Velocity of the mass after 11 seconds = ( value of the gravitational acceleration) * ( time )
velocity = ( 9.81 m / s^2 ) ( 11)
velocity = 107.91 meters per second
Answer:x(t)= Acos(wt)
Explanation:
According to Newton's 2nd law,a particle of mass m acted on by a force is given by:Fs=-kx
Where x is displacement from equilibrium
K = spring constant
Therefore X(t) = Acos(2pit/T)
X(t)= Acos(wt)
To solve this problem we will use the concepts related to Magnification. Magnification is the process of enlarging the apparent size, not physical size, of something. This enlargement is quantified by a calculated number also called "magnification".
The overall magnification of microscope is

Where
N = Near point
l = distance between the object lens and eye lens
= Focal length
= Focal of eyepiece
Given that the minimum distance at which the eye is able to focus is about 25cm we have that N = 25cm
Replacing,


Therefore the correct answer is C.
Answer
Time period T = 1.50 s
time t = 40 s
r = 6.2 m
a)
Angular speed ω = 2π/T
=
= 4.189 rad/s
Angular acceleration α = 
= 
= 0.105 rad/s²
Tangential acceleration a = r α = 6.2 x 0.105 = 0.651 m/s²
b)The maximum speed.
v = 2πr/T
= 
= 25.97 m/s
So centripetal acceleration.
a = 
= 
= 108.781 m/s^2
= 11.1 g
in combination with the gravitation acceleration.


I believe the answer is B, a real and inverted image is formed on the side of the lens opposite the rubber ducky. The focal length is 15 cm and therefore the center of curvature (2F) will be 30 cm. When the object is placed between F and 2F (in this case 20 cm) in front of a convex lens, an inverted, real image is formed on the other side of the lens.