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3 years ago

Describe a solar eclipse. Be sure to include the positions of the sun, moon, and earth.

1 answer:

7 0

It’s the type of eclipse that occurred when the moon passes between the sun and earth, and when the moon fully or partially blocks the sun.

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A puddle of water has a thin film of gasoline floating on it. A beam of light is shining perpendicular on the film. If the wavel

erastovalidia [21]

Answer:

Explanation:

To solve the problem it is necessary to apply the concepts of Destructive and constructive interference. The constructive interference in tin film is given by

$2t = (m+\frac{1}{2})\frac{\lambda}{n}$

Where,

t = thickness

$\lambda=$ Wavelenght

m= is an integer

n= film/refractive index

We use this equaton because phase change is only present for gasoline air interface, but not at the gasoline-water interface. The minimum t only would be when the value of m=0 then

$2nt = \frac{\lambda}{2}$

$t = \frac{560nm}{4*1-4}$

$t = 100nm$

Therefore the correct answer is D. The minimum thickness of the film to see ab right reflection is 100nm

4 0

3 years ago

The grant that considered the foundation of financial aid is the:

navik [9.2K]

Answer:

I think it is the Federal Pell Grant Program.

Explanation:

4 0

3 years ago

Read 2 more answers

An object of irregular shape has a characteristic length of L = 0.5 m and is maintained at a uniform surface temperature of Ts =

goblinko [34]

Answer:

The value of the average convection coefficient is 20 W/Km².

Explanation:

Given that,

For first object,

Characteristic length = 0.5 m

Surface temperature = 400 K

Atmospheric temperature = 300 K

Velocity = 25 m/s

Air velocity = 5 m/s

Characteristic length of second object = 2.5 m

We have same shape and density of both objects so the reynold number will be same,

We need to calculate the value of the average convection coefficient

Using formula of reynold number for both objects

$R_{1}=R_{2}$

$\dfrac{u_{1}L_{1}}{\eta_{1}}=\dfrac{u_{2}L_{2}}{\eta_{2}}$

$\dfrac{h_{1}L_{1}}{k_{1}}=\dfrac{h_{2}L_{2}}{k_{2}}$

Here, $k_{1}=k_{2}$

$h_{2}=h_{1}\times\dfrac{L_{1}}{L_{2}}$

$h_{2}=\dfrac{q}{T_{2}-T_{1}}\times\dfrac{L_{1}}{L_{2}}$

Put the value into the formula

$h_{2}=\dfrac{10000}{400-300}\times\dfrac{0.5}{2.5}$

$h_{2}=20\ W/Km^2$

Hence, The value of the average convection coefficient is 20 W/Km².

7 0

3 years ago

A ship 1200m off shore fires a gun. how long after the gun is fired will it be heard on the shore?

ryzh [129]

Answer:

We know that the speed of sound is 343 m/s in air

we are also given the distance of the boat from the shore

From the provided data, we can easily find the time taken by the sound to reach the shore using the second equation of motion

s = ut + 1/2 at²

since the acceleration of sound is 0:

s = ut + 1/2 (0)t²

s = ut (here, u is the speed of sound , s is the distance travelled and t is the time taken)

Replacing the variables in the equation with the values we know

1200 = 343 * t

t = 1200 / 343

t = 3.5 seconds (approx)

Therefore, the sound of the gun will be heard at the shore, 3.5 seconds after being fired

6 0

3 years ago

Three identical very dense masses of 3500 kg each are placed on the x axis. One mass is at x1 = -100 cm , one is at the origin,

larisa86 [58]

Answer:

A) 7.37 x 10^-4 N

B) The resultant force will be towards the -x axis

Explanation:

The three masses have mass = 3500 kg

For the force of attraction between the mass at the origin and the mass -100 cm away:

distance r = 100 cm = 1 m

gravitational constant G= 6.67×10^−11 N⋅m^2/kg^2

Gravitational force of attraction $F_{g}$ = $\frac{Gm^{2} }{r^{2} }$

where G is the gravitational constant

m is the mass of each of the masses

r is the distance apart = 1 m

substituting, we have

$F_{g}$ = $\frac{6.67*10^{-11}*3500^{2} }{1^{2} }$ = 8.17 x 10^-4 N

For the force of attraction between the mass at the origin and the mass 320 cm away

distance r = 320 cm = 3.2 m

$F_{g}$ = $\frac{Gm^{2} }{r^{2} }$

substituting, we have

$F_{g}$ = $\frac{6.67*10^{-11}*3500^{2} }{3.2^{2} }$ = 7.98 x 10^-5 N

Resultant force = (8.17 x 10^-4 N) - (7.98 x 10^-5 N) = 7.37 x 10^-4 N

B) The resultant force will be towards the -x axis

7 0

3 years ago