This type of listening response is called back-channel signal. This allows the speaker to know that the listener is attentive or willing to engage a conversation between them. It is shown through short utterances, facial expressions, head nods and others.
Answer:
1.38*10^18 kg
Explanation:
According to the Newton's law of universal gravitation:

where:
G= Gravitational constant (6.674×10−11 N · (m/kg)2)
ma= mass of the astronaut
mp= mass of the planet

so:

Answer:
y= 240/901 cos 2t+ 8/901 sin 2t
Explanation:
To find mass m=weighs/g
m=8/32=0.25
To find the spring constant
Kx=mg (given that c=6 in and mg=8 lb)
K(0.5)=8 (6 in=0.5 ft)
K=16 lb/ft
We know that equation for spring mass system
my''+Cy'+Ky=F
now by putting the values
0.25 y"+0.25 y'+16 y=4 cos 20 t ----(1) (given that C=0.25 lb.s/ft)
Lets assume that at steady state the equation of y will be
y=A cos 2t+ B sin 2t
To find the constant A and B we have to compare this equation with equation 1.
Now find y' and y" (by differentiate with respect to t)
y'= -2A sin 2t+2B cos 2t
y"=-4A cos 2t-4B sin 2t
Now put the values of y" , y' and y in equation 1
0.25 (-4A cos 2t-4B sin 2t)+0.25(-2A sin 2t+2B cos 2t)+16(A cos 2t+ B sin 2t)=4 cos 20 t
So by comparing the coefficient both sides
30 A+ B=8
A-30 B=0
So we get
A=240/901 and B=8/901
So the steady state response
y= 240/901 cos 2t+ 8/901 sin 2t
The weight of the meterstick is:

and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance

from the pivot.
The torque generated by the weight of the meterstick around the pivot is:

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:

from which we find the value of d2:

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.