Accurate forecasts require careful study of the location, size, movement, and characteristics of large ____

Charging a balloon by rubbing it on wool is an example of

Electronic friction ^.^

3 0

3 years ago

A low orbit satellite at 100 km altitude had a camera that resolved a car location to within 0.3 meter. Find the minimum diamete

NeX [460]

Answer:

Minimum diameter of the camera lens is 22.4 cm

The focal length of the camera's lens is 300cm

Explanation:

y = Resolve distance = 0.3 m

h = Height of satellite = 100 km

λ = Wavelength = 550 nm

Angular resolution

$tan\theta\approx \theta =\frac{y}{h}\\\Rightarrow \theta=\frac{0.3}{100\times 10^3}=3\times 10^{-6}$

From Rayleigh criteria

$sin\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{sin\theta}\\\Rightarrow D=1.22\frac{550\times 10^{-9}}{sin3\times 10^{-6}}=0.2236\ m=22.4\ cm$

Minimum diameter of the camera lens is 22.4 cm

Relation between resolvable feature, focal length and angular resolution

$d=f\Delta \theta\\\Rightarrow f=\frac{d}{\Delta \theta}\\\Rightarrow f=\frac{9\times 10^{-6}}{3\times 10^{-6}}=3\ m=300\ cm$

The focal length of the camera's lens is 300cm

3 0

2 years ago

HELP PLS i need this right now

N76 [4]

Answer:

V = 44.4 units.

Explanation:

In order to solve this problem, we must use the Pythagorean theorem. Which is defined by the following expression.

$V = \sqrt{(V_{x})^{2} +(V_{y})^{2} }$

where:

Vx = - 43 units

Vy = 11.1 units

Now replacing:

$V=\sqrt{(-43)^{2} +(11.1)^{2} }\\V=44.4 units$

7 0

3 years ago

If the moon were 2 times closer to earth than it is now, the gravitational force between earth and the moon would be

il63 [147K]

So much brighter and the moon would be so much darker than it is now because the moon is further away from the moon than it is now...

7 0

3 years ago

An ideal monatomic gas initially has a temperature of 300 K and a pressure of 5.79 atm. It is to expand from volume 420 cm3 to v

maxonik [38]

Answer:

a) The final pressure is 1.68 atm.

b) The work done by the gas is 305.3 J.

Explanation:

a) The final pressure of an isothermal expansion is given by:

$T = \frac{PV}{nR}$

$T_{i} = T_{f}$

$\frac{P_{i}V_{i}}{nR} = \frac{P_{f}V_{f}}{nR}$

Where:

$P_{i}$ : is the initial pressure = 5.79 atm

$P_{f}$ : is the final pressure =?

$V_{i}$ : is the initial volume = 420 cm³

$V_{f}$ : is the final volume = 1450 cm³

n: is the number of moles of the gas

R: is the gas constant

$P_{f} = \frac{P_{i}V_{i}}{V_{f}} = \frac{5.79 atm*420 cm^{3}}{1450 cm^{3}} = 1.68 atm$

Hence, the final pressure is 1.68 atm.

b) The work done by the isothermal expansion is:

$W = P_{i}V_{i}ln(\frac{V_{f}}{V_{i}}) = 5.79 atm*\frac{101325 Pa}{1 atm}*420 cm^{3}*\frac{1 m^{3}}{(100 cm)^{3}}ln(\frac{1450 cm^{3}}{420 cm^{3}}) = 305.3 J$

Therefore, the work done by the gas is 305.3 J.

I hope it helps you!

3 0

3 years ago

Accurate forecasts require careful study of the location, size, movement, and characteristics of large _____.