A well-designed experiment
1 answer:
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Answer:
0.2s
Explanation:
Given parameters:
Speed = 25m/s
Distance = 5m
Unknown:
Time taken for Juan to get to his work = ?
Solution:
To solve this problem we use the relationship below:
Speed =
Time =
So;
Time = = 0.2s
Answer:
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
Explanation:
The additional information to the question is embedded in the diagram attached below:
The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m
Balancing the equilibrium about point A;
F(1.1) - mg (1.25) =
- 1200(9.8)(1.25) = 1200a(0.35)
- 14700 = 420 a ------- equation (1)
--------- equation (2)
Replacing equation 2 into equation 1 ; we have :
1320 a - 14700 = 420 a
1320 a - 420 a =14700
900 a = 14700
a = 14700/900
a = 16.33 m/s²
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
Answer:
<h2>3</h2>Explanation:
Using the efficiency formula. Efficiency = MA/VR * 100%
MA = Mechanical Advantage
VR = velocity ratio =
Distance moved by effort = 4.5m
distance moved by load = 1.5m
VR = 4.5/1.5 =3
Assuming efficiency is 100% (since friction can be ignored)
100% = MA/3 * 100%
1 = MA/3
MA = 3*1
MA = 3
Mechanical Advantage of the ramp is 3