Answer:
im pretty sure it is C. Insulation
Explanation:
because it says reduced, and basically ur insulating the fire.
If a circuit has a current of 3.6 Amps and resistance of 5 Ohms, then Ohm's law can be used to find the voltage. Ohm's law states that the voltage is equal to the product of current and resistance (V=IR). In this case the voltage is equal to 3.6 Amps x 5 Ohms = 18.0 Volts. The law can also be used with the rearranged equation to obtain current or resistance.
Here, you need to use your "Protractor" as it is given in the question, but we can calculate the value with the help of our mathematical calculation too:
[ Protractor can be use only in real life, not here ]
Draw an imaginary line from initial position to final position.
Now, In that triangle, tan x = P/B
tan x = 1.4 / 2
tan x = 0.70
x = tan⁻¹ (0.70)
x = 35 [ tan 35 = 0.70 ]
In short, Your Answer would be 35 degrees
Hope this helps!
Answer:
.
Explanation:
When the ball is placed in this pool of water, part of the ball would be beneath the surface of the pool. The volume of the water that this ball displaced is equal to the volume of the ball that is beneath the water surface.
The buoyancy force on this ball would be equal in magnitude to the weight of water that this ball has displaced.
Let
denote the mass of this ball. Let
denote the mass of water that this ball has displaced.
Let
denote the gravitational field strength. The weight of this ball would be
. Likewise, the weight of water displaced would be
.
For this ball to stay afloat, the buoyancy force on this ball should be greater than or equal to the weight of this ball. In other words:
.
At the same time, buoyancy is equal in magnitude the the weight of water displaced. Thus:
.
Therefore:
.
.
In other words, the mass of water that this ball displaced should be greater than or equal to the mass of of the ball. Let
denote the density of water. The volume of water that this ball should displace would be:
.
Given that
while
:
.
In other words, for this ball to stay afloat, at least
of the volume of this ball should be under water. Therefore, the volume of this ball should be at least
.
Referring to Compton scattering
Δλ = h/m₀c (I- cos Ф)
λ' =λ = (0,0242×10⁻¹⁰) (1- cos 60°)
λ= λ' -(0.0242 × 10⁻¹⁰) (1- cos 60°)
7.19 ˣ 10⁻¹²m
The increased potential is given by
Vₐc = hc/eλ = 6.625 × 10 ⁻³⁴ J,s) ( 3× 10⁸ m/s ( 1.6 ˣ 10 ⁻¹⁰C)
(7.19 ˣ 10⁻¹²m)
173kV.