Ask question

Ask question

All categories

3 years ago

6

The intensity of the Sun's light in the vicinity of the Earth is about 1000W/m^2. Imagine a spacecraft with a mirrored square sa

il of dimension 5.5 km. Estimate how much thrust (in newtons) this craft will experience due to collisions with the Sun's photons. [Hint: assume the photons bounce off the sail with no change in the magnitude of their momentum.}

1 answer:

NeX [460]3 years ago

8 0

Explanation:

Formula to represent thrust is as follows.

F = $\frac{dP}{dt}$

= $\frac{2p}{dt}$

or, p = $\frac{E}{c}$

$\frac{p}{dt} = \frac{W}{c}$

F = $\frac{2IA}{c}$

= $\frac{2 (1000 W/m^{2})(5.5 \times 10^{3} m)^{2}}{3 \times 10^{8} m/s}$

= 201.67 N

Thus, we can conclude that the thrust is 201.67 N.

You might be interested in

While trying out for the position of pitcher on your high school baseball team, you throw a fastball at 87.6 mi/h toward home pl

JulijaS [17]

The ball drops about 1 m.

8 0

2 years ago

The volume of a cylindrical tin can with a top and a bottom is to be 16π cubic inches. If a minimum amount of tin is to be used

sergiy2304 [10]

Answer:

h = 4 in

Explanation:

GIVEN DATA:

volume of tin $= 16 \pi$

we know that

volume of cylinder is $v = \pi r^2 h$

so,

$16 \pi = \pi r^2 h$

$16 = r^2 h$

$r = \sqrt{\frac{16}{h}}$

construct formula for surface area

$S = 2\pi r^2 + 2\pi rh$

$S = \frac{2v}{h} + 2 \sqrt{v \pi h}$

minimize the function wrt h

$S' = \frac{2v}{h^2} + \sqrt{\frac{v \pi}{h} = 0$

solving for h we have

$h = [\frac{4 v}{\pi}]^{1/3}$

we kow $v = 16 \pi$ so

h = 4 in

8 0

2 years ago

A theory was not originally a hypothesis. true or false?

Serhud [2]

False.

Theories are hypotheses that have been shown to accurately and predictably explain results obtained through repeated experimentation, to the point where the hypothesis can be assumed to be true. Then, such a hypothesis would be considered a theory.

8 0

3 years ago

Two parallel-plate capacitors have the same plate area. Capacitor 1 has a plate separation twice that of capacitor 2, and the qu

Luba_88 [7]

Answer:

$V_1=8 V_2$

Explanation:

Given that:

Area of the plate of capacitor 1= Area of the plate of capacitor 2=A

separation distance of capacitor 2, $d_2=d$

separation distance of capacitor 1, $d_1=2d$

quantity of charge on capacitor 2, $Q_2=Q$

quantity of charge on capacitor 1, $Q_1=4Q$

We know that the Capacitance of a parallel plate capacitor is directly proportional to the area and inversely proportional to the distance of separation.

Mathematically given as:

$C=\frac{k.\epsilon_0.A}{d}$ .....................................(1)

where:

k = relative permittivity of the dielectric material between the plates= 1 for air

$\epsilon_0 = 8.85\times 10^{-12}\,F.m^{-1}$

From eq. (1)

For capacitor 2:

$C_2=\frac{k.\epsilon_0.A}{d}$

For capacitor 1:

$C_1=\frac{k.\epsilon_0.A}{2d}$

$C_1=\frac{1}{2} [ \frac{k.\epsilon_0.A}{d}]$

We know, potential differences across a capacitor is given by:

$V=\frac{Q}{C}$ ..........................................(2)

where, Q = charge on the capacitor plates.

for capacitor 2:

$V_2=\frac{Q}{\frac{k.\epsilon_0.A}{d}}$

$V_2=\frac{Q.d}{k.\epsilon_0.A}$

& for capacitor 1:

$V_1=\frac{4Q}{\frac{k.\epsilon_0.A}{2d}}$

$V_1=\frac{4Q\times 2d}{k.\epsilon_0.A}$

$V_1=8\times [\frac{Q.d}{k.\epsilon_0.A}]$

$V_1=8 V_2$

6 0

2 years ago

What is number 14 please tell me

UNO [17]

<span>a thin fibrous cartilage between the surfaces of some joints, e.g., the knee.</span>

3 0

3 years ago