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3 years ago

I need help on this one please

Physics

1 answer:

klio [65]3 years ago

4 0

Lithium shares more properties with Sodium, because they have the same number of valence electrons (one). Choice 1 is correct. Hope this helps!

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An LED with total power P tot = 960 mW emits UV light of wavelength 360 nm. Assuming the LED is 55% efficient and acts as an iso

Anit [1.1K]

Answer:

$E_0=225.09N/C$

Explanation:

We are given that

Power,Ptot=960mW= $960\times 10^{-3}W$

$1mW=10^{-3} W$

Wavelength, $\lambda=360 nm=360\times 10^{-9} m$

$1nm=10^{-9} m$

Distance,r=2.5 cm= $2.5\times 10^{-2} m$

1m=100 cm

Efficiency=55%

Power radiation emitted= $\frac{55}{100}\times 960\times 10^{-3}=0.528W$

Intensity,I= $\frac{P}{4\pi r^2}$

$I=\frac{0.528}{4\pi(2.5\times 10^{-2})^2}=67.26W/m^2$

Intensity, $I=\frac{1}{2}c\epsilon_0E^2_0$

$E^2_0=\frac{2I}{c\epsilon_0}$

Where $\epsilon_0=8.85\times 10^{-12}$

$c=3\times 10^8 m/s$

$E_0=\sqrt{\frac{2I}{c\epsilon_0}}$

$E_0=\sqrt{\frac{2\times 67.26}{3\times 10^8\times 8.85\times 10^{-12}}}$

$E_0=225.09N/C$

4 0

3 years ago

A 190 mH inductor is connected to an emf given by

Bas_tet [7]

(a) The reactance of the inductor is 25.46 ohms.

(b) The expression for the current through the inductor is I(t) = (6.32 A) sin(134t)

<h3>Ractance of the inductor</h3>

The reactance of the inductor is calculated as follows;

XL = ωL

where;

ω is angular frequency
L is 190 mH

v(t) = (161 V) sin(134t)

v(t) = V sin(ωt)

The reactance of the inductor is calculated as follows;

XL = (134) x (190 x 10⁻³)

XL = 25.46 ohms

<h3>Peak current in the circuit</h3>

I₀ = V₀/XL

I₀ = (161) / (25.46)

I₀ = 6.32 A

<h3>Expression for the current through the inductor</h3>

I(t) = (6.32 A) sin(134t)

Learn more about inductance here: brainly.com/question/16765199

7 0

2 years ago

A 4.2 kg parachutist is moving straight downward with a speed of 3.85 m/s

statuscvo [17]

Kinetic energy= .5 x m x v^2
KE=.5 x 4.2 x 3.85^2
KE=31.13

4 0

3 years ago

Two satellites, A and B are in different circular orbits

jek_recluse [69]

Answer:

The ratio of the orbital time periods of A and B is $\frac{1}{2}$

Solution:

As per the question:

The orbit of the two satellites is circular

Also,

Orbital speed of A is 2 times the orbital speed of B

$v_{oA} = 2v_{oB}$ (1)

Now, we know that the orbital speed of a satellite for circular orbits is given by:

$v_{o} = \farc{2\piR}{T}$

where

R = Radius of the orbit

Now,

For satellite A:

$v_{oA} = \farc{2\piR}{T_{a}}$

Using eqn (1):

$2v_{oB} = \farc{2\piR}{T_{a}}$ (2)

For satellite B:

$v_{oB} = \farc{2\piR}{T_{b}}$ (3)

Now, comparing eqn (2) and eqn (3):

$\frac{T_{a}}{T_{b}} = \farc{1}{2}$

6 0

3 years ago

Select the correct answer.

erica [24]

<span>The correct answer is C:Waves transfer energy, but not matter. A wave does not move matter in the direction of its propagation. It only transfers energy just like the ocean wave traveling many miles away with the water just moving up and down.</span>

6 0

4 years ago