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3 years ago

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Two examples of ways that physicists model the physical world

2 answers:

aliya0001 [1]3 years ago

6 0

They build small-scale replicas . physicists build computer simulations.

AVprozaik [17]3 years ago

4 0

Answering the question, three examples of ways that physicists model the physical world include:

Diagrams
Equations
Computer simulations

Computer simulations refer to the reproduction of behavior of a system with the use of a computer to simulate the results of a mathematical model that is linked with the system.

<h2>Further Explanation</h2>

Computer simulations are a tool that is very useful for mathematical modeling in various natural systems in physics.

Physicists write the rule of our universe using equations. Equations are an aspect of one of the models used by physicists to explain a particular situation.

Physicists use a diagram to convey information. It is a symbolic representation of information with the use of visualization techniques. Diagram provides other means for people to better understand information and process its importance.

Physics predictions about the physical world mainly depend on numerical stimulation, mathematics and empirical observations and Physicists models the physical world based on invariances and laws which are built on mathematical models.

The purpose of physics is to explain or describe everything around us. It described virtually everything around us accurately by deploring the law of physics.

However, physics is considered a natural science because it is based on mathematical analysis which is geared towards findings qualitative physical law for the physical world.

LEARN MORE:

examples of ways that physicists model the physical world brainly.com/question/10869755

KEYWORDS:

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physics
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computer simulations
diagrams
equations

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A cannon is mounted on a cart which sits on the ground, supported by frictionless wheels. The mass of the cannon and cart is 4.6

Kitty [74]

Answer:

The velocity of the launcher after the projectile is launched is -5.011 m/s

Explanation:

Here we have the mass of the cannon and cart, m₁ = 4.65 kg

Velocity of cannon and cart, v₁ = 2.00 m/s

Mass of projectile, m₂ = 50.0 g = 0.05 kg

Velocity of projectile, v₂ = 647 m/s

Velocity of the launcher, v₃ = Required

Mass of cannon and cart, launcher after launching projectile m₃ = 4.65-0.05

= 4.6 kg

Therefore, from the principle of the conservation of linear momentum, we have

Total initial momentum = Total final momentum

m₁ × v₁ = m₂ × v₂ + m₃ × v₃

Substituting gives

4.65 kg × 2.00 m/s = 0.05 kg × 647 m/s + 4.6 kg × v₃

4.65 kg × 2.00 m/s - 0.05 kg × 647 m/s = 4.6 kg × v₃

-23.05 kg·m/s = 4.6 kg × v₃

$v_3 = \frac{-23.05 \, kg\cdot m/s}{4.6 \, kg} = \frac{-461}{92} m/s$

v₃ = -5.011 m/s.

3 0

3 years ago

Read 2 more answers

What magnification will be produced by a lens of power –4.00 D (such as might be used to correct myopia) if an object is held 43

kiruha [24]

Answer:

The magnification is $m = 0.3674$

Explanation:

From the question we are told that

The power of the lens is $P = -4.00 D(dioptre)$

Generally $1 dioptre = 1 \ meter$

The object distance is $u = -43 \ cm$ the negative sign is because the distance is measured in the opposite direction of incident light (i.e away )

Generally the focal length is mathematically represented as

$f = \frac{1}{P}$

=> $f = \frac{1}{4.00 }$

=> $f = 0.25 \ m$

converting to cm

=> $f = 0.25 \ m = 0.25 * 100 = 25 \ cm$

Generally from lens equation we have that

$\frac{1}{f} +\frac{1}{v} -\frac{1}{u}$

=> $\frac{1}{25} +\frac{1}{v} -\frac{1}{-43}$

=> $v = -15.8 \ cm$

Generally the magnification is mathematically represented as

$m = \frac{v}{u}$

=> $m = \frac{- 15.8}{-43}$

=> $m = 0.3674$

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Explanation:

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Answer:

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Answer:

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