When light passes the

The space program contributes a great deal to enrich the lives of Floridians and beyond. One way that NASA brings an income to F

Ivan

Answer: Through tourism.

Explanation:

Since NASA space centre is located in Florida where NASA launches spaceflight, this is a huge tourist attraction for viewers in the United States and world entirely.

Florida generates income through people coming to view the launch of the spaceflight as tourism.

The tourists help in many ways like lodging into hotels, buying one or two things, taking transport from one place to another which will eventually help Florida economy.

6 0

3 years ago

Determine the slit width that produces a diffraction pattern with the 2nd dark fringe at 6.2mm from the central fringe. The scre

Elanso [62]

Answer:

d= 0.242 mm

Explanation:

Slit width (d ) = ?

Screen distance ( D ) = 1.25 m

Wave length of light λ = 600 nm

Distance of n the dark fringe from centre

= n λ D / d

Here n = 2

so

$6.2\times10^{-3}=\frac{2\times600\times10^{-9}}{d}$

$d=\frac{1500\times10^{-6}}{6.2}$

d= 0.242 mm

4 0

3 years ago

a solid metal sphere of radius 3.00m carries a total charge of -5.50. what is the magnitude of the electric field at a distance

aivan3 [116]

Answer:

(a) Electric field at 0.250 m is zero.

(b) Electric field at 2.90 m is zero.

(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.

(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.

Explanation:

Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.

Electric field inside and outside the metal sphere is :

E = 0 for r ≤ R ( inside )

= $\frac{KQ}{r^{2} }$ for r > R ( outside )

Here K is electric constant and r is the distance from the center of the metal sphere.

(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.

(b) Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.

(c) Electric field at 3.10 m is given by the relation as r > R :

E = $\frac{KQ}{r^{2} }$

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.

E = - $\frac{9\times10^{9}\times5.50\times10^{-6} }{3.10^{2} }$

E = - 5.15 x 10³ V/m

(d) Electric field at 8.00 m is given by the relation as r > R :

E = $\frac{KQ}{r^{2} }$

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.

E = - $\frac{9\times10^{9}\times5.50\times10^{-6} }{8^{2} }$

E = - 0.77 x 10³ V/m

8 0

3 years ago

If the wave represents a sound wave, explain how increasing amplitude will affect the loudness of the sound? If we decrease the

Viktor [21]

Answer:

Explanation:

Think of a sound wave like a wave on the ocean, or lake... It's not really water moving, as much as it's energy moving through the water. Ever see something floating on the water, and notice that it doesn't come in with the wave, but rides over the top and back down into the trough between them? Sound waves are very similar to that. If you looked at a subwoofer speaker being driven at say... 50 cycles a second, you'd actually be able to see the speaker cone moving back and forth. The more power you feed into the speaker, the more it moves back and forth, not more quickly, as that would be a higher frequency, but further in and further out, still at 50 cycles per second. Every time it pushed out, it's compressing the air in front of it... the compressed air moves away from the speaker's cone, but not as a breeze or wind, but as a wave through the air, similar to a wave on the ocean

More power, more amplitude, bigger "wave", louder ( to the human ear) sound.

If you had a big speaker ( subwoofer ) and ran a low frequency signal with enough power in it, you could hold a piece of paper in front of it, and see the piece of paper move in and out at exactly the same frequency as the speaker cone. The farther away from the speaker you got, the less it'd move as the energy of the sound wave dispersed through the room.

Sound is a wave

We hear because our eardrums resonates with this wave I.e. our ear drums will vibrate with the same frequency and amplitude. which is converted to an electrical signal and processed by our brain.

By increasing the amplitude our eardrums also vibrate with a higher amplitude which we experience as a louder sound.

Of course when this amplitude is too high the resulting resonance tears our eardrums so that they can't resonate with the sound wave I.e. we become deaf

6 0

2 years ago

Why echo sound cannot be heard in small room

Art [367]

They can't hear an echo in small room because in it the sound can't be reflected back. For an echo of a sound to be heard,the minimum distance between the source of sound and the walls of the roomshould be 17.2 m. Obviously,in asmall room echoes cannot beheard.

7 0

3 years ago