Water vapor and carbon dioxide
A motorboat accelerates uniformly from a velocity of 6.5m/s
to the west to a velocity of 1.5m/s to the west. if its accelerate was 2.7m/s2
to the east ,
how far did it travel during the accelration? Give your
answer in units of kilometers per hour/sec. To find the acceleration of the car
we have to
<span>
1. First determine
the suitable formula for this word problem.
Which is a. A=vf-vi/t</span>
which will be
Given are: Vi= 6.5 m/s Vf= 1.5 m/s a= 2.7 m/sec2 t=1.85s
Solution:
<span>
x = v0t + ½at2</span>
<span>x = <span>16.645375 m </span></span>
Answer:
128 m
Explanation:
From the question given above, the following data were obtained:
Horizontal velocity (u) = 40 m/s
Height (h) = 50 m
Acceleration due to gravity (g) = 9.8 m/s²
Horizontal distance (s) =?
Next, we shall determine the time taken for the package to get to the ground.
This can be obtained as follow:
Height (h) = 50 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
50 = ½ × 9.8 × t²
50 = 4.9 × t²
Divide both side by 4.9
t² = 50 / 4.9
t² = 10.2
Take the square root of both side
t = √10.2
t = 3.2 s
Finally, we shall determine where the package lands by calculating the horizontal distance travelled by the package after being dropped from the plane. This can be obtained as follow:
Horizontal velocity (u) = 40 m/s
Time (t) = 3.2 s
Horizontal distance (s) =?
s = ut
s = 40 × 3.2
s = 128 m
Therefore, the package will land at 128 m relative to the plane
Thermometer-temperature
Wind vane -wind direction
Anemometer- both wind speed (and direction)
Hygrometer- humidity