Define loudness as L, distance as d, and k as the variation constant. The appropriate equation representing the relationship is:
L = k / d^2
For Situation 1,
L1 = k / d1^2
For Situation 2,
L2 = k / (d1 / 4)^2
To have equal k, it follows that L2 = 16 L1.
Thus, the loudness your friend experiences is 16 times greater than yours.
There's an absence of circuit diagrams.
Initially, this causes worry for a moment, until we remember that we have no understanding of the experiment mentioned in the problem either, rendering such worries unnecessary.
Discharge refers to the volume of water flowing down a river or stream within a specific timeframe, typically measured in cubic feet per second or gallons per day. Generally, the discharge of a river is calculated by taking the product of the cross-sectional area of water in the channel and the average velocity of water at that section: discharge = area * velocity. In this instance, the result is 0.2 m/s.
<span>The partial pressure of A = 1.06 atm and the partial pressure of B = 0.53 atm</span>
The time required for the water balloon to reach the ground is given as
Here we understand that
Now applying the earlier mentioned formula
Now in the same time frame, we can conclude the distance covered will be
Thus, it will land at a distance of 15.7 m from where it started
