The work done will be equal to the potential energy of the piano at the final position
P.E=m.g.h
.consider the plank the hypotenuse of the right triangle formed with the ground
.let x be the angle with the ground=31.6°
.h be the side opposite to the angle x (h is the final height of the piano)
.let L be the length of the plank
sinx=opposite side / hypotenuse
= h/L
then h=L.sinx=3.49×sin31.6°=0.638m
weight w=m.g
m=w/g=3858/10=385.8kg
Consider Gravity g=10m/s2
then P.E.=m.g.h=385.8kg×10×0.638=2461.404J
then Work W=P.E.=2451.404J
Answer:
m = 105.37 kg
Explanation:
We are given;
Mass of man; m = 113 kg
Length of boat = 6.3m
Now, The position of the center of mass will not change during the motion of the man.
Thus,
X_g,i = X_g,f
So,
[113(6.3) + ma]/(113 + m) = [113(3.26) + m(a +3.26)]/(113 + m)
113 + m will cancel on both sides to give;
113(6.3) + ma = [113(3.26) + m(a +3.26)]
711.9 + ma = 368.38 + ma + 3.26m
ma will cancel out to give;
711.9 - 368.38 = 3.26m
343.52/3.26 = m
m = 105.37 kg
One double bond consists 4 electrons, so 2 double bonds means 8 electrons
Answer:
10.6cm
Explanation:
We are given 5.3cm below the starting point (spring extension).
Therefore, to find static vertical equilibrium, we use the equation:
kx = mg
Where:
k = spring constant =
=mg/5.3 kg/s²
We are told the object was dropped from rest.
Therefore:
loss in potential energy = gain in spring p.e
Let's use the expression:
mgx = ½kx²
We are asked to find the stretch at maximum elongation x.
To find x, we make x subject of the formula.
Therefore, we have:
x = 2mg/k (after rearranging the equation above)
x = (2mg) / (mg/5.3)
x = 10.6cm
