The position at time t is
x(t) = 0.5t³ - 3t² + 3t + 2
When the velocity is zero, the derivative of x with respect to t is zero. That is,
x' = 1.5t² - 6t + 3 = 0
or
t² - 4t + 2 = 0
Solve with the quadratic formula.
t = (1/2) [ 4 +/- √(16 - 8)] = 3.4142 or 0.5858 s
When t =0.5858 s, the position is
x = 0.5(0.5858³) - 3(0.5858²) + 3(0.5858) + 2 = 2.828 m
When t=3.4142 s, the position is
x = 0.5(3.4142³) - 3(3.4142²) + 3(3.4142) + 2 = -2.828 m
Reject the negative answer.
Answer:
The velocity is zero when t = 0.586 s, and the distance is 2.83 m
When the acceleration is zero, the second derivative of x with respect to t is zero. That is,
3t - 6 = 0
t = 2
The distance traveled is
x = 0.5(2³) - 3(2²) + 3(2) + 2 = 0
Answer:
When the acceleration is zero, t = 2 s, and the distance traveled is zero.
Answer:
Explanation:
Work done on the lever ( input energy ) = force applied x input distance
= 24 N x 2m = 48 J
Work done by the lever ( output energy ) = load x output distance
= 72 N x 0.5m = 36 J
efficiency = output energy / input energy
= 36 J / 48 J
= 3 / 4 = .75
In percentage terms efficiency = 75 % .
The correct answer is 99.06 .
Since we know that,
1 ft = 12 inches, (1ft)3 = (12)3inches3
& 1 inch = 2.54 cm , (1 inch)3 = (2.54)3cm3
& 1 cm3 = 1ml,
&(1 ml)/1000 = 1L, so (1cm3)/1000 = 1L.
& 1 min. = 60 sec.
Now, 1ft3 = (12)3x(2.54)3/(1000) L
and 1 min = 60 sec
So, 1ft3/min = (12)^3x(2.54)^3/(1000)x(60) L/s
Hence 210 ft3/min = 210x(12^)3x(2.54)^3/(1000)x(60).
= 99.06.
Learn more about rate of the vacuum's air flow here :-
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A transformer is used to increase or decrease a voltage. BUT ... it has to be an AC voltage, or the transformer doesn't work.
