When a crest-trough meet the interference produced will be destructive in nature hence they both will cancel out and the amplitude produced will be equal to zero hence the loudness will reduce to zero.
Answer:
12.2 m
Explanation:
Given:
v₀ = 15.6 m/s
v = 0 m/s
a = -10 m/s²
Find: Δy
v² = v₀² + 2aΔy
(0 m/s)² = (15.6 m/s)² + 2 (-10 m/s²) Δy
Δy = 12.2 m
Hypothesis: Enriched environments will have a positive effect and better performance of the experimental group of animals exposed to toys in comparison to the control group of animals in running in mazes.
Dependent Variable: Control Group
Independent Variable: Exposed/Experimental Group
Answer:
The relative density of the second liquid is 7.
Explanation:
From archimede's principle we know that the force that a liquid exerts on a object equals to the weight of the liquid that the object displaces.
Let us assume that the volume of the object is 'V'
Thus for the liquid in which the block is completely submerged
The buoyant force should be equal to weight of liquid
Mathematically
Thus for the liquid in which the block is 1/7 submerged
The buoyant force should be equal to weight of liquid
Mathematically
Comparing equation 'i' and 'ii' we see that
Since the first liquid is water thus 
Thus the relative density of the second liquid is 7.
Then the force of gravity between them would be quadrupled and so on but the gravitational force is inversely proportional to the square of the separation distance between the two interacting objects which makes more separation distance will result in weaker gravitational forces
