Answer:
a) 1.2*10^-7 m
b) 1.0*10^-7 m
c) 9.7*10^-8 m
d) ultraviolet region
Explanation:
To find the different wavelengths you use the following formula:

RH: Rydberg constant = 1.097 x 10^7 m^−1.
(a) n=2

(b)

(c)

(d) The three lines belong to the ultraviolet region.
<h3>Question:</h3>
A 2.0-cm length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnetic field at the point x = 5.0m on the x-axis.
<h3>
Answer:</h3>
1.6nT [in the negative z direction]
<h2>
Explanation:</h2>
The magnetic field, B, due to a distance of finite value b, is given by;
B = (μ₀IL) / (4πb
) -----------(i)
Where;
I = current on the wire
L = length of the wire
μ₀ = magnetic constant = 4π × 10⁻⁷ H/m
From the question,
I = 20A
L = 2.0cm = 0.02m
b = 5.0m
Substitute the necessary values into equation (i)
B = (4π × 10⁻⁷ x 20 x 0.02) / (4π x 5.0
)
B = (10⁻⁷ x 20 x 0.02) / (5.0
)
B = (10⁻⁷ x 20 x 0.02) / (5.0
)
B = (10⁻⁷ x 20 x 0.02) / (25.0)
B = 1.6 x 10⁻⁹T
B = 1.6nT
Therefore, the magnetic field at the point x = 5.0m on the x-axis is 1.6nT.
PS: Since the current is directed in the positive y direction, from the right hand rule, the magnetic field is directed in the negative z-direction.
Answer:
Los fusibles están diseñados de tal forma que estos se "rompen" o se funden, cuando la demanda eléctrica supera un dado valor (cuando demasiada electricidad pasa a través de el).
Una vez el filamento se rompe, la corriente ya no puede circular por el (podes pensar en esta situación como un cable roto, la electricidad no puede circular por este cable)
Entonces, al romperse el filamento, en caso de una sobrecarga eléctrica, el flujo de electricidad se corta, y de esta forma se protege al computador de posibles sobrecargas.
Answer:
v₂ = v/1.5= 0.667 v
Explanation:
For this exercise we will use the conservation of the moment, for this we will define a system formed by the two students and the cars, for this isolated system the forces during the contact are internal, therefore the moment conserves.
Initial moment before pushing
p₀ = 0
Final moment after they have been pushed
= m₁ v₁ + m₂ v₂
p₀ = 
0 = m₁ v₁ + m₂ v₂
m₁ v₁ = - m₂ v₂
Let's replace
M (-v) = -1.5M v₂
v₂ = v / 1.5
v₂ = 0.667 v
If an object is projected with vertical speed given as

now the time of flight of the object that time in which it comes back on ground

now here we will have


now the range of projectile is given as


now here we know that


now the range is given as


now in order to have maximum range we can say


so we will have

so now we can say

so both speed must be same to have maximum horizontal range