By definition,
q = 1.22y/D
Where,
q = min. angle
y = wavelength
D = Aperture diameter = diameter of the antenna
At distance "x" from the antenna,
L =xq = 1.22xy/D
Where, L = Min. distance
But, y =c/f = (3*10^8)/(16*10^9) = 0.01875 m
Substituting;
L = 1.22*5*10^3*0.01875/2.1 = 54.46 m
Answer:
Explanation:
Given a particle of mass
M = 1.7 × 10^-3 kg
Given a potential as a function of x
U(x) = -17 J Cos[x/0.35 m]
U(x) = -17 Cos(x/0.35)
Angular frequency at x = 0
Let find the force at x = 0
F = dU/dx
F = -17 × -Sin(x/0.35) / 0.35
F = 48.57 Sin(x/0.35)
At x = 0
Sin(0) =0
Then,
F = 0 N
So, from hooke's law
F = -kx
Then,
0 = -kx
This shows that k = 0
Then, angular frequency can be calculated using
ω = √(k/m)
So, since k = 0 at x = 0
Then,
ω = √0/m
ω = √0
ω = 0 rad/s
So, the angular frequency is 0 rad/s
Answer:
I = M R^2 is the moment of inertia about a point that is a distance R from the center of mass (uniform distributed mass).
The moment of inertia about the center of a sphere is 2 / 5 M R^2.
By the parallel axis theorem the moment of inertia about a point on the rim of the sphere is I = 2/5 M R^2 + M R^2 = 7/5 M R^2
I = 7/5 * 20 kg * .2^2 m = 1.12 kg m^2
A. 1/9
Explanation:
The gravitational force between two objects is given by

where
G is the gravitational constant
m1 and m2 are the two masses
r is the distance between the two masses
From the formula, we see that the magnitude of the force is inversely proportional to the square of the distance: therefore, if the distance is tripled (increased by a factor 3), the magnitude of the force changes by a factor

Velocity. Net force causes acceleration and acceleration causes a change in direction and/or magnitude of velocity