Which natural hazards are most likely to affect Florida, due to its geography? Check all that apply. PLS HELP! I WILL GIVE BEST
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9. An object which is in circular motion (moving along a circle) is said to be accelerating because it changes it's direction constantly even if it is moving with a constant speed. cuz acceleration is change in either magnitude or direction of an object with respect to time.
therefore, it's still acceleration as change in direction with time.
10. Average speed of an object can be calculated by dividing the total distance covered by an object by time taken to cover that distance.
i.e
it can be re- arranged to find the distance as :
11. speed = 20 m/s : conversion into km/h
distance covered : 4 km = 4000 m
time taken = 200 seconds
12. let's use the first equation of motion to find the acceleration :
(a) 30.9 m
Let's analyze the motion of the parachutist. Its vertical position above the ground is given by
where
h = 45 m is the initial height
u = 10 m/s is the initial velocity (upward)
t is the time
g = -9.8 m/s^2 is the acceleration of gravity (downward)
Substituting t=3 s , we find the height of the parachutist when it opens the parachute:
(b) 44.1 m
Here we have to find first the height of the balloon 3 seconds after the parachutist has jumped off from it. The vertical position of the balloon is given by
y = h + ut
where
h = 45 m is the initial height
u = 10 m/s is the initial velocity (upward)
t is the time
Substituting t = 3 s, we find
y = 45 m + (10 m/s)(3 s) = 75 m
So the distance between the balloon and the parachutist after 3 s is
d = 75 m - 30.9 m = 44.1 m
(c) 8.2 m/s downward
The velocity of the parachutist at the moment he opens the parachute is:
v = u +gt
where
u = 10 m/s is the initial velocity (upward)
t is the time
g = -9.8 m/s^2 is the acceleration of gravity (downward)
Substituting t = 3 s,
v = 10 m/s + (-9.8 m/s^2)(3 s)= -19.4 m/s
where the negative sign means it is downward
After t=3 s, the parachutist open the parachute and it starts moving with a deceleration of
a =+5 m/s^2
where we put a positive sign since this time the acceleration is upward.
The total distance he still has to cover till the ground is
d = 30.9 m
So we can find the final velocity by using
where this time we have u = 19.4 m/s as initial velocity. Taking the downward direction as positive, the deceleration must be considered as negative:
a = -5 m/s^2
Solving for v,
(d) 5.24 s
We can find the duration of the second part of the motion of the parachutist (after he has opened the parachute) by using
where
a = -5 m/s^2 is the deceleration
v = 8.2 m/s is the final velocity
u = 19.4 m/s is the initial velocity
t is the time
Solving for t, we find
And added to the 3 seconds between the instant of the jump and the moment he opens the parachute, the total time is
t = 3 s + 2.24 s = 5.24 s
Answer:
H vaporization = 100.0788 kJ/mol
Explanation:
Use clausius clapyron's adaptation for the calculation of Hvap as:
Where,
P₂ and P₁ are the pressure at Temperature T₂ and T₁ respectively.
R is the gas constant.
T₂ = 823°C
T₁ = 633°C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So, the temperature,
T₂ = (823 + 273.15) K = 1096.15 K
T₁ = (633 + 273.15) K = 906.15 K
P₂ = 400.0 torr , P₁ = 40.0 torr
R = 8.314 J/K.mol
Applying in the formula to calculate heat of vaporization as:
Solving for heat of vaporization, we get:
H vaporization = 100078.823 J/mol
Also, 1 J = 10⁻³ kJ
So,
<u>H vaporization = 100.0788 kJ/mol</u>