Answer:
correct option is a. True
Explanation:
solution
the noise floor is AWGN ( additive white Gaussian noise )
and when viewed in the frequency domain, it is the continuous noise level
because as they have a uniform power over all the frequency.
so that it is additive white Gaussian noise
as we can say given statement is True
correct option a true
That depends on how far it is from the nearest planet. If it's on the surface of Earth, it weighs (19 kg) x (9.8 m/s^2) = 186.2 newtons.
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.
Answer:
Explanation:
= Permittivity of free space = 
A = Area = 
d = Thickness = 
k = Dielectric constant = 5.4
V = Voltage = 86.2 mV
Charge is given by
The charge on the outer surface is
B. F<em>spring = k(triangle)</em> x
