As per kinematics equation we are given that

now we are given that
a = 2.55 m/s^2


now we need to find x
from above equation we have



so it will cover a distance of 93.2 m
Answer:
The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J
Explanation:
The given variables are
Work done = 550 J
Volume change = V₂ - V₁ = -0.5V₁
Thus the product of pressure and volume change = work done by gas, thus
P × -0.5V₁ = 500 J
Hence -PV₁ = 1000 J
also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂
Also to compress the gas by a factor of 11 we have
P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11 = 909.091 J of work
I really wish I could be helping you. I don't know.
We have that The ratio U1/U2 of their potential energies due to their interactions with Q is
From the question we are told that
Question 1
Charge q1 is distance r from a positive point charge Q.
Question 2
Charge q2=q1/3 is distance 2r from Q.
Charge q1 is distance s from the negative plate of a parallel-plate capacitor.
Charge q2=q1/3 is distance 2s from the negative plate.
Generally the equation for the potential energy is mathematically given as

Therefore
The Equations of U1 and U2 is
For U1

For U2

Since
U is a function of q and q2=q1/3
Therefore

For Question 2
For U1

Therefore

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