Answer:
y = 80.2 mille
Explanation:
The minimum size of an object that can be seen is determined by the diffraction phenomenon, if we use the Rayleigh criterion that establishes that two objects can be distinguished without the maximum diffraction of a body coincides with the minimum of the other body, therefore so much for the pupil of the eye that it is a circular opening
θ = 1.22 λ/ d
in a normal eye the diameter of the pupils of d = 2 mm = 0.002 m, suppose the wavelength of maximum sensitivity of the eye λ = 550 nm = 550 10⁻⁹ m
θ = 1.22 550 10⁻⁹ / 0.002
θ = 3.355 10⁻⁴ rad
Let's use trigonometry to find the distance supported by this angle, the distance from the moon to the Earth is L = 238900 mille = 2.38900 10⁵ mi
tan θ = y / L
y = L tan θ
y = 2,389 10⁵ tan 3,355 10⁻⁴
y = 8.02 10¹ mi
y = 80.2 mille
This is the smallest size of an object seen directly by the eye
Answer:
(a)

(b) 

Explanation:
Let us take the north direction to be the positive y-axis and the east to be positive x-axis.
First day:
25.0 km southeast, which implies
south of east. The y-component will be negative and the x-component will be positive.


Second day:
She starts off at the stopping point of last day. This time, both the y- and x-components are positive.


Therefore, total displacements:


Magnitude of displacements,

Direction,

Both cells are formed in bone marrow.....but t cells matures into thymus....and b cells matures into bone marrow ! both helps in defense !
B cells can connect to antigens right on the surface of the invading virus or bacteria.
T- cells can only connect to virus antigens on the outside of infected cells.
for more info , comment !
Answer:
1.38 x 10^-18 J
Explanation:
q = - 1.6 x 10^-19 C
d = 5 x 10^-10 m
the potential energy of the system gives the value of work done
The formula for the potential energy is given by

So, the total potential energy of teh system is

As all the charges are same and the distance between the two charges is same so the total potential energy becomes

K = 9 x 10^9 Nm^2/C^2
By substituting the values

U = 1.38 x 10^-18 J
Answer:

Explanation:
1) The moment of inertia of the grindstone is:


