L = illuminance
A = surface
i = intensity
L = i / A ==: i = L * A
i = 6 lux * 4 m^2 = 24 lumen
-- volume = (length)(width)(height)
-- Since the cube is a cube, its three dimensions are all the same number.
Volume = (2.5cm)(2.5cm)(2.5cm)
Volume = 15.625 cubic cm
-- density = (mass) / (volume)
Density = (1129.56g) / (15.625cm^3)
Density = 72.3 g/cm^3
(roughly 3.2 TIMES the density of the most dense naturally occurring substance on Earth)
Answer:
t=17.838s
Explanation:
The displacement is divided in two sections, the first is a section with constant acceleration, and the second one with constant velocity. Let's consider the first:
The acceleration is, by definition:
So, the velocity can be obtained by integrating this expression:
The velocity is, by definition: 
, so
.
Do x=11 in order to find the time spent.
At this time the velocity is: 
This velocity remains constant in the section 2, so for that section the movement equation is:
The left distance is 89 meters, and the velocity is 
, so:
So, the total time is 14.303+3.5355s=17.838s
Answer: 2.94×10^8 J
Explanation:
Using the relation
T^2 = (4π^2/GMe) r^3
Where v= velocity
r = radius
T = period
Me = mass of earth= 6×10^24
G = gravitational constant= 6.67×10^-11
4π^2/GMe = 4π^2 / [(6.67x10^-11 x6.0x10^24)]
= 0.9865 x 10^-13
Therefore,
T^2 = (0.9865 × 10^-13) × r^3
r^3 = 1/(0.9865 × 10^-13) ×T^2
r^3 = (1.014 x 10^13) × T^2
To find r1 and r2
T1 = 120min = 120*60 = 7200s
T2 = 180min = 180*60= 10800s
Therefore,
r1 = [(1.014 x 10^13)7200^2]^(1/3) = 8.07 x 10^6 m
r2 = [(1.014 x 10^13)10800^2]^(1/3) = 10.57 x 10^6 m
Required Mechanical energy
= - GMem/2 [1/r2 - 1/r1]
= (6.67 x 10^-11 x 6.0 x 10^24 * 50)/2 * [(1/8.07 × 10^-6 )- (1/10.57 × 10^-6)]
= (2001 x 10^7)/2 * (0.1239 - 0.0945)
= (1000.5 × 10^7) × 0.0294
= 29.4147 × 10^7 J
= 2.94 x 10^8 J.
