Answer:
A 30 lb weight is attached to the end of a spring. The spring is stretched 6 in. Find the equation of motion if the weight is released from rest a point 3 inches above equilibrium position 。x(,) =-2 sin(81) 32 x(t) =-32 cos(80 O x(r) =-icos(81)
Explanation:
Answer: Impulse = 4 kgm/s
Explanation:
From the question, you're given the following parameters:
Momentum P1 = 12 kgm/s
Momentum P2 = 16 kgm/s
Time t = 0.2 s
According to second law of motion,
Force F = change in momentum ÷ time
That is
F = (P2 - P1)/t
Cross multiply
Ft = P2 - P1
Where Ft = impulse
Substitute P1 and P2 into the formula
Impulse = 16 - 12 = 4 kgm/s
The magnitude of the impulse is therefore 4 kgm/s.
How much work in J does the string do on the boy if the boy stands still?
<span>answer: None. The equation for work is W = force x distance. Since the boy isn't moving, the distance is zero. Anything times zero is zero </span>
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<span>How much work does the string do on the boy if the boy walks a horizontal distance of 11m away from the kite? </span>
<span>answer: might be a trick question since his direction away from the kite and his velocity weren't noted. Perhaps he just set the string down and walked away 11m from the kite. If he did this, it is the same as the first one...no work was done by the sting on the boy. </span>
<span>If he did walk backwards with no velocity indicated, and held the string and it stayed at 30 deg the answer would be: </span>
<span>4.5N + (boys negative acceleration * mass) = total force1 </span>
<span>work = total force1 x 11 meters </span>
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<span>How much work does the string do on the boy if the boy walks a horizontal distance of 11m toward the kite? </span>
<span>answer: same as above only reversed: </span>
<span>4.5N - (boys negative acceleration * mass) = total force2 </span>
<span>work = total force2 x 11 meters</span>
Answer:
1km = o.621371 mile
Explanation:
1.609 kilometers equal 1 mile. The kilometer is a unit of measurement, as is the mille. However, a mile is longer than a kilometer.
