Power can be calculate through the equation,
Power = Force x velocity
It should be noted that velocity is calculated by dividing displacement by time. Thus, from the given in this item we can calculate for the power.
Power = (120 lb) x (12 ft/9 s)
<em> </em><span><em>Power = 160 lb.ft/s</em></span>
Use pythagorean theorem
to find the opposite side, which is 7.3
so then you can just use inverse sinA=7.3/10 which equals 46.9 degrees
I think it "Death rate" but I am not very sure though.
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: A.AB
Explanation:
This Velocity vs Time graph shows the acceleration of a body or object, since acceleration is the variation of velocity in time.
As we can see in the attached image, the graph can be divided in four segments:
OA: In this segment the acceleration is changing at a uniform rate. In addition we can see it has a positive slope, hence we are dealing with a positive uniform acceleration.
AB: In this segment the acceleration is changing at a nonuniform rate, since in this part it is not possible to calculate the slope. However if this were uniform, the slope woul be positive. This means the <u>acceleration is nonuniform and positive.</u>
BC: In this segment the acceleration is changing at a nonuniform rate, since in this part it is not possible to calculate the slope. However if this were uniform, the slope woul be negative. This means the acceleration is nonuniform and negative.
CD: In this segment the acceleration is changing at a uniform rate. In addition we can see it has a negative slope, hence we are dealing with a negative uniform acceleration.
From all these segments, the only one that fulfils the nonuniform positive acceleration condition is option A:
Segment AB
