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vitfil [10]
3 years ago
6

Research paper on tsunamis (4-5 paragraphs minimum) 3) Tsunamis how they work….evacuation routes, structures built in oceans to

protect, early detection, etc
Expectations:

A. Must be at least 4 full paragraphs. (probably will need to be more)

B. Must have an introductory paragraph that explains what the problem is, and the general overview of the paper.

C. Must Define Problem, including details and how it relates to Irvine

D. Main Body of Paper (probably couple of paragraphs) This is where you will show what you found out in your research. Make sure to include things like current use/techniques, as well as what technological advancements are being researched to improve your specific topic/issue.

E. Conclusion: Short summary of main points. Your Opinion on what is working and why (or what is not working and why)

F. Citations: Must use at LEAST 3 sources. Must site works at end, in either MLA or APA format, AND must site them where used in the paper (also MLA or APA format).

Make sure all work is yours and original. Copying is cheating. Using information from a source without citing is plagiarism.
Physics
1 answer:
elena55 [62]3 years ago
6 0
A must be at least 4 full paragraphs probably will need more
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How will the electrostatic force between two electric charges change if the first charge is doubled and the second charge is onl
Vinvika [58]

Answer:

B) \frac{2}{3}

Explanation:

The electric force between charges can be determined by;

F = \frac{kq_{1} q_{2} }{r^{2} }

Where: F is the force, k is the Coulomb's constant, q_{1} is the value of the first charge, q_{2} is the value of the second charge, r is the distance between the centers of the charges.

Let the original charge be represented by q, so that;

q_{1} = 2q

q_{2} = \frac{q}{3}

So that,

F = q_{1}q_{2} x \frac{k}{r^{2} }

  = 2q x \frac{q}{3} x \frac{k}{r^{2} }

  = \frac{2q^{2} }{3} x \frac{k}{r^{2} }

  = \frac{2}{3} x \frac{kq}{r^{2} }

F = \frac{2}{3} x \frac{kq}{r^{2} }

The electric force between the given charges would change by \frac{2}{3}.

4 0
3 years ago
1) A car is moving to the right with the constant speed of 1.2 m/s. If the car starts
Marina CMI [18]
(100, 108)
Due to
1.2x90=108

100, 108
7 0
3 years ago
Coherent light of wavelength 525 nm passes through two thin slits that are 4.15×10^(−2) mm apart and then falls on a screen 7
IRINA_888 [86]

A) 4.7 cm

The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is

sin \theta=\frac{n \lambda}{d}

where

n is the order of the maximum

\lambda is the wavelength

a is the distance between the slits

In this problem,

n = 5

\lambda=525 nm =5.25\cdot 10^{-7} m

a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m

So we find

\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}

And given the distance of the screen from the slits,

D=75.0 cm = 0.75 m

The distance of the 5th  bright fringe from the central bright fringe will be given by

y=D tan \theta = (0.75 m)tan 3.62^{\circ}=0.047 m = 4.7 cm

B) 8.1 cm

The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}

And the distance of the 8th dark fringe from the central bright fringe will be given by

y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm

5 0
3 years ago
Which statement accurately describes a relationship between parts of the
sammy [17]

<u>Answer:</u>

C. There are trillions of galaxies in the universe.

<u>Explanation:</u>

A. is wrong as nebulae are found inside galaxies and inside the universe, not inside stars.

B. is wrong because there are trillions of galaxies in the universe, not the latter.

D. The solar system consists of the eight planets, the Sun, comets, meteors, dwarf planets, and is inside the Milky Way galaxy and thus cannot have galaxies inside it.

<em>Please give Brainliest</em>

8 0
2 years ago
Read 2 more answers
A ranger in a national park is driving at 56 km/h when a deer jumps onto the road 65 m ahead of the vehicle. After a reaction ti
vagabundo [1.1K]

Answer:

 t = 1.58 s

Explanation:

given,

Speed of ranger, v = 56 km/h

                            v = 56 x 0.278 = 15.57 m/s

distance, d = 65 m

deceleration,a = 3 m/s²

reaction time = ?

using stopping distance formula

d = v. t + \dfrac{v^2}{2a}

t = \dfrac{d}{v} -\dfrac{v}{2a}

t is the reaction time

t = \dfrac{65}{15.57} -\dfrac{15.57}{2\times 3}

 t = 1.58 s

hence, the reaction time of the ranger is equal to 1.58 s.

3 0
3 years ago
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