On the sunlit surface of Venus, the atmospheric pressure is 9.0 106 Pa, and the temperature is 740 K. On the earth's surface the
atmospheric pressure is 1.0 105 Pa, while the surface temperature can reach 320 K. These data imply that Venus has a "thicker" atmosphere at its surface than does the earth, which means that the number of molecules per unit volume (N/V) is greater on the surface of Venus than on the earth. Find the ratio (N/V)Venus / (N/V)Earth.
2 answers:
Answer:
(1) Venus
(2) Earth
Explanation:
Ideal Gas Equation which relates these quantities of thermodynamics is.
where P is pressure, V is volume , N is number of molecules, R is ideal Gas Constant and has value of 8.3144621J/Mol*K.
Rearranging the above equation, by first solving for N in terms of P, V, R and T and then dividing by V gives us.
Now it is only matter of Plugging in numbers so for Venus planet we can plug in P = 9.0106Pa, T = 740K R is constant in every situation = 8.3144621J/mol K. and get (1) and similarly for Earth we will plug in P = 1.0105Pa , T = 320K and same R, and will get (2).
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Answer:
The minimum average speed the opponent must move so that he is in position to hit the ball is approximately 5.79 m/s
Explanation:
The given parameters of the ball are;
The initial speed of the ball = 15 m/s
The direction in which the ball is launched = 50° above the horizontal
The location of the other tennis player when the ball is launched = 10 m from the ball
The time at which the other tennis player begins to run = 0.3 seconds after the ball is launched
The height at which the ball is hit back = 2.1 m above the height from which the ball is launched
The vertical position, 'y', at time, 't', of a projectile motion is given as follows;
y = (u·sinθ)·t - 1/2·g·t²
When y = 2.1 m, we have;
2.1 = (15·sin(50°))·t - 1/2·9.8·t²
∴ 4.9·t² - (15·sin(50°))·t + 2.1 = 0
Solving with the aid of a graphing calculator function, we get;
t = 0.199776187257 s or t = 2.14525782198 s
Therefore, the ball is at 2.1 m above the start point on the other side of the court at t ≈ 2.145 seconds
The horizontal distance, 'x', the ball travels at t ≈ 2.145 seconds is given as follows;
x = u × cos(50°) × t = 15 × cos(50°) × 2.145 ≈ 20.682 m
The horizontal distance the ball travels at t ≈ 2.145 seconds, x ≈ 20.682 m
Therefore, we have;
The time the other player has to reach the ball, t₂ =2.145 s - 0.3 s ≈ 1.845 s
The distance the other player has to run, d = 20.682 m - 10 m = 10.682 m
The minimum average speed the other player has to move with, = d/t₂
∴ = 10.682 m/(1.845 s) ≈ 5.78970189702 m/s ≈ 5.79 m/s
The minimum average speed the opponent must move so that he is in position to hit the ball, ≈ 5.79 m/s.