It's important to know that diffraction gratings can be identified by the number of lines they have per centimeter. Often, more lines per centimeter is more useful because the images separation is greater when this happens. That is, the distance between lines increases.
<h2>Therefore, the answer is 2.</h2>
Answer:
a) total moment of inertia is 1359.05 kg m^2
b) angular acceleratio is 0.854rad/sec^2
Explanation:
Given data:
m1=6.9 kg
L=4.88 m
m2=34.5 kg
R=1.22 m
we klnow that moment of inertia for rod is given as
J1=(1/12) ×m×L^2
![J1 = (1/12) \times 6.9 \times 4.88^2 = 13.693 kg m^2](https://tex.z-dn.net/?f=J1%20%3D%20%281%2F12%29%20%5Ctimes%206.9%20%5Ctimes%204.88%5E2%20%3D%2013.693%20kg%20m%5E2)
moment of inertia for sphere is given as
J1=(2/5) ×m×r^2
![J1 = (2/5) \times 34.5 \times 1.22^2 = 20.539 kg m^2](https://tex.z-dn.net/?f=J1%20%3D%20%282%2F5%29%20%5Ctimes%2034.5%20%5Ctimes%201.22%5E2%20%3D%2020.539%20kg%20m%5E2)
As object rotates around free end of rod then for sphere the axis around what it rotates is at a distance of d2=L+R
For rod distance is d1=0.5*L
By Steiner theorem
for the rod we get ![J_1'=J_1 + m_1\times d_1^2](https://tex.z-dn.net/?f=J_1%27%3DJ_1%20%2B%20m_1%5Ctimes%20d_1%5E2)
![J_1' = 13.693 + 2.44^2\times 6.9 = 54.77 kg m^2](https://tex.z-dn.net/?f=J_1%27%20%3D%2013.693%20%2B%202.44%5E2%5Ctimes%206.9%20%3D%2054.77%20kg%20m%5E2)
for the sphere we get ![J_2' = J_2 + m_2\times d_2^2](https://tex.z-dn.net/?f=J_2%27%20%3D%20J_2%20%2B%20m_2%5Ctimes%20d_2%5E2)
![J2' = 20.539 + 34.5\times 6.1^2 = 1304.28 kg^m2](https://tex.z-dn.net/?f=J2%27%20%3D%2020.539%20%2B%2034.5%5Ctimes%206.1%5E2%20%3D%201304.28%20kg%5Em2)
And the total moment of inertia for the first case is
![J_{t1} = J_1'+J_2' = 54.77 + 1304.28 = 1359.05kg.m^2](https://tex.z-dn.net/?f=J_%7Bt1%7D%20%3D%20J_1%27%2BJ_2%27%20%3D%2054.77%20%2B%201304.28%20%3D%201359.05kg.m%5E2)
b) F=476 N
The torque for system is given as
![M = F\times d\times sin(a)](https://tex.z-dn.net/?f=M%20%3D%20F%5Ctimes%20d%5Ctimes%20sin%28a%29)
where a is angle between Force and distance d
and where d represent distance from rotating axis.
In this case a = 90 degree
![M = F\times L/2](https://tex.z-dn.net/?f=M%20%3D%20F%5Ctimes%20L%2F2)
M=476*2.44 = 1161.44 Nm
The acceleration is calculated as
![= \frac{1161.44}{1359.05}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B1161.44%7D%7B1359.05%7D)
= 0.854 rad/sec^2
X=bt + ct3, b=1.50m/s, c=0.640m/s3.
Average velocity = distance / time
ti = 1.00, tf = 3.00
xi = (1.50)(1.00) + (0.640 )(1.00)^3 = 1.5 + 0.640 = 2.14 m
xf = = (1.50)(3.00) + (0.640)(3.00)^3 = 4.5 + 27(17.28) = 21.78 m
Ave Velocity = (21.78 - 2.14) / (3 - 1) = 9.82 m/s
Answer:
wavelength = 2m
Speed = 342m/s
Explanation:
The length of the pipe in terms of the wavelength is expressed as;
L = v/4 (For closed pipe)
v = 4L
Given
L = 50cm
L = 0.5m
Get the wavelength v;
v = 4(0.5)
v = 2.0m
Hence the wavelength is 2.0m
The speed of the sound in pipe is expressed as;
Fo = V/4L
Fo is the fundamental frequency
V is the speed of sound
L is the length of the pipe
171 = V/4(0.5)
V = 171 * 2
V = 342m/s
Hence the speed of the sound in pipe is 342m/s
B is correct makes more sense