<span>The terminal velocity of a sand grain in the air is greater than the terminal velocity of the same sand grain in seawater. The terminal velocity is greater in the air because it is less dense compared to a much denser sea water. Terminal velocity takes lace when an object is constantly free-falling.</span>
The "missing" 4 kg may have escaped the scene of the fire in the
form of hot gases, and as particles of soot and ash that were seen
leaving in the form of "smoke".
7. First write down all the known variables while separating the values for each direction:
x-direction:
vix = 20m/s
vfx = 20m/s
x = 39.2m
y-direction:
viy = 0m/s
ay = -9.8m/s^2
y = ?
Based on the knowns, the first step is to calculate the time of flight from the x-direction as it will be the same as value for the y-direction. Find the correct kinematic equation to do so:
x = (1/2)(vix+vfx)t
(39.2) = (1/2)(20+20)t
1.96s = t
Now that we have the time of flight, we can use the kinematic equation that will relate the known variables in the y-direction:
y = viy*t + (1/2)ay*t^2
y = (0)(1.96) + (1/2)(-9.8)(1.96)^2
y = -18.82m (Value is negative because gravity constant was negative. It is the height reference that from the top of the building down, which is why it is negative. The sign can be ignored for this question.)
8. First write down all the known variables while separating the values for each direction:
x-direction:
x = 12m
vfx = 0m/s
vix = ?
y-direction:
y1 = 1.2m
y2 = 0.6m
viy = 0m/s;
ay = -9.8m/s^2
First find time in the y-direction as it would be the same value for the x-direction.
(y2 - y1) = viy*t + (1/2)ay*t^2
(-0.6) = (0)t + (1/2)(-9.8)t^2
t = 0.35s
Now that we have the time of flight, we can use the kinematic equation that will relate the known variables in the x-direction:
x = (1/2)(vix+vfx)t
(12) = (1/2)(vix+(0))(0.35)
68.6m/s = vix
Answer:
i don't understand the hw
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