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il63 [147K]
3 years ago
9

Consider the hypothetical observation "a planet beyond Saturn rises in west, sets in east." This observation is not consistent w

ith a Sun-centered model, because in this model __________.
a. there are no planets beyond Saturn
b. all objects in space must orbit the Sun in the same direction
c. planets beyond Saturn must orbit the Sun more slowly than closer-in planets
d. the rise and set of all objects depends only on Earth’s rotation
Physics
1 answer:
pickupchik [31]3 years ago
5 0

This observation is not consistent with a Sun-centered model, because in this model "the rise and set of all objects depends only on Earth’s rotation".

<u>Answer:</u> Option D

<u>Explanation:</u>

Copernican heliocentrism, the astronomical model's title which was created by Nicolaus Copernicus and released in 1543. This model placed the Sun in motionless position near the Universe's center, with Earth and the planets of other which orbiting it in a circular orbits, altered by epicycles, and also at constant speeds.

Out of several planets in a determined order the Earth orbits around a stationary sun having three motions like annual revolution, daily rotation, and its axis is tilted annually. The planets' retrograde motion is understood by motion from the Earth.

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When a 25000-kg fighter airplane lands on the deck of the aircraft carrier, the carrier sinks 0.23cm deeper into the water.
Genrish500 [490]

Answer:

10604 square meters

Explanation:

0.23 cm = 0.0023m

Assume the carrier has a shape of a rectangular box. When the carrier sinks 0.0023m deeper into water, the extra volume submerged into water is the same as the extra water volume being replaced. This extra volume would add an additional buoyant force to counter balance the extra weight created by the 25000 kg fighter.

With g being constant, the mass of the extra water displaced is the same as the mass of the fighter airplane m_f. And mass of the water displaced is its volume V times its density \rho

V\rho = m_f

1025V = 25000

V = 25000/1025 = 24.39 m^3

We assume the carrier has a shape of a rectangular box, so this extra displaced volume is the extra depth d = 0.0023 m times cross-section area A

V = dA

24.39 = 0.0023A

A = 24.39 / 0.0023 = 10604 m^2

4 0
4 years ago
A skier starts from rest at the top of a hill that is inclined 10.5° with respect to the horizontal. The hillside is 200 m long,
Svetlanka [38]

Answer:

d) 289.31 m

Explanation:

Energy provided by potential energy = mgh = m x 9.8x 200 sin10.5 = 357.18m

Energy used by friction = μmgcos 10.5 x 200 = .075 x m x 9.8 x cos 10.5 x200 = 144.54 m .

Energy used by friction on plain surface = μmg x d.( dis distance covered on plain ) =.075x m x 9.8 xd = .735 m d

To equate

357.18 m -144.54 m = .735 m d

d = 289.31 m .

4 0
3 years ago
¿Cuál es la frecuencia de una ola con una velocidad de 14 m / s y una longitud de onda de 20 metros?
Margaret [11]

Responder:

<h2>0.7Hertz </h2>

Explicación:

Usando la fórmula para calcular la velocidad de onda que se expresa como se muestra.

Velocidad de una onda = frecuencia * longitud de onda

v = fλ

Dada la velocidad de onda = 14 m / sy longitud de onda = 20 metros

frecuencia f = v / λ

f = 14/20

f = 0.7Hertz

La frecuencia de la onda es de 0.7 Hertz.

7 0
3 years ago
3.00 textbook rests on a frictionless, horizontal tabletop surface. A cord attached to the book passes over a pulley whose diame
sammy [17]

Answer:

a1 = 3.56 m/s²

Explanation:

We are given;

Mass of book on horizontal surface; m1 = 3 kg

Mass of hanging book; m2 = 4 kg

Diameter of pulley; D = 0.15 m

Radius of pulley; r = D/2 = 0.15/2 = 0.075 m

Change in displacement; Δx = Δy = 1 m

Time; t = 0.75

I've drawn a free body diagram to depict this question.

Since we want to find the tension of the cord on 3.00 kg book, it means we are looking for T1 as depicted in the FBD attached. T1 is calculated from taking moments about the x-axis to give;

ΣF_x = T1 = m1 × a1

a1 is acceleration and can be calculated from Newton's 2nd equation of motion.

s = ut + ½at²

our s is now Δx and a1 is a.

Thus;

Δx = ut + ½a1(t²)

u is initial velocity and equal to zero because the 3 kg book was at rest initially.

Thus, plugging in the relevant values;

1 = 0 + ½a1(0.75²)

Multiply through by 2;

2 = 0.75²a1

a1 = 2/0.75²

a1 = 3.56 m/s²

6 0
3 years ago
A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15
Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

T-mg = m\frac{v^2}{R}

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

m \frac{v^2}{R} is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

T=m\frac{v^2}{R}

And by substituting the numerical values, we find

T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

T+mg=m\frac{v^2}{R}

And substituting the numerical values and re-arranging it, we find

T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

T=0\\mg = m\frac{v^2}{R}

and re-arranging the equation, we find

v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s

7 0
3 years ago
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