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Evgen [1.6K]
3 years ago
12

Which is affected more by an earthquake, : deep sea or less deep sea???

Physics
2 answers:
lisabon 2012 [21]3 years ago
6 0

Answer:

deep sea will obviously be more affected because of the sea floor shaking

DaniilM [7]3 years ago
3 0

Answer:

deep see

because of the earth inside part highly pressure

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HELP ME! PLEASE! SOMEONE! THIS QUESTION IS VERY HARD! THERE ARE 3 ANSWERS! HELP ME PLEASE! MAKE SURE TO EXPALIN
blondinia [14]

Answer:

search up the kinetic energy and potential energy etc. then take them and look at the characteristica are they the same? What makes them similar? Why are they different ? How? Then add the chemical nuclear and electrical changes it creates. Now the rest! There you’ve got this! If you need support I’m here! Hope this helped!

Explanation:

6 0
3 years ago
A group of college students eager to get to Florida on a spring break drove the 710-mi trip with only minimum stops. They comput
Elis [28]

Answer:

Time taken for trip = 12.74 hour (Approx)

Explanation:

Given:

Distance of trip = 710-mi

Average speed for the trip = 55.7 mi/h

Find:

Time taken for trip = ?

Computation:

⇒ Time = Distance / Speed

⇒ Time taken for trip = Distance of trip / Average speed for the trip

⇒ Time taken for trip = 710-mi / 55.7 mi/h

⇒ Time taken for trip = 12.74 hour (Approx)

7 0
3 years ago
Due to the wave nature of light, light shined on a single slit will produce a diffraction pattern? Green light (520 nm) is shine
TiliK225 [7]

Answer:

Yes, it will produce a diffraction pattern.

a. 3.9 mm b. 1.95 mm

Explanation:

The light shined from a single slit will produce a diffraction pattern because,  the wavefront act as wavelets which generates its own wave according to Huygens principle. This therefore causes the diffraction pattern.

Given

wavelength of green light, λ = 520 nm = 520 × 10⁻⁹ m = 5.20 × 10⁻⁷ m

width of slit, d = 0.440 mm = 0.44 × 10⁻³ m = 4.4 × 10⁻⁴ m

Distance of slit from central maximum , D = 1.65 m

Distance of first minimum from central maximum, y = ?

a. The relationship between the slit width and wavelength is given by [tex} dsinθ = mλ [/tex]where d = slit width, θ = angular distance from central maximum, λ = wavelength of light and m = ±1, ±2, ±3...

The relationship between y and D is given by tanθ = y/D

Since θ is small, sinθ ≈ θ ≈ tanθ

so, dθ = mλ ⇒ θ = mλ/d = y/D

Therefore, y = mλD/d

Now, for the first minimum above the slit, m = +1 and for the first minimum below the slit, m = -1. So, y₁ =  λD/d and y₋₁ =  -λD/d. So, the width of the central maximum Δy is the difference between the first minima below and above the central maximum. So, Δy = y₁ - y₋₁ = λD/d -(-λD/d) = 2λD/d

Substituting the values from above, Δy= 2 × 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴ =  3900 × 10⁻⁶ m = 3.9 × 10⁻³ m = 3.9 mm

b. The first order fringe is the fringe located between the first minimum and the second minimum. From dsinθ = mλ and tanθ = y/D when θ is small, sinθ ≈ θ ≈ tanθ. So, y = mλD/d. Let m= 1 and m=2 be the first and second minima respectively. So,y₁ =  λD/d and y₂ =  2λD/d. The difference Δy₁ = y₂ - y₁ is the width of the first order fringe. Therefore, Δy₁ = 2λD/d - λD/d= λD/d. Substituting the values from above, we have

λD/d= 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴= 1.95 × 10⁻³ m = 1.95 mm

7 0
3 years ago
What is the mass of an object that has a weight of 2867 N?
Gnom [1K]

Answer:

F = M a

W = M g     equivalent equation to express weight of object of mass M

M = W / g = 2867 N / 9.8 m/s^2 = 292.6 kg

7 0
3 years ago
In a certain region of space, a uniform electric field has a magnitude of 4.30 x 104 n/c and points in the positive x direction.
denis23 [38]
The magnetic force exerted by a field E to a charge q is given by F=Eq. In this case, F=4.30*10^4*(6.80mu C). 1mu C=10^-6C, so F=4.30*6.80=10^-2=0.29N. The direction is in the x direction, the direction that the field is applied because the charge is positive.
5 0
4 years ago
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