initial velocity respectively.
H is the height.
For the object to have a sustain a maximum height it means the final velocity of the object is zero.
By computing the height of the object sustain by A, we have:
0^2 = 2^2 -2×10×H
0= 4 -20H
4 = 20H;
H= 0.2m
For object B we have;
0^2 = 1^2 -2×10×H
0 = 1 -20H
H = 1/20= 0.05m
From computing the height sustain by both objects, we see object B is projected at a shorter height into atmosphere than A.
Hence object B will return to the ground first.
Answer: The correct answers are (A) and (C).
Explanation:
The expression from electrostatic force is as follows;

Here, F is the electrostatic force, k is constant, r is the distance between the charges and
are the charges.
The electrostatic force follows inverse square law. It is inversely proportional to the square of the distance between the charges. It is directly proportional to the product of the charges.
Like charges repel each other. There is a force of electrostatic repulsion between the like charges. Unlike charges attract each other. There is a force of electrostatic attraction between unlike charges.
The charges are induced on the neutral object when it is placed nearby the charged object without actually touching it.
Therefore, the true statements from the given options are as follows;
Like charges repel.
Unlike charges attract.
Answer:
Implicit memory
Explanation:
Implicit memory stores skill-related data by repeating an activity that always follows the same pattern. It includes all motor, sensory, and intellectual skills, as well as every form of conditioning. The capacity thus acquired does not depend on consciousness. We are able to perform sometimes complex tasks with our thinking turned to something completely different. This explains why Adam types so fast without even remembering the exact location of the letters on the keyboard.
Explanation:
(a) Given:
Δx = 150 m
v₀ = 27 m/s
v = 54 m/s
Find: a
v² = v₀² + 2aΔx
(54 m/s)² = (27 m/s)² + 2a (150 m)
a = 7.29 m/s²
(b) Given:
Δx = 150 m
v₀ = 0 m/s
a = 7.29 m/s²
Find: t
Δx = v₀ t + ½ at²
150 m = (0 m/s) t + ½ (7.29 m/s²) t²
t = 6.42 s
(c) Given:
v₀ = 0 m/s
v = 27 m/s
a = 7.29 m/s²
Find: t
v = at + v₀
27 m/s = (7.29 m/s²) t + 0 m/s
t = 3.70 s
(d) Given:
v₀ = 0 m/s
v = 27 m/s
a = 7.29 m/s²
Find: Δx
v² = v₀² + 2aΔx
(27 m/s)² = (0 m/s)² + 2 (7.29 m/s²) Δx
Δx = 50 m