With the use of electric force formula, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x 
N and 66 degrees
ELECTRIC FORCE (F)
F = 
Where K = 9 x 
N
/
The distance between 
and 
can be calculated by using Pythagoras theorem.
d = 
d = 46.7 cm = 0.467 m
For force 
, substitute all the parameters into the formula above
= (9 x x 3 x 1)/
= 2.7 x 
/0.218
= 1.24 x N
For force 
, substitute all the parameters into the formula above
= (9 x x 3 x 4)/
= 1.08 x /0.1089
= 9.92 x N
For force 
, substitute all the parameters into the formula above
= (9 x x 3 x 2)/
= 5.4 x /0.1089
= 4.96 x N
Summation of forces on Y component will be
= - Sin 45
= 9.92 x - 1.24 x Sin 45
= 9.04 x N
Summation of forces on X component will be
= - Cos 45
= 4.96 x - 1.24 x Sin 45
= 4.08 x N
Net Force = 
Net force = 
Net force = 9.9 x N
The direction will be
Tan ∅ = /
Tan ∅ = 9.04 x / 4.08 x
Tan ∅ = 2.216
∅ = 
(2.216)
∅ = 65.7 degrees
Therefore, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x N and 66 degrees approximately.
Learn more about electric Force here: brainly.com/question/4053816
1) 0.0011 rad/s
2) 7667 m/s
Explanation:
1)
The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:
where
is the angular displacement of the object
t is the time elapsed
is the angular velocity
In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is
rad
And the time taken is
Therefore, the angular velocity of the telescope is
2)
For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation
where
v is the linear velocity
is the angular velocity
r is the radius of the circular orbit
In this problem:
is the angular velocity of the Hubble telescope
The telescope is at an altitude of
h = 600 km
over the Earth's surface, which has a radius of
R = 6370 km
So the actual radius of the Hubble's orbit is
Therefore, the linear velocity of the telescope is:
Three complete orders on each side of the m=0 order can be produced in addition to the m=0 order.
The ruling separation is d=1/(470mm-1)
Diffraction lines occurs at an angle θ such that dsin=mλ,when λ is the wavelength and m is an integer.
Notice that for a given order,the line associated with a long wavelength is produced at a greater angle than the line associated with shorter wavelength.
we take λ to be the longest wavelength in the visible spectrum (538nm) and find the greatest integer value of m such that θ is less than 90°.
That is,find the greater integer value of m for which mλ<d.
since,d/λ
There are three complete orders on each side of the m=0 order.
The second and third orders overlap.
learn more about diffraction from here: brainly.com/question/28168352
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IM sure there is C, D, and E in kuiper belts, but not really sure of silicon and iron
