Answer: The volume of an ideal gas will triple in value if the pressure is reduced to one-third of its initial value
Explanation:
We can determine this from the gas laws. Using Boyle's law, which states that "the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature"
Mathematically, P ∝ (1/V)
Since P ∝ (1/V), we can then write that
P = k(1/V)
Where P is the pressure, V is the volume and k is the proportionality constant
PV = k
We can then write that
P1V1 = P2V2 = P3V3 = ...
Hence, P1V1 = P2V2
Where P1 is the initial pressure of the gas
P2 is the final pressure of the gas
V1 is the initial volume of the gas
and V2 is the final volume of the gas
From the question, we want to determine what will make the new volume be thrice the initial volume.
Hence,
P1 = P
V1 = V
P2= ??
V2 = 3V
Therefore,
P × V = P2 × (3V)
P2 = PV/3V
P2 = P/3 = 1/3(P)
This means the volume of an ideal gas will triple in value if the pressure is reduced to one-third of its initial value
Answer:
well actually that depends
Explanation:
that depends on the force of the launch of throw but a common would be like in 2-3secobds
Answer:
Explanation:
Given that,
A point charge is placed between two charges
Q1 = 4 μC
Q2 = -1 μC
Distance between the two charges is 1m
We want to find the point when the electric field will be zero.
Electric field can be calculated using
E = kQ/r²
Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.
Then, the magnitude of the electric at point x is zero.
E = kQ1 / r² + kQ2 / r²
0 = kQ1 / x² - kQ2 / (1-x)²
kQ1 / x² = kQ2 / (1-x)²
Divide through by k
Q1 / x² = Q2 / (1-x)²
4μ / x² = 1μ / (1 - x)²
Divide through by μ
4 / x² = 1 / (1-x)²
Cross multiply
4(1-x)² = x²
4(1-2x+x²) = x²
4 - 8x + 4x² = x²
4x² - 8x + 4 - x² = 0
3x² - 8x + 4 = 0
Check attachment for solution of quadratic equation
We found that,
x = 2m or x = ⅔m
So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.