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2 years ago

How might a theory relate to a model in science?

2 answers:

8 0

These two things are related but can be independent, because it is possible to... of theories is studied formally in mathematical logic, especially in model theory.

MrRissso [65]2 years ago

5 0

A theory may be a set of statements that are developed through a method of continued abstractions. A theory is aimed toward a generalized statement aimed toward explaining a development. A model, on the opposite hand, maybe a purposeful illustration of reality. As you'll be able to see, each shares common parts in their definitions.

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A long solenoid with 1.65 103 turns per meter and radius 2.00 cm carries an oscillating current I = 6.00 sin 90πt, where I is in

Leno4ka [110]

Answer:

The electric field is $35\cos(90\pi t)\ mV/m$

Explanation:

Given that,

Radius = 2.00 cm

Number of turns per unit length $n= 1.65\times10^{3}$

Current $I = 6.00\sin 90\pi t$

We need to calculate the induced emf

$\epsilon =\mu_{0}nA\dfrac{dI}{dt}$

Where, n = number of turns per unit length

A = area of cross section

$\dfrac{dI}{dt}$ =rate of current

Formula of electric field is defined as,

$E=\dfrac{\epsilon}{2\pi r}$

Where, r = radius

Put the value of emf in equation (I)

$E=\dfrac{\mu_{0}nA\dfrac{dI}{dt}}{2\pi r}$ ....(II)

We need to calculate the rate of current

$I=6.00\sin 90\pi t$ ....(III)

On differentiating equation (III)

$\dfrac{dI}{dt}=90\pi\times6.00\cos(90\pi t)$

Now, put the value of rate of current in equation (II)

$E=\dfrac{4\pi\times10^{-7}\times1.65\times10^{3}\times\pi\times(2.00\times10^{-2})^2\times90\pi\times6.00\cos(90\pi t)}{2\pi\times 2.00\times10^{-2}}$

$E=35\cos(90\pi t)\ mV/m$

Hence, The electric field is $35\cos(90\pi t)\ mV/m$

7 0

2 years ago

Read 2 more answers

To get employees to work longer hours, employers often offer __________ in the form of extra pay.

Dovator [93]

C- Incentives it’s like a reward ~!

6 0

2 years ago

A mass weighting 48 lbs stretches a spring 6 inches. The mass is in a medium that exerts a viscous resistance of 27 lbs when the

Mademuasel [1]

Answer:

$u(t)=0.499ft.e^{-\frac{144.76lb/s}{2(48lb)}t}cos(\omega t)\\\\u(t)=0.499ft.e^{-1.5t}cos(\omega t)$

m = 48lb

b = 144.76lb

Explanation:

The general equation of a damping oscillate motion is given by:

$u(t)=u_oe^{-\frac{b}{2m}t}cos(\omega t-\alpha)$ (1)

uo: initial position

m: mass of the block

b: damping coefficient

w: angular frequency

α: initial phase

a. With the information given in the statement you replace the values of the parameters in (1). But first, you calculate the constant b by using the information about the viscous resistance force:

$|F_{vis}|=bv\\\\b=\frac{|F_{vis}|}{v}\\\\|F_{vis}|=27lbs=27*32.17ft.lb/s^2=868.59ft.lb/s^2\\\\b=\frac{868.59}{6}lb/s=144.76lb/s$

Then, you obtain by replacing in (1):

6in = 0.499 ft

$u(t)=0.499ft.e^{-\frac{144.76lb/s}{2(48lb)}t}cos(\omega t)\\\\u(t)=0.499ft.e^{-1.5t}cos(\omega t)$

mass, m = 48lb

b = 144.76 lb/s

8 0

2 years ago

In a tug of war , two teams pull on the rope with a certain force. If team A pulls on it with a force of 225 N and team B pulls

kherson [118]

Answer:

135 N

Explanation:

Given that:

In a tug of war, team A pulls the rope with a force of 225 N; &

Team B also pulls with a force of 360 N

The magnitude of the resultant force is the difference between the two forces since they are positioned in the opposite direction.

Thus;

The magnitude = 360 N - 225 N

So, magnitude of resultant force on the rope = 135 N

8 0

2 years ago

A curve of radius 55.1 m is banked so that a car of mass 1.6 Mg traveling with uniform speed 61 km/hr can round the curve withou

svp [43]

Answer:

θ = 28°

Explanation:

For this exercise We will use the second law and Newton, let's set a System of horizontal and vertical.

X axis

Fₓ = m a

Nₓ = m a

Where the acceleration is centripetal

a = v² / r

The only force that we must decompose is normal, let's use trigonometry

sin θ = Nₓ / N

cos θ = $N_{y}$ / N

Nₓ = N sin θ

$N_{y}$ = N cos θ

Let's replace

N sin θ = m v² / r

Y Axis

$N_{y}$ - W = 0

N cos θ = mg

Let's divide the two equations of Newton's second law

Sin θ / cos θ = v² / g r

tan θ = v² / g r

θ = tan⁻¹ (v² / g r)

We reduce the speed to the SI system

v = 61 km / h (1000 m / 1 km) (1h / 3600 s) = 16.94 m / s

Let's calculate

θ = tan⁻¹ (16.94 2 / (9.8 55.1)

θ = tan⁻¹ (0.5317)

θ = 28°

3 0

2 years ago