Which of the following forms of electromagnetic radiation are considered non-ionizing? Select all that apply.

An RC circuit takes t = 2.5 ms to charge to 35% of its full charge after it weas connected to a battery, What is the total time

marin [14]

Answer: time taken to charge to 95%

t = -5.80[ln(1-0.95)]

t = 17.38ms

Explanation:

For an RC Charging circuit

Where Vs

Vc = Vs (1 - e^(-t/RC))

Vc/Vs = 1 - e^(-t/RC)

-t/RC = ln(1 - Vc/Vs)

t = -RC[ln(1 - Vc/Vs)] and RC = k = -t/ln(1 - Vc/Vs)

Where ;

Vc = voltage across the capacitor

Vs = voltage supply

t = charging time = 2.5ms

k = RC = time constant.

Vc/Vs = 0.35

To calculate the time constant k;

k = -t/ln(1- Vc/Vs)

k = -2.5/ln(1-0.35)

k = 5.80ms

time taken to charge to 95%

t = -5.80[ln(1-0.95)]

t = 17.38ms

4 0

3 years ago

Plz help me thank you

jarptica [38.1K]

Area is how much matter somethin has hypothetically speaking if u take one cycinder and put it in a squad purism with lengths of .52 the area will be equal to pi

3 0

2 years ago

Unpolarized light of intensity Io is incident on a stack of 7 polarizing filters, each with its axis rotated 17Â°cw with respect

mash [69]

The ratio of the intensity between light intensity that emerges from the last filter and unpolarized light of intensity, I₀ is It/I₀ = 0.2925

To answer the question, we need to know what polarization of light is.

<h3>What is polarization of light?</h3>

This is when the electric field vector of light is oscillating in one plane.

Now for light of intensity I' which is initially unpolarized, its intensity after polarization is I = 1/2I'.
Also, for light initially polarized, its intensity after polarization is I"' = I"cos²Ф where Ф is the angle between the initial direction and the direction of polarization.

<h3>Intensity of light through each polarized filter</h3>

Given that we have 7 polarizing filters, each rotated 17° cw with respect to the previous filter.

So, since the light is initially unpolarized,

The intensity through the first polarizing filter is I₁ = 1/2I₀ where I₀ is the initial intensity.

The intensity through the second polarizing filter is I₂ = I₁cos²17°= 1/2I₀cos²17°
The intensity through the third polarizing filter is I₃ = I₂cos²17° = 1/2I₀cos⁴17°
The intensity through the fourth polarizing filter is I₄ = I₃cos²17° = 1/2I₀cos⁶17°
The intensity through the fifth polarizing filter is I₅ = I₄cos²17° = 1/2I₀cos⁸17°
The intensity through the sixth polarizing filter is I₆ = I₅cos²17° = 1/2I₀cos¹⁰17°
The intensity through the seventh polarizing filter is I₇ = I₆cos²17° = 1/2I₀cos¹²17°.

<h3>The ratio of the intensity between light intensity that emerges from the last filter and unpolarized light of intensity</h3>

Since I₇ is the last intensity I₇ = It = 1/2I₀cos¹²17°.

So, It/I₀ = 1/2cos¹²17°

= 1/2(0.9563)¹²

= 1/2 × 0.5850

= 0.2925

So, the ratio of the intensity between light intensity that emerges from the last filter and unpolarized light of intensity, I₀ is It/I₀ = 0.2925

Learn more about intensity of polarized light here:

brainly.com/question/25402491

5 0

2 years ago

After a model rocket reached its maximum height, it then took 5.0 seconds to return to the launch site. what is the approximate

kifflom [539]

Air resistance is ignored.
g = 9.8 m/s².
At maximum height, the vertical velocity is zero.

Let h = the maximum height reached.
Let u = the vertical launch velocity.

Because ot takes 5.0 seconds to reach maximum height, therefore
(u m/s) - (9.8 m/s²)*(5 s) = 0
u = 49 m/s

The maximum height reached is
h = (49 m/s)*(5 s) - (1/2)*(9.8 m/s²)*(5 s)²
= 122.5 m

Answer: 122.5 m

3 0

3 years ago

A merry-go-round of radius 2 m is rotating at one revolution every 5 s. A

galben [10]

Answer:

a) The angular speed of the child is approximately 1.257 rad/s

b) The angular speed of the teenager is approximately 1.257 rad/s

c) The tangential speed of the child is approximately 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager is approximately 2.513 m/s

Explanation:

The revolutions per minute, r.p.m. of the merry-go-round = 1 revolution/(5 s)

The radius of the merry-go-round = 2 m

The location of the child = 1 m from the axis

The location of the teenager = 2 m from the axis

1 revolution = 2·π radians

Therefore, we have;

The angular speed, ω = (Angle turned)/(Time elapsed) = (2·π radians)/(5 s)

∴ The angular speed of the merry-go-round, ω = 2·π/5 radians/second

a) The angular speed of the child = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

b) The angular speed of the teenager = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

c) The tangential speed, v = r × The angular speed, ω

Where;

r = The radius of rotation of the object

For the child, r = 1 m

The tangential speed of the child = 1 m × 2·π/5 radians/second = 2·π/5 m/s ≈ 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager = 2 m × 2·π/5 radians/second = 4·π/5 m/s ≈ 2.513 m/s

8 0

2 years ago