Answer:
I would say that the answer is B because it is the most reasonable one
Explanation:
Answer:
n = 1.89 x 10¹⁹ photons/s
Explanation:
given,
Power of bulb = 100 W
Visible light = 5 W
wavelength of the visible light = 600 nm
number of photons emitted per second
using formula of wavelength




energy of photon
U = h υ
U = 6.63 x 10⁻³⁴ x 4 x 10¹⁴
U = 2.65 x 10⁻¹⁹ J
number of photon


n = 1.89 x 10¹⁹ photons/s
Answer:
a) R₂/R₁ = 0.3049
b) I₂ / I₁ = 1.81
c) (230)² / ( 127)² = 3.28
Explanation:
By definition the power consumption is:
P = I²*R ⇒ as I = V/R I² = (V/R)² and P = (V²/R²)*R
P = V²/R
P = power consumtion
I = current
R = is the resistor
Case Device 1 Voltage 127 (V)
P₁ = V₁² /R₁ P₁ = (127)²/ R₁
Case Device 2 Voltage 230 (V)
P₂ = ( 230)²/ R₂
As P₁ = P₂ (127)²/R₁ = (230)²/R₂
(127)²/(230)² = R₂/R₁
a) R₂/R₁ = 0.3049
b) P₁ = I₁²*R₁ P₂ = I₂²*R₂
I₂² *R₂ = I₁²*R₁
I₂² / I₁² = R₁/R₂
I₂² / I₁² = 1/ 0.3049
I₂² / I₁² = 3.2797
√ (I₂² / I₁² ) = √3.2797
b) I₂ / I₁ = 1.81
c) P₁ = V₁*I₁ ⇒ P₁ = V₁²/R₁ if we "use" the device in 230 (V)
P₁´ = V₂² / R₁
P₁ = (127)²/R₁ and P₁´ = (230)²/ R₁
Then the increasing factor is:
c) (230)² / ( 127)² = 3.28
Answer: There are
protons present in one coulomb of charge.
Explanation:
Given: Charge on proton = 
It is known that the charge on a proton is equal to the charge present on an electron.
Formula used to calculate number of protons is as follows.
Q = ne
where,
Q = charge considered
n = number of protons
e = charge on electron or proton
Substitute the values into above formula as follows.

Thus, we can conclude that there are
protons present in one coulomb of charge.