To solve this problem we will apply the principles of conservation of energy, for which we have to preserve the initial kinetic energy as elastic potential energy at the end of the movement. If said equality is maintained then we can affirm that,


Here,
m = mass
k = Spring constant
x = Displacement
v = Velocity
Rearranging to find the velocity,



Our values are,



Replacing our values we have,


Therefore the velocity is 
Answer: Option (b) is correct.
Explanation:
Since we know that,
P = VI
where;
P = power
V= Voltage
I = Current
Since it's given that,
P = 600W
I = 2.5 A
equating these values in the above equation, we get;
<em>V =
</em>
<em>V = 240 V</em>
Answer:
a) I₁ = 11.2 Lux
, vertical direction
, b) I₂ = 1.44 Lux
Explanation:
a) A polarized is a system that absorbs light that is not polarized in the direction of its axis, therefore half of the non-polarized light must be absorbed
consequently the above the processed light has half of the incident intensity and the directional of the polarized
I₁ = I₀ / 2
I₁ = 22.4 / 2
I₁ = 11.2 Lux
is polarized in the vertical direction
b) The polarized light falls on a second polarizer, therefore it must comply with the law of Malus
I₂ = I₁ cos² θ
I₂ = 11.2 cos² 69
I₂ = 1.44 Lux
Answer:
THEY WOULD FIND DED FISH:)))))
Explanation:
Answer:

The Magnitude of electric field is in the upward direction as shown directly towards the charge
.
Explanation:
Given:
- side of a square,

- charge on one corner of the square,

- charge on the remaining 3 corners of the square,

<u>Distance of the center from each corners</u>


∴Distance of center from corners, 
Now, electric field due to charges is given as:

<u>For charge
we have the field lines emerging out of the charge since it is positively charged:</u>

<u>Force by each of the charges at the remaining corners:</u>

<u> Now, net electric field in the vertical direction:</u>


<u>Now, net electric field in the horizontal direction:</u>


So the Magnitude of electric field is in the upward direction as shown directly towards the charge
.