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3 years ago

What happens to beaches over time?

Physics

2 answers:

Rama09 [41]3 years ago

7 0

the answer is d i took the test be cause i go to a k12 onlie school

vitfil [10]3 years ago

3 0

Answer:

Option (D)

Explanation:

The beaches are defined as the landform that is oriented along the side of the large water bodies such as seas and oceans. They are dominantly comprised of sand and other rock particles such as pebbles and gravels.

These beaches are often affected by factors such as, the change in the sea level, that are caused due to the tidal effect, and the occurrence of storms and tsunamis.

The shoreline of the beach shifts towards the inland areas when the sea level rises and the opposite happens when the sea level drops. They are suddenly affected when a storm occurs, eroding the sand particles towards or away from the water body.

Thus, the beaches or shorelines are continuously affected due to the above factors.

Hence, the correct answer is option (D).

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What is your speed (in m/s) if you walk 1000m in 20 minutes? (Hint: how many seconds are in a minute)

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Answer:

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3 years ago

Compare the gravitational acceleration on the following objects compared to the Sun using:

arsen [322]

The gravitational acceleration of White dwarf compared to Sun is 13,675.86.

The gravitational acceleration of Neutron star compared to Sun is 6.79 x 10⁻²⁴.

The gravitational acceleration of Star Betelgeuse compared to Sun is 8.5 x 10¹⁰.

<h3>Mass of the planets</h3>

Mass of sun = 2 x 10³⁰ kg

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Mass of Neutron star = 5.5 x 10¹² kg

Mass of star Betelgeuse = 2.188 x 10³¹ kg

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Radius of sun = 696,340 km

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Radius of Neutron star = 11 km

Radius of star Betelgeuse = 617.1 x 10⁶ km

<h3>Gravitational acceleration of White dwarf compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.765 \times 10^{30}}{2\times 10^{30}} \times [\frac{696,340,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 13,675.86$

<h3>Gravitational acceleration of Neutron star compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{5.5 \times 10^{12}}{2\times 10^{30}} \times [\frac{11,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 6.79\times 10^{-24}$

<h3>Gravitational acceleration of Star Betelgeuse compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.188 \times 10^{31}}{2\times 10^{30}} \times [\frac{617.1 \times 10^9}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 8.5\times 10 ^{10}$

Learn more about acceleration due to gravity here: brainly.com/question/88039

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2 years ago

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3 years ago

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3 years ago

At noon, ship A is 110 km west of ship B. Ship A is sailing east at 20 km/h and ship B is sailing north at 15 km/h. How fast is

Katyanochek1 [597]

Answer:

$4.47\ \text{km/h}$

Explanation:

$\dfrac{da}{dt}$ = Rate at which the distance between A and starting point of B is changing = -20 km/h

$\dfrac{db}{dt}$ = Rate at which the distance of B is changing = 15 km/h

$\dfrac{dc}{dt}$ = Rate at which the distance between A and B is changing

Time after which the rate at which the distance between A and B is changing is 4 hours

Distance covered by A in 4 hours = $20\times 4=80\ \text{km}$

a = Distance remaining to the start point of B = $110-80=30\ \text{km}$

b = Distance covered by B in 4 hours = $15\times 4=60\ \text{km}$

Distance between A and B after 4 hours

$c=\sqrt{a^2+b^2}\\\Rightarrow c=\sqrt{30^2+60^2}\\\Rightarrow c=67.08\ \text{km}$

$c^2=a^2+b^2$

Differentiating with respect to time we get

$c\dfrac{dc}{dt}=a\dfrac{da}{dt}+b\dfrac{db}{dt}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{a\dfrac{da}{dt}+b\dfrac{db}{dt}}{c}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{30\times -20+60\times 15}{67.08}\\\Rightarrow \dfrac{dc}{dt}=4.47\ \text{km/h}$

The rate at which the distance between the ships is changing at 4 PM is $4.47\ \text{km/h}$ .

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2 years ago