Please help, thank you!

It has been proposed that a spaceship might be propelled in the solar system by radiation pressure, using a large sail made of f

Anestetic [448]

Answer:

A = 8.34 x 10^(5) m²

Explanation:

The intensity of the solar radiation is the average solar power per unit area. Thus,

I = P/4A = P/(4(πr²))

Where;

P is average solar power.

r is it's distance from centre of sun

A is area

Now, The rate at which the Sun emits energy has a standard value of 3.90 × 10^(26) W

Thus;

I = [3.90 × 10^(26)]/(4πr²)

I = [3 x 10^(25)]/r²

Now, the formula for radiation pressure is;

P_rad = 2I/c

Where c is speed of light and has a value of 3 x 10^(8) m/s

Thus,

P_rad = 2([3 x 10^(25)]/r²)/(3 x 10^(8))

P_rad = [2.07 x 10^(17)]/r² N/m²

Also, Radiation pressure on ship; P_rad = F/A

Where Force on ship and A is area.

Thus;

F = P_rad x A

So, F = [2.07 x 10^(17)•A]/r²

Now,

F_grav = GMm/r²

Where;

G is gravitational constant with a value = 6.67 x 10^(-11) Nm²/kg²

M is mass of sun with a value of 1.99 x 10^(30)

m is mass of ship and sail = 1300 kg

Thus, plugging in the relevant values to obtain;

F_grav = (6.67×10^(-11) × 1.99 x 10^(30) × 1300)/r²

F_grav = [17.255 x 10^(22)]/r²

Now, equating F to F_grav, we get;

[2.07 x 10^(17)•A]/r² = [17.255 x 10^(22)]/r²

r² will cancel out to give;

2.07 x 10^(17)•A= [17.255 x 10^(22)]

A = [17.255 x 10^(22)]/2.07 x 10^(17)

A = 8.34 x 10^(5) m²

7 0

3 years ago

Which of the following scientists successfully combined human understanding of forces

Studentka2010 [4]

His name his James Maxwell

4 0

3 years ago

Object A moves with a constant velocity of -10 m/s and object B moves with a constant velocity of 5 m/s. Which object has the la

KonstantinChe [14]

8 0

3 years ago

Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J =

Sloan [31]

Answer:

The current is $I = 8.9 *10^{-5} \ A$

Explanation:

From the question we are told that

The radius is $r = 3.17 \ mm = 3.17 *10^{-3} \ m$

The current density is $J = c\cdot r^2 = 9.00*10^{6} \ A/m^4 \cdot r^2$

The distance we are considering is $r = 0.5 R = 0.001585$

Generally current density is mathematically represented as

$J = \frac{I}{A }$

Where A is the cross-sectional area represented as

$A = \pi r^2$

=> $J = \frac{I}{\pi r^2 }$

=> $I = J * (\pi r^2 )$

Now the change in current per unit length is mathematically evaluated as

$dI = 2 J * \pi r dr$

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows

$I = 2\pi \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr$

$I = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr$

$I = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ] | \left 0.001585} \atop 0}} \right.$

$I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ]$

substituting values

$I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ]$

$I = 8.9 *10^{-5} \ A$

5 0

4 years ago

How does the electron-cloud model describe electrons?

Dominik [7]

Answer:

Yo, you just kind of answered it yourself. The Electron Cloud Model is the informal way of describing an atomic orbital.

Explanation:

The analogy of the cloud of electrons is really describing the groups of electrons orbiting around said atom. Depending on the atom, there will be many or few electrons orbiting around it on all sides which can resemble an all-encompassing cloud.

5 0

3 years ago