Image from a far away object formed by a concave mirror
I have no idea but this is my best guess as a sophomore in college
B. Aluminum is possibly correct
Answer:

Explanation:
We are given that
Mass,
Radius,r=0.8 m

Height,h=2.9 m
We have to find the angular acceleration of the cylinder.
According to question


Where



Substitute the value


Where 


Angular acceleration,
Answer:
a) x = 4.33 m
, b) w = 2 rad / s
, f = 0.318 Hz
, c) a = - 17.31 cm / s²,
d) T = 3.15 s, e) A = 5.0 cm
Explanation:
In this exercise on simple harmonic motion we are given the expression for motion
x = 5 cos (2t + π / 6)
they ask us for t = 0
a) the position of the particle
x = 5 cos (π / 6)
x = 4.33 m
remember angles are in radians
b) The general form of the equation is
x = A cos (w t + Ф)
when comparing the two equations
w = 2 rad / s
angular velocity and frequency are related
w = 2π f
f = w / 2π
f = 2 / 2pi
f = 0.318 Hz
c) the acceleration is defined by
a == d²x / dt²
a = - A w² cos (wt + Ф)
for t = 0
, we substitute
a = - 5,0 2² cos (π / 6)
a = - 17.31 cm / s²
d) El period is
T = 1/f
T= 1/0.318
T = 3.15 s
e) the amplitude
A = 5.0 cm
Oxygen is diatomic, so its degree of freedom, (f1)= 5,
also its number of moles, n1= 1
Helium is monoatomic, so its degree of freedom (f2)= 3
and its number of moles given is, n2=2
Now using formula of effective degree of freedom of mixture, (f), we have:
f= (f1n1+f2n2)/(n1+n2)
= (5*1 + 3*2)/ (1+3)
=11/3
Also, from first law of thermodynamics;
U= n Cv. T = nRT(f2)
or, Cv = R. (f/2) (n & T cancel)
We know f=11/6,
substituting the value in above relation, we have:
Cv= R. 11/3*2
= R. 11/6
Also, Cp-Cv = R
or, Cp- R.(11/6)= R
or, Cp= R(11/6 )+1
= 17/6 R
Therefore, Cp/Cv = 17/11