A. How much work is being done to hold the beam in place?
Work is the product of Force and Displacement. Since there
is no Displacement involved in just holding the beam in place, hence the work
is zero.
B. How much work was done to lift the beam?
In this case, force is simply equal to weight or mass
times gravity. Hence the work is:
Work = weight * displacement
Work = 500 lbf * 100 ft
Work = 50,000 lbf * ft
C. How much work would it take if the steel beam were
raised from 100 ft to 200ft?
The displacement is still 100 ft since 200 – 100 = 100 ft,
hence the work done is still similar in B which is:
<span>Work = 50,000 lbf * ft</span>
Answer:
<em><u>Assuming that the vertical speed of the ball is 14 m/s</u></em> we found the given values:
a) V₀ = 23.4 m/s
b) h = 27.9 m
c) t = 0.96 s
d) t = 4.8 s
Explanation:
a) <u>Assuming that the vertical speed is 14 m/s</u> (founded in the book) the initial speed of the ball can be calculated as follows:
<u>Where:</u>
: is the final speed = 14 m/s
: is the initial speed =?
g: is the gravity = 9.81 m/s²
h: is the height = 18 m
b) The maximum height is:
c) The time can be found using the following equation:
d) The flight time is given by:
I hope it helps you!
Answer:
the third one is incorrect
Explanation:
10 x 10³= 10^1 x 10^3 = 10^4
The become more negative hope this helps!:))
Answer:
364.4 J
Explanation:
I = Moment of inertia of the forearm = 0.550 kgm²
v = linear velocity of the ball relative to elbow joint = 17.1 m/s
r = distance from the joint = 0.470 m
w = angular velocity
Using the equation
v = r w
17.1 = (0.470) w
w = 36.4 rad/s
Rotational kinetic energy of the forearm is given as
RKE = (0.5) I w²
RKE = (0.5) (0.550) (36.4)²
RKE = 364.4 J
