Qualitative data would be taken in this investigation. This is because the effect of the cold temperature on the leaves can not be measured quantitatively, but can be observed in the leaves' appearances. The qualitative data may include how green the leaf looks or how upright and rigid it is. The readings would be taken for leaves with and without snow.
Electric current is measured in Ampere
Haven't done one like this in awhile but I see no one is answering so I gave it a try. I think it's right but let me know if you see something fishy...
Answer:
1. 19.28 secs
2. 154.22 m
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 16 m/s
Final velocity (v) = 0
Force (F) = 1000 N
Mass (m) = 1200 Kg
Time (t) =..?
Distance (s) =...?
Next, we shall determine the acceleration of the car. This can be obtained as follow:
Force (F) = 1000 N
Mass (m) = 1200 Kg
Acceleration (a) =.?
Force (F) = mass (m) x acceleration (a)
F = ma
1000 = 1200 x a
Divide both side by 1200
a = 1000/1200
a = 0.83 m/s²
Since the car is coming to rest, it means it is decelerating. Therefore, the acceleration is – 0.83 m/s²
1. Determination of time taken for the car to halt i.e stop. This can be obtained as follow:
Initial velocity (u) = 16 m/s
Final velocity (v) = 0
acceleration (a) = – 0.83 m/s²
Time (t) =.?
v = u + at
0 = 16 + (–0.83 x t)
0 = 16 – 0.83t
Rearrange
0.83t = 16
Divide both side by 0.83
t = 16/0.83
t = 19.28 secs.
Therefore, the time taken for the car to halt is 19.28 secs.
2. Determination of the distance travelled by the car before coming to rest. This can be obtained as follow:
Initial velocity (u) = 16 m/s
Final velocity (v) = 0
acceleration (a) = – 0.83 m/s²
Distance (s) =..?
v² = u² + 2as
0 = 16² + (2 x –0.83 x s)
0 = 256 – 1.66s
Rearrange
1.66s = 256
Divide both side by 1.66
s = 256/1.66
s = 154.22 m
Therefore, the distance travelled by the car before coming to rest is 154.22 m.