The work done in pushing the box is 64000 lb-ft²/s²
<h3>How to calculate the work done by the person?</h3>
Since person pushing a 40 pound box up an incline that is 30 degrees above the horizontal and 100 feet long, the work done, W is
W = mgh where
- m = mass of box = 40 lb,
- g = acceleration due to gravity = 32 ft/s² and
- h = height of incline = LsinФ where
- L = length of incline = 100 ft and
- Ф = angle of incline = 30°
So, W = mgh
W = mgLsinФ
So, substituting the values of the variables into the equation, we have
W = mgLsinФ
W = 40 lb × 32 ft/s² × 100 ftsin30°
W = 1280 lb-ft/s² × 100 ft × 0.5
W = 1280 lb-ft/s² × 50 ft
W = 64000 lb-ft²/s²
So, the work done in pushing the box is 64000 lb-ft²/s²
Learn more about work done here:
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Remember that a parallel line has the same slope...
So we already know the slope of the line containing point C is -4.
Using point slope equation, we can do...
y-y₁=m(x-x₁)
y-7= -4(x-4)
y-7- -4x+16
*Add seven to both sides...* y=-4x+23
There is you answer... hope this helps!
ANSWER
B. 1
EXPLANATION
The given function is
This is a piece-wise defined function.
We want to find f(1)
We substitute x=1 into f(x)=x because, 1 belongs to the interval,
1≤x≤5
f(1)=1
The correct answer is B.
Answer:
y= -2x-1
Step-by-step explanation:
Write in slope-intercept form, y=mx+by=mx+b.
