The final volume of the gas is 144.25 L
Explanation:
For an ideal gas kept at constant pressure, the work done by the gas on the surroundings is given by
where
p is the pressure of the gas
is the initial volume
is the final volume
For the gas in the cylinder in this problem,
p = 2.00 atm
And we also know the work done,
W = 288 J
So we can solve the equation for , the final volume:
Learn more about ideal gases:
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Answer:
38 cm from q1(right)
Explanation:
Given, q1 = 3q2 , r = 60cm = 0.6 m
Let that point be situated at a distance of 'x' m from q1.
Electric field must be same from both sides to be in equilibrium(where EF is 0).
=> k q1/x² = k q2/(0.6 - x)²
=> q1(0.6 - x)² = q2(x)²
=> 3q2(0.6 - x)² = q2(x)²
=> 3(0.6 - x)² = x²
=> √3(0.6 - x) = ± x
=> 0.6√3 = x(1 + √3)
=> 1.03/2.73 = x
≈ 0.38 m = 38 cm = x
<span>5. Dry ice is an example of _________, which is the process of a solid turning directly into a gas. (1 point)
sublimation
6. The ____ is a unit of force. (1 point)
</span>n<span>ewton
7. Which of the following is the boiling point of water? (1 point)
100°C
8. Which of the following describes the molecular structure of water at 40°C? (1 point)
water molecules are close together and moving freely around each other </span>
Answer:
natural motion and violent motion
Explanation:
Answer:
Explanation:
given
T = 3months = 7.9 × 10⁶s
orbital speed = 88 × 10³m/s
V= 2πr÷T
∴ r = (V×T) ÷ 2π
r = (88km × 7.9 × 10⁶s) ÷ 2π
r = 1.10 × 10⁸km
using kepler's 3rd law
mass of both stars = (seperation diatance)³/(orbital speed)²
M₁ + M₂ = (2r)³/(
year)²
= (1.06 × 10²⁵)/(6.2×10¹³)
1.71×10¹²kg
since M₁ = M₂ =1.71×10¹²kg ÷ 2
M₁ = M₂ = 8.55×10¹¹kg
