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3 years ago

14

The sun delivers an average power of 1.499 w/m2 to the top of neptune's atmosphere. find the magnitudes of vector e max and vect

or b max for the electromagnetic waves at the top of the atmosphere.

1 answer:

sesenic [268]3 years ago

4 0

See attachment for answer:

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A typical cell phone charger is rated to transfer a maximum of 1.0 Coulomb of charge per second. Calculate the maximum number of

Alexandra [31]

Answer:

The maximum no. of electrons- $2.25\times 10^{22}$

Solution:

As per the question:

Maximum rate of transfer of charge, I = 1.0 C/s

Time, t = 1.0 h = 3600 s

Rate of transfer of charge is current, I

Also,

$I = \frac{Q}{t}$

Q = ne

where

n = no. of electrons

Q = charge in coulomb

I = current

Thus

Q = It

Thus the charge flow in 1. 0 h:

$Q = 1.0\times 3600 = 3600\ C$

Maximum number of electrons, n is given by:

$n = \frac{Q}{e}$

where

e = charge on an electron = $1.6\times 10^{- 19}\ C$

Thus

$n = \frac{3600}{1.6\times 10^{- 19}} = 2.25\times 10^{22}$

3 0

3 years ago

What is the technical name for super glue type adhesives

Murrr4er [49]

The technical name for super glue type adhesive is Cyanoacrylate glue

5 0

3 years ago

Read 2 more answers

Any help is appreciated please

stiv31 [10]

Answer:

C. It speeds up, and the angle increases

Explanation:

We can answer by using the Snell's law:

$n_i sin \theta_i = n_r sin \theta_r$

where

$n_i, n_r$ are the refractive index of the first and second medium

$\theta_i$ is the angle of incidence (measured between the incident ray and the normal to the surface)

$\theta_r$ is the angle of refraction (measured between the refracted ray and the normal to the surface)

In this problem, light moves into a medium that has lower index of refraction, so

$n_r < n_i$

We can rewrite Snell's law as

$sin \theta_r =\frac{n_i}{n_r}sin \theta_i$

and since

$\frac{n_i}{n_r}>1$

this means that

$sin \theta_r > sin \theta_i$

which implies

$\theta_r > \theta_i$

so, the angle increases.

Also, the speed of light in a medium is given by

$v=\frac{c}{n}$

where c is the speed of light and v the refractive index: we see that the speed is inversely proportional to n, therefore the lower the index of refraction, the higher the speed. So, in this problem, the light will speed up, since it moves into a medium with lower index of refraction.

4 0

3 years ago

An air filter can remove dust particles from air but will reach capacity (saturation) at 50.0 mg. Of air containing 225 µg dust

zubka84 [21]

Answer:

Time to Reach Saturation = 0.0146 day

Explanation:

In order to solve this problem, we first need to calculate the dust filtered by the filter per cubic meter of air:

Filtered Dust per m³ = Dust Particles Entering per m³ - Dust Particles Leaving per m³

Filtered Dust per m³ = 225 μg/m³ - 15 μg/m³

Filtered Dust per m³ = 210 μg/m³ = 210 x 10⁻³ mg/m³

Now, we find volume flow rate of air through filter:

Volume Flow Rate of Air = (400 ft³/min)(0.3048 m/1 ft)³

Volume Flow Rate of Air = 11.33 m³/min

Now, we calculate rate of dust filtered:

Rate of Dust Filtered = (Filtered Dust per m³)(Volume Flow Rate of Dust)

Rate of Dust Filtered = (210 x 10⁻³ mg/m³)(11.33 m³/min)

Rate of Dust Filtered = 2.38 mg/min

Now, for the time required to reach saturation:

Time to Reach Saturation = (Saturation Capacity)/(Rate of Dust Filtered)

Time to Reach Saturation = (50 mg)/(2.38 mg/min)

Time to Reach Saturation = (21.02 min)(1 day/24 h)(1 h/60 min)

<u>Time to Reach Saturation = 0.0146 day</u>

3 0

3 years ago

At a certain location, Earth's magnetic field of 34 µT is horizontal and directed due north. Suppose the net field is zero exact

zheka24 [161]

Answer:

a) I = 13.77 A

b) 0 ° or to the East

Explanation:

Part a

The magnetic field by properties would be 0 at the radius on this case r =8.1 cm.Analyzing the situation the wirde would produce a magnetic field equals in magnitude to the magnetic field on Earth by with the inverse direction.

The formula for the magnetic field due to a wire with current is:

$B = \frac{\mu_0 I}{2 \pi r}$

In order to have a value of 0 for the magnetic field at the radius then we need to have this balance

B (r=8.1) = B (Earth)

Replacing:

$B = \frac{\mu_0 I}{2 \pi r)}= B_{Earth}$

Solving from I, from the last equation we got:

$I = \frac{2 \pi r B_{earth}}{\mu_0}$

$I=\frac{2 \pi 0.081 m (34 x 10^{-6} T)}{4 \pi x 10^{-7} Tm/A}$ = 13.77 A

Part b

We can use the right hand rule for this case.

The magnetic field of the wire would point to the South, because the magnetic field of the earth given points to the North. Based on this the current need's to flow from West to East in order to create a magnetic field pointing to the south, because the current would be perpendicular to the magnetic field created.

8 0

3 years ago