Explanation:
The velocity of sound depends on the density of the medium. So we need to find the density of air at each set of conditions. The density of air is:
ρ = (Pd / (Rd T)) + (Pv / (Rv T))
where Pd and Pv are the partial pressures of dry air and water vapor,
Rd and Rv are the specific gas constants of dry air and water vapor,
and T is the absolute temperature.
At the first condition:
Pv = 31.7 mmHg = 4226.3 Pa
Pd = 650 mmHg - 31.7 mmHg = 618.3 mmHg = 82433 Pa
Rv = 461.52 J/kg/K
Rd = 287.00 J/kg/K
T = 30°C = 303.15°C
ρ = (82433 / 287.00 / 303.15) + (4226.3 / 461.52 / 303.15)
ρ = 0.94746 + 0.03021
ρ = 0.97767 kg/m³
At the second condition:
Pv = 0 Pa
Pd = 650 mmHg = 86660 Pa
Rv = 461.52 J/kg/K
Rd = 287.00 J/kg/K
T = 0°C = 273.15°C
ρ = (86660 / 287.00 / 273.15) + (0 / 461.52 / 273.15)
ρ = 1.1054 + 0
ρ = 1.1054 kg/m³
The square of the velocity of sound is proportional to the ratio between pressure and density:
v² = k P / ρ
Since the atmospheric pressure is constant, we can say it's proportional to just the density:
v² = k / ρ
Using the first condition to find the coefficient:
(340)² = k / 0.97767
k = 113018.652
Now finding the velocity of sound at the second condition:
v² = 113018.652 / 1.1054
v = 319.75
Answer:
X(t) = 9.8 *t - 4.9 * t^2
Explanation:
We set a frame of reference with origin at the hand of the girl the moment she releases the ball. We assume her hand will be in the same position when she catches it again. The positive X axis point upwards.The ball will be subject to a constant gravitational acceleration of -9.81 m/s^2.
We use the equation for position under constant acceleration:
X(t) = X0 + V0 * t + 1/2 * a *t^2
X0 = 0 because it is at the origin of the coordinate system.
We know that at t = 2, the position will be zero.
X(2) = 0 = V0 * 2 + 1/2 * -9.81 * 2^2
0 = 2 * V0 - 4.9 * 4
2 * V0 = 19.6
V0 = 9.8 m/s
Then the position of the ball as a function of time is:
X(t) = 9.8 *t - 4.9 * t^2
Answer:
Explanation:
The answer is D- The number of flies decreased, then leveled off.
I believe it is acceleration!
Answer:
Maximum height reached by the ball is 32 meters.
Explanation:
It is given that,
If a baseball is project upwards from the ground level with an initial velocity of 32 feet per second, then it's height is a function of time. The equation is given as :
...........(1)
t is the time taken
s is the height attained as a function of time.
Maximum height achieved can be calculated as :
-16 t + 32 = 0
t = 2 seconds
Put the value of t in equation (1) as :
s = 32 meters
So, the maximum height reached by the ball is 32 meters. Hence, this is the required solution.
