Answer:
5. -24 m/s²
Explanation:
Acceleration: This can be defined as the rate of change of velocity.
The S.I unit of acceleration is m/s².
mathematically,
a = dv/dt ............................ Equation 1
Where a = acceleration, dv/dt = is the differentiation of velocity with respect to time.
But
v = dx(t)/dt
Where,
x(t) = 27t-4.0t³...................... Equation 2
Therefore, differentiating equation 2 with respect to time.
v = dx(t)/dt = 27-12t²............. Equation 3.
Also differentiating equation 3 with respect to time,
a = dv/dt = -24t
a = -24t .................... Equation 4
from the question,
At the end of 1.0 s,
a = -24(1)
a = -24 m/s².
Thus the acceleration = -24 m/s²
The right option is 5. -24 m/s²
Lighting gives a sense of scale (The sky)
A. Translucent because it’s shows a little light on the inside and looks dark on the outside
In the double-slit interference experiment, the distance of the nth-maximum from the center of the screen is given by

where

is the wavelength
D is the distance between the screen and the slits
d is the distance between the slits
In our problem,



By applying the previous formula, we can calculate the distance of the 4th maximum from the center of the screen:

Similarly, the distance of the 8th- maximum is

Therefore, the distance between the two maxima is