Work needed = 23,520 J
<h3>
Further explanation
</h3>
Given
height = 12 m
mass = 200 kg
Required
work needed by the crane
Solution
Work is the transfer of energy caused by the force acting on a moving object
Work is the product of force with the displacement of objects.
Can be formulated
W = F x d
W = Work, J, Nm
F = Force, N
d = distance, m
F = m x g
Input the value :
W = mgd
W = 200 kg x 9.8 m/s²x12 m
W = 23520 J
When the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².
<h3>
Frictional force between the block and the horizontal surface</h3>
The frictional force between the block and the horizontal surface is determined by applying Newton's law;
∑F = ma
F - Ff = ma
Ff = F - ma
Ff = 4 - 2(1.2)
Ff = 4 - 2.4
Ff = 1.6 N
When the applied force increases to 5 N, the magnitude of the block's acceleration is calculated as follows;
F - Ff = ma
5 - 1.6 = 2a
3.4 = 2a
a = 3.4/2
a = 1.7 m/s²
Thus, when the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².
Learn more about frictional force here: brainly.com/question/4618599
a) At a position of 2.0m, the Initial energy is
all made up of the potential energy=m*g*hi<span>
and meanwhile at 1.5 all its energy is also potential energy=m*g*hf
The percentage of energy remaining is E=m*g*hi/m*g*hf x 100
and since mass and gravity are constant so it leaves us with
just E=hi/hf
which 1.5/2.0 x100= 75% so we see that we lost 25% of the
energy or 0.25 in fraction
b) Here use the equation vf^2=vi^2+2gd
<span>where g is gravity, vf is the final velocity and vi is the
initial velocity while d is the distance travelled
so in here we are looking for the vi so let us isolate that
variable
we know that at maximum height or peak, the velocity is 0 so
vf is 0
therefore,</span></span>
vi =sqrt(-2gd) <span>
vi =sqrt(-2x-9.81x1.5) </span>
<span>vi =5.4 m/s
<span>c) The energy was converted to heat due to friction with the
air and the ground.</span></span>
So base on your question that as if the vapors volume were to incorrectly recorded as 125ml, the effect of the error to calculate the molar mass is the same as the error in measuring the volume of the vapor. I hope you are satisfied with my answer and feel free to ask for more
