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3 years ago

Tides are a result of

Physics

2 answers:

Virty [35]3 years ago

6 0

The answer should be D but if its not then im sorrry, idk

Mrrafil [7]3 years ago

6 0

Tides are a result of <u>"the gravitational attraction of the Sun and Moon".</u>

An especially high tide (Spring tide) happens when the sun and moon are in line and both pull in a similar way.

Tides may appear to be straightforward at first glance, yet the intricate details of tides jumbled extraordinary logical masterminds for a considerable length of time they even driven Galileo astray into a bunk theory.

Today individuals realize that the gravitational pulls between the earth, moon and sun manage the tides. The moon, however, impacts tides the most.

The moon's gravitational pull on the earth is solid enough to pull the seas into bulge. On the off chance that no different powers were impacting everything, shores would encounter one high tide a day as the earth pivoted on its hub and coasts kept running into the seas' bulge confronting the moon.

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Solve the rational and radical equation

yawa3891 [41]

2v + 10 = 11 - v
3v = 1
v = 1/3

3 0

3 years ago

Read 2 more answers

While practicing the trumpet you notice that every time you play a particular note a window in the room rattles. How can you exp

kherson [118]

The sound waves travel because of the pitch of the note so the window rattles.

7 0

3 years ago

A steel bar that is at 10 ° c is 5 meters long, a bar for heated to 120 ° c, how long is that bar? Α = 1.2.10- ° c

svet-max [94.6K]

Answer:

1) 5.0066 m

2A) β = 3×10⁻⁷ / °C

2B) 2500.045 cm²

3A) γ = 8.1×10⁻⁵ / °C

3B) 1618.144 cm³

Explanation:

1) Linear thermal expansion is:

ΔL = α L₀ ΔT

where ΔL is the change in length,

α is the linear thermal expansion coefficient,

L₀ is the original length,

and ΔT is the change in temperature.

Given L₀ = 5 m, ΔT = 110°C, and α = 1.2×10⁻⁵ / °C:

ΔL = (1.2×10⁻⁵ / °C) (5 m) (110°C)

ΔL = 0.0066 m

The length increases by , so the new length is:

L = L₀ + ΔL

L = 5 m + 0.0066 m

L = 5.0066 m

2A) The surface expansion coefficient is:

β = 2α

β = 2 (1.5×10⁻⁷ / °C)

β = 3×10⁻⁷ / °C

2B) The change in area is:

ΔA = β A₀ ΔT

ΔA = (3×10⁻⁷ / °C) (50 cm × 50 cm) (60°C)

ΔA = 0.045 cm²

So the new area is:

A = A + ΔA

A = 2500 cm² + 0.045 cm²

A = 2500.045 cm²

3A) The volumetric expansion coefficient is:

γ = 3α

γ = 3 (2.7×10⁻⁵ / °C)

γ = 8.1×10⁻⁵ / °C

3B) The change in volume is:

ΔV = γ V₀ ΔT

ΔV = (8.1×10⁻⁵ / °C) (1600 cm³) (140°C)

ΔV = 18.144 cm³

So the new area is:

V = V + ΔV

V = 1600 cm³ + 18.144 cm³

V = 1618.144 cm³

6 0

3 years ago

In 2000, a 20-year-old astronaut left Earth to explore the galaxy; her spaceship travels at 2.5 x 10^8 m/s. She returns in 2040.

lina2011 [118]

Answer:

42.11 years old

Explanation:

Given that:

In 2000, a 20-year-old astronaut left Earth to explore the galaxy; her spaceship travels at 2.5 x 10^8 m/s. She returns in 2040

To find her age we use:

$\Delta t_m=\frac{\Delta t_s}{\sqrt{1-\frac{v^2}{c^2} } }\\$

Δtm is time interval for the observer stationary relative to the sequence of

events = 2040 - 2000 = 40 years

Δts is is the time interval for an observer moving with a speed v relative to the sequence of event

v = velocity = 2.5 x 10^8 m/s

c = speed of light = 3 x 10^8 m/s

$\Delta t_s=\Delta t_m}{\sqrt{1-\frac{v^2}{c^2} } }\\\Delta t_s=40\sqrt{1-\frac{(2.5*10^8)^2}{(3*10^8)^2}}\\\Delta t_s=22.11\ yr$

Here age in 2000 is 20 year, therefore when she appear she would be 20 year + 22.11 year = 42.11 years old

7 0

3 years ago

Based on the information presented in the graph, what is the velocity of the object?

Sladkaya [172]

Answer:

3 m/s

Explanation:

<u>The velocity of a position-time graph is the slope of the line.</u> Slope is rise over run, or rise divided by run. The rise (how many units the line goes up) is 3 units and the run is 1 unit. 3/1 is 3, so the velocity is 3 m/s.

7 0

3 years ago