If two variables are inversely proportional, then when one increases, the other decreases, and vice versa. If a variable, y, is inversely proportional to a variable, x, then y = k/x, where k is the proportionality constant.
<u>Answer
</u>
D. Salt water is denser than freshwater.
<u>Explanation</u>
A boat is able to float on water because it experiences an upthrust upwards.
The magnitude of the upthrust depends on the density of the liquid.
When the liquid is denser the boat will experience a great upthrust as compared to when in a less dense liquid.
If the boat sinks lower in the freshwater than in salty water, then Salt water is more dense than freshwater.
Number 20 is B and 21 is C
The speed of a proton after it accelerates from rest through a potential difference of 350 V is
.
Initial velocity of the proton ![u = 0](https://tex.z-dn.net/?f=u%20%3D%200)
Given potential difference ![\Delta V = 350V](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%20350V)
let's assume that the speed of the proton is
,
Since the proton is accelerating through a potential difference, proton's potential energy will change with time. The potential energy of a particle of charge
when accelerated with a potential difference
is,
![U = q \Delta V](https://tex.z-dn.net/?f=U%20%3D%20q%20%5CDelta%20V)
Due to Work-Energy Theorem and Conservation of Energy - <em>If there is no non-conservative force acting on a particle then loss in Potential energy P.E must be equal to gain in Kinetic Energy K.E</em> i.e
![\Delta K = \Delta V](https://tex.z-dn.net/?f=%5CDelta%20K%20%3D%20%5CDelta%20V)
If the initial and final velocity of the proton is
and
respectively then,
change in Kinetic Energy ![\implies \Delta K = \frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}mv^2 - 0](https://tex.z-dn.net/?f=%5Cimplies%20%20%5CDelta%20K%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20-%5Cfrac%7B1%7D%7B2%7Dmu%5E2%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20-%200)
change in Potential Energy ![\implies \Delta U = q\Delta V](https://tex.z-dn.net/?f=%5Cimplies%20%5CDelta%20U%20%3D%20q%5CDelta%20V)
from conservation of energy,
![v= \sqrt{\frac{2q\Delta V}{m}}](https://tex.z-dn.net/?f=v%3D%20%5Csqrt%7B%5Cfrac%7B2q%5CDelta%20V%7D%7Bm%7D%7D)
so, ![v = \sqrt{\frac{2\times 350 \times 1.6\times 10^{-19}}{1.67 \times 10^{-27}}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5Cfrac%7B2%5Ctimes%20350%20%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D%7D%7B1.67%20%5Ctimes%2010%5E%7B-27%7D%7D)
![= 25.86 \times 10^4 ~m/s](https://tex.z-dn.net/?f=%3D%2025.86%20%5Ctimes%2010%5E4%20~m%2Fs)
To read more about the conservation of energy, please go to brainly.com/question/14668053
Answer:
A book on its side exerts a greater force.
Explanation:
Pressure = Force / Area
Assuming that 1kg = 10N
2kg = 20N
Area of book lying flat = 0.3m × 0.2m
= 0.6m²
Pressure of book lying flat = 20N / 0.6m²
= 30Pa (1 s.f.)
Area of book on its side = 0.2m × 0.05m
= 0.01m²
Pressure of book on its side = 20N / 0.01m²
= 2000Pa (1 s.f.)
Since 2000Pa (1 s.f.) > 30Pa (1 s.f.), a book on its side applies greater pressure than lying flat.