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3 years ago

What is an experience from everyday life that appears to support the geocentric model

1 answer:

fomenos3 years ago

3 0

<span>One everyday life experience that seems to support the geocentric model is the rising and setting of the Sun and Moon. The Moon rises and falls because it does revolve around the Earth and so it is easy to assume the same is true for the Sun.</span>

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All atoms of an element have the same number of__But within the same element there may exist atoms that have__numbers of__.Atoms

Elza [17]

Answer:

isotopes

Explanation:

3 0

3 years ago

A circular loop of wire with 10 turns, radius 0.241 m and resistance 0.235 Ohms is connected to a 13.1 V power supply. The magne

SVETLANKA909090 [29]

Given Information:

Resistance of circular loop = R = 0.235 Ω

Radius of circular loop = r = 0.241 m

Number of turns = n = 10

Voltage = V = 13.1 V

Required Information:

Magnetic field = B = ?

Answer:

Magnetic field = 0.00145 T

Explanation:

In a circular loop of wire with n number of turns and radius r and carrying a current I induces a magnetic field B

B = μ₀nI/2r

Where μ₀= 4πx10⁻⁷ is the permeability of free space and current in the loop is given by

I = V/R

I = 13.1/0.235

I = 55.74 A

B = 4πx10⁻⁷*10*55.74/2*0.241

B = 0.00145 T

Therefore, the magnetic field at the center of this circular loop is 0.00145 T

8 0

3 years ago

The height of a projectile t seconds after it is launched straight up in the air is given by f (t )equals negative 16 t squared

velikii [3]

Answer:

$\displaystyle a(5)=-32$

Explanation:

<u>Instant Acceleration</u>

The kinetic magnitudes are usually related as scalar or vector equations. By doing so, we are assuming the acceleration is constant over time. But when the acceleration is variable, the relations are in the form of calculus equations, specifically using derivatives and/or integrals.

Let f(t) be the distance traveled by an object as a function of the time t. The instant speed v(t) is defined as:

$\displaystyle v(t)=\frac{df}{dt}$

And the acceleration is

$\displaystyle a(t)=\frac{dv}{dt}$

Or equivalently

$\displaystyle a(t)=\frac{d^2f}{d^2t}$

The given height of a projectile is

$f(t)=-16t^2 +238t+3$

Let's compute the speed

$\displaystyle v(t)=-32t+238$

And the acceleration

$\displaystyle a(t)=-32$

It's a constant value regardless of the time t, thus

$\boxed{\displaystyle a(5)=-32}$

3 0

3 years ago

It has been suggested that rotating cylinders several miles in length and several miles in diameter be placed in space and used

stepladder [879]

Answer:

the required revolution per hour is 28.6849

Explanation:

Given the data in the question;

we know that the expression for the linear acceleration in terms of angular velocity is;

$a_{c}$ = rω²

ω² = $a_{c}$ / r

ω = √( $a_{c}$ / r )

where r is the radius of the cylinder

ω is the angular velocity

given that; the centripetal acceleration equal to the acceleration of gravity a $a_{c}$ = g = 9.8 m/s²

so, given that, diameter = 4.86 miles = 4.86 × 1609 = 7819.74 m

Radius r = Diameter / 2 = 7819.74 m / 2 = 3909.87 m

so we substitute

ω = √( 9.8 m/s² / 3909.87 m )

ω = √0.002506477 s²

ω = 0.0500647 ≈ 0.05 rad/s

we know that; 1 rad/s = 9.5493 revolution per minute

ω = 0.05 × 9.5493 RPM

ω = 0.478082 RPM

1 rpm = 60 rph

ω = 0.478082 × 60

ω = 28.6849 revolutions per hour

Therefore, the required revolution per hour is 28.6849

7 0

3 years ago

A thin spherical shell of radius R has a total charge +Q uniformly distributed over its surface. Of the following distance r fro

grigory [225]

Answer:

The correct answer is B

Explanation:

Let's calculate the electric field using Gauss's law, which states that the electric field flow is equal to the charge faced by the dielectric permittivity

Φ $._{E}$ = ∫ E. dA = $q_{int}$ / ε₀

For this case we create a Gaussian surface that is a sphere. We can see that the two of the sphere and the field lines from the spherical shell grant in the direction whereby the scalar product is reduced to the ordinary product

∫ E dA = $q_{int}$ / ε₀

The area of a sphere is

A = 4π r²

E 4π r² = $q_{int}$ / ε₀

E = (1 /4πε₀ ) q / r²

Having the solution of the problem let's analyze the points:

A ) r = 3R / 4 = 0.75 R.

In this case there is no charge inside the Gaussian surface therefore the electric field is zero

E = 0

B) r = 5R / 4 = 1.25R

In this case the entire charge is inside the Gaussian surface, the field is

E = (1 /4πε₀ ) Q / (1.25R)²

E = (1 /4πε₀ ) Q / R2 1 / 1.56²

E₀ = (1 /4π ε₀ ) Q / R²

$E_{B}$ = Eo /1.56 ²

$E_{B}$ = 0.41 Eo

C) r = 2R

All charge inside is inside the Gaussian surface

$E_{B}$ =(1 /4π ε₀ ) Q 1/(2R)²

$E_{B}$ = (1 /4π ε₀ ) q/R² 1/4

$E_{B}$ = Eo 1/4

$E_{B}$ = 0.25 Eo

D) False the field changes with distance

The correct answer is B

4 0

3 years ago