Answer:
KE = 1/2mv^2
KE = 1/2(1120)(40^2)
KE = 560(1600)
KE = 896000
Let me know if this helps!
This question is incomplete. The complete question is given below:
Question 3 Both the angle and the magnitude of the force have a certain uncertainty: εF = 28 N and εθ = 0.8°. Using the propagation methods described in the video you watched at the beginning of this prelab, calculate the corresponding propagated uncertainty for Fx, in N. For this question, round up your final answer to two significant figures. Do not include the ± sign in your answer. Example: If the x component of F is 200±14 N, you should enter “14”.
Both the force and the angle are measured, and the results are quoted as a central value plus/minus an uncertainty:
F = F0 ± εF
θ = θ0 ± εθ
We would like to evaluate the component of the force in the x direction.
Question 2
Let us first concentrate on the central value. Take F0 = 325 N and θ0 = 57°.
The answer & explanation for this question is given in the attachment below.
<span>Answer:
The moments of inertia are listed on p. 223, and a uniform cylinder through its center is:
I = 1/2mr2
so
I = 1/2(4.80 kg)(.0710 m)2 = 0.0120984 kgm2
Since there is a frictional torque of 1.20 Nm, we can use the angular equivalent of F = ma to find the angular deceleration:
t = Ia
-1.20 Nm = (0.0120984 kgm2)a
a = -99.19 rad/s/s
Now we have a kinematics question to solve:
wo = (10,000 Revolutions/Minute)(2p radians/revolution)(1 minute/60 sec) = 1047.2 rad/s
w = 0
a = -99.19 rad/s/s
Let's find the time first:
w = wo + at : wo = 1047.2 rad/s; w = 0 rad/s; a = -99.19 rad/s/s
t = 10.558 s = 10.6 s
And the displacement (Angular)
Now the formula I want to use is only in the formula packet in its linear form, but it works just as well in angular form
s = (u+v)t/2
Which is
q = (wo+w)t/2 : wo = 1047.2 rad/s; w = 0 rad/s; t = 10.558 s
q = (125.7 rad/s+418.9 rad/s)(3.5 s)/2 = 952.9 radians
But the problem wanted revolutions, so let's change the units:
q = (5528.075087 radians)(revolution/2p radians) = 880. revolutions</span>
In physical chemistry, the terms body-centered cubic (BCC) and face-centered cubic (FCC) refer to the cubic crystal system of a solid. Each solid is made up simple building blocks called lattice units. There are different layouts of a lattice unit.
It is better understood using 3-D models shown in the picture. A BCC unit cell has one lattice point in the center, together with eight corner atoms which represents 1/8 of an atom. Therefore, there are 1+ 8(1/8) = 2 atoms in a BCC unit cell. On the other hand, a FCC unit cell is composed of half of an atom in each of its faces and 1/8 of an atom in its corners. Therefore, there are (1/2)6 + (1/8)8 = 4 atoms in a FCC unit cell.
Explanation:
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