Violet i took the test and it gave me the answer
Answer:
66.02m/s
Explanation:
the equation describing the distance covered in the horizontal direction is
but the acceleration in the horizontal path is zero, hence we have
Since the horizontal distance covered is 155m at 7.6secs, we have 
Also from the vertical path, the distance covered is expressed as
since the horizontal distance covered in 7.6secs is 195m, then we have
Hence if we divide both equation 1 and 2 we arrive at
Hence if we substitute the angle into the equation 1 we have
Hence the initial velocity is 66.02m/s
Given:
Speed = 3.5 m/s
Distance = 20 m
To find:
Time taken
Solution:
Speed = Distance/Time
so
Time = Distance/Speed
Substituting given values, we get
Time = 20/3.5
<em>Time = 5.714 seconds</em>
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Answer:
As a science fair project, you want to launch an 950g model rocket straight up and hit a horizontally moving target as it passes 33.0m above the launch point. The rocket engine provides a constant thrust of 20.0N . The target is approaching at a speed of 18.0m/s . At what horizontal distance between the target and the rocket should you launch?
= 43.56m
Explanation:
acceleration =
(20 - (0.95 * 9.8) )/ (0.95)
= 10.68 / 0.95
= 11.24 m/s²
we use
s = ut + (1/2) at²
Given that
s= 40
u =0
s = 0 * t + (1/2) (11.24)t²
t = √(66/1.24)
t = √5.87
t = 2.42sec
hence
Horizontal distance = 18 * 2.42
= 43.56m
A crystalline substance that contains water molecules is a ___. <em><u>hydrate</u></em>
