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3 years ago

What was Copernicus’s model of the solar system?

1 answer:

4 0

The Heliocentric System. In a book called On the Revolutions of the Heavenly Bodies (that was published as Copernicus lay on his deathbed), Copernicus proposed that the Sun, not the Earth, was the center of the Solar System. Such a model is called a heliocentric system.

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How are speed and velocity different?

Law Incorporation [45]

Basically velocity is a vector quantity and is specified in m/s (meters/second). Speed is the distance traveled by an object where as, velocity is distance traveled by an object per unit time in a particular direction. Speed is a scalar quantity where as velocity is a vector quantity

7 0

3 years ago

Read 2 more answers

Plz answer this question

Lady_Fox [76]

The answer will be pressure.

Pressure is force per unit area.

8 0

3 years ago

Calculate the ratio of the resistance of 12.0 m of aluminum wire 2.5 mm in diameter, to 30.0 m of copper wire 1.6 mm in diameter

alukav5142 [94]

Answer: 0.258

Explanation:

The resistance $R$ of a wire is calculated by the following formula:

$R=\rho\frac{l}{s}$ (1)

Where:

$\rho$ is the resistivity of the material the wire is made of. For aluminium is $\rho_{Al}=2.65(10)^{-8}m\Omega$ and for copper is $\rho_{Cu}=1.68(10)^{-8}m\Omega$

$l$ is the length of the wire, which in the case of aluminium is $l_{Al}=12m$ , and in the case of copper is $l_{Cu}=30m$

$s$ is the transversal area of the wire. In this case is a circumference for both wires, so we will use the formula of the area of the circumference:

$s=\pi{(\frac{d}{2})}^{2}$ (2) Where $d$ is the diameter of the circumference.

For aluminium wire the diameter is $d_{Al}=2.5mm=0.0025m$ and for copper is $d_{Cu}=1.6mm=0.0016m$

So, in this problem we have two transversal areas:

For aluminium:

$s_{Al}=\pi{(\frac{d_{AL}}{2})}^{2}=\pi{(\frac{0.0025m}{2})}^{2}$

$s_{Al}=0.000004908m^{2}$ (3)

For copper:

$s_{Cu}=\pi{\frac{(d_{Cu}}{2})}^{2}=\pi{(\frac{0.0016m}{2})}^{2}$

$s_{Cu}=0.00000201m^{2}$ (4)

Now we have to calculate the resistance for each wire:

Aluminium wire:

$R_{Al}=2.65(10)^{-8}m\Omega\frac{12m}{0.000004908m^{2}}$ (5)

$R_{Al}=0.0647\Omega$ (6) Resistance of aluminium wire

Copper wire:

$R_{Cu}=1.68(10)^{-8}m\Omega\frac{30m}{0.00000201m^{2}}$ (6)

$R_{Cu}=0.250\Omega$ (7) Resistance of copper wire

At this point we are able to calculate the ratio of the resistance of both wires:

$Ratio=\frac{R_{Al}}{R_{Cu}}$ (8)

$\frac{R_{Al}}{R_{Cu}}=\frac{0.0647\Omega}{0.250\Omega}$ (9)

Finally:

$\frac{R_{Al}}{R_{Cu}}=0.258$ This is the ratio

3 0

4 years ago

Question #4

olga2289 [7]

Answer:

Distance 5 km, Displacement 3 km east

Explanation:

6 0

3 years ago

Need help!!! Physics due tomorrow help, help ,help

Sergio [31]

1. Due to the first law of Newton: there is a force that accelerate the body,

then the net force for 2 forces in opposite directions can calculated as

$F_{net}=F_{1}-F_{2}=26-8=14N$

2. For a force make an angle with the horizontal plane, we need to analysis the force into 2 perpendicular forces:

- In horizontal plane:

$F_{x}=Fsin ($ θ $)= 67sin(73)=64N$

then, as Q(1),

$F_{net}=F_{1}-F_{2}=64-33=31 N$

3. Because of the non change in the velocity, there is no acceleration

Then, there is no force is acted from the body

Hence all forces is according to the tension in the cable

$F_{Tension}=420 N$

4. A) As we solve the Q(2), we will analysis the un horizontal or vertical forces into 2 forces

$F_{x}=Fsin ($ θ $)= 600sin(45)=424.264N$

$F_{y} = F sin ($ θ $)= 600cos45)=424.264N$

Hence we have 2 opposite horizontal forces. also, 2 opposite vertical forces.

now, we calculate the net forces in each direction

$F_{net in x}=F_{1}-F_{2}=424.264-250=174.264 N$ in east

$F_{net in y}=F_{1}-F_{2}=-500-424.264=75.736 N$ in south

The total net force for perpendicular forces is calculated from the relation:

$F_{total}=\sqrt{F_{x} ^{2}+F_{y}^{2}}=\sqrt{174.264^{2}+75.736^{2}}=190N$

B) To find the direction, we need to calculate the angle which the total force is made with the east direction of the x-axis (horizontal)

θ = $tan^{-1}(\frac{F_{y}}{F_{x} })= tan^{-1}(\frac{-75.736}{174.264})=-23.48^{o}$

That meas that the total force is made with the east direction of the x-axis angle equal to 23.48 for the south (the 4th quarter)

5. According to the third law of Newton, to make the balance the net force equal to zero (for each action there is a reaction with the same amplitude and in opposite direction)

Then, $F_{right}=F_{left}=55N$

7 0

4 years ago