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3 years ago

What is the difference between a low tide and a high tide

Physics

2 answers:

Savatey [412]3 years ago

6 0

A high tide means when the water has risen and is higher up(closer to high up land). Low tide is when it’s receded

Oxana [17]3 years ago

5 0

Answer:

High water level during a tide is called High tide.

Low water level during a tide is called Low tide.

You might be interested in

When voltage sources are connected in series, the total voltage is equal to the algebraic sum of the individual voltages?

Vesnalui [34]

According to the third rule of series circuit, the total voltage is equal to the sum of individual voltages. These proves that the statement above "When voltage sources are connected in series, the total voltage is equal to the algebraic sum of the individual voltages?" is TRUE.

7 0

4 years ago

An inner city revitalization zone is a rectangle that is twice as long as it is wide. The width of the region is growing at a ra

lukranit [14]

Answer:

$\frac{dA}{dt} = 28800 \ m^2/year$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Geometry

Area of a Rectangle: A = lw

Algebra I

Exponential Property: $w^n \cdot w^m = w^{n + m}$

Calculus

Derivatives

Differentiating with respect to time

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Explanation:

Step 1: Define

Area is A = lw

2w = l

w = 300 m

$\frac{dw}{dt} = 24 \ m/year$

Step 2: Rewrite Equation

Substitute in l: A = (2w)w
Multiply: A = 2w²

Step 3: Differentiate

Differentiate the new area formula with respect to time.

Differentiate [Basic Power Rule]: $\frac{dA}{dt} = 2 \cdot 2w^{2-1}\frac{dw}{dt}$
Simplify: $\frac{dA}{dt} = 4w\frac{dw}{dt}$

Step 4: Find Rate

Use defined variables

Substitute: $\frac{dA}{dt} = 4(300 \ m)(24 \ m/year)$
Multiply: $\frac{dA}{dt} = (1200 \ m)(24 \ m/year)$
Multiply: $\frac{dA}{dt} = 28800 \ m^2/year$

3 0

3 years ago

Read 2 more answers

PLZ! PLZ! PLZ! HELP! WILL GIVE BRAINLIEST! Scientific Claim Engaging in scientific argument is a critical piece to the applicati

Nastasia [14]

Answer:

Explanation:

1The study of science and engineering should produce a sense of the process of argument necessary for advancing and defending a new idea or an explanation of a phenomenon and the norms for conducting such arguments. In that spirit, students should argue for the explanations they construct, defend their interpretations of the associated data, and advocate for the designs they propose. (NRC Framework, 2012, p. 73)

Argumentation is a process for reaching agreements about explanations and design solutions. In science, reasoning and argument based on evidence are essential in identifying the best explanation for a natural phenomenon. In engineering, reasoning and argument are needed to identify the best solution to a design problem. Student engagement in scientific argumentation is critical if students are to understand the culture in which scientists live, and how to apply science and engineering for the benefit of society. As such, argument is a process based on evidence and reasoning that leads to explanations acceptable by the scientific community and design solutions acceptable by the engineering community.

Argument in science goes beyond reaching agreements in explanations and design solutions. Whether investigating a phenomenon, testing a design, or constructing a model to provide a mechanism for an explanation, students are expected to use argumentation to listen to, compare, and evaluate competing ideas and methods based on their merits. Scientists and engineers engage in argumentation when investigating a phenomenon, testing a design solution, resolving questions about measurements, building data models, and using evidence to evaluate claims.

Compare and critique two arguments on the same topic and analyze whether they emphasize similar or different evidence and/or interpretations of facts.

Respectfully provide and receive critiques about one’s explanations, procedures, models and questions by citing relevant evidence and posing and responding to questions that elicit pertinent elaboration and detail.

Construct, use, and/or present an oral and written argument supported by empirical evidence and scientific reasoning to support or refute an explanation or a model for a phenomenon or a solution to a problem.

Make an oral or written argument that supports or refutes the advertised performance of a device, process, or system, based on empirical evidence concerning whether or not the technology meets relevant criteria and constraints.

Evaluate competing design solutions based on jointly developed and agreed-upon design criteria.

8 0

3 years ago

The National Aeronautics and Space Administration (NASA) studies the physiological effects of large accelerations on astronauts.

m_a_m_a [10]

Answer:

$v = 23.95 m/s$

Explanation:

As we know that when astronaut is revolving in circular path then the acceleration of the astronaut is due to centripetal acceleration

so it is given as

$a_c = \frac{v^2}{R}$

here we know that

$a_c = 4.50 g$

also we know that

$R = 13 m$

now we have

$4.50 \times 9.81 = \frac{v^2}{13}$

$v = 23.95 m/s$

3 0

4 years ago

Please help me

qaws [65]

-- We know that the y-component of acceleration is the derivative of the
y-component of velocity.

-- We know that the y-component of velocity is the derivative of the
y-component of position.

-- We're given the y-component of position as a function of time.

So, finding the velocity and acceleration is simply a matter of differentiating
the position function ... twice.

Now, the position function may look big and ugly in the picture. But with the
exception of 't' , everything else in the formula is constants, so we don't even
need any fancy processes of differentiation. The toughest part of this is going
to be trying to write it out, given the text-formatting capabilities of the wonderful
envelope-pushing website we're working on here.

From the picture . . . . . y (t) = (1/2) (a₀ - g) t² - (a₀ / 30t₀⁴ ) t⁶

First derivative . . . y' (t) = (a₀ - g) t - 6 (a₀ / 30t₀⁴ ) t⁵ = (a₀ - g) t - (a₀ / 5t₀⁴ ) t⁵

There's your velocity . . . /\ .

Second derivative . . . y'' (t) = (a₀ - g) - 5 (a₀ / 5t₀⁴ ) t⁴ = (a₀ - g) - (a₀ /t₀⁴ ) t⁴

and there's your acceleration . . . /\ .
That's the one you're supposed to graph.

a₀ is the acceleration due to the model rocket engine thrust
combined with the mass of the model rocket
'g' is the acceleration of gravity ... 9.8 m/s² or 32.2 ft/sec²
t₀ is how long the model rocket engine burns

Pick, or look up, some reasonable figures for a₀ and t₀
and you're in business.

The big name in model rocketry is Estes. Their website will give you
all the real numbers for thrust and burn-time of their engines, if you
want to follow it that far.

6 0

3 years ago