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3 years ago

Select all that apply. Which of the following astronomers supported the Earth-centered system?

Physics

2 answers:

Vilka [71]3 years ago

8 0

Answer:

Ptolemy and partially Tycho Brahe

Explanation:

Both Ptolemy and Tycho Brahe believed that the earth was the center of the orbit of the sun, the first also of all the other planets and stars, the later believed the sun was in exchange the center of all the other celestial bodies, then Brahe's posture was an intermediary between that of Ptolemy and the heliocentrism(Sun as center of the solar system, where the planets orbit around it) of Copernicus, that proved to be the right observation later with galileo galilei.

Stella [2.4K]3 years ago

6 0

Among the options, the two astronomers who supported the Earth-centered system were Tycho Brahe and Ptolemy.

Both astronomers developed in fact a system where the Sun was located at the center, while the Earth and the other planets were orbiting around the Sun.

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uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is station

Alex787 [66]

Answer:

The magnitude of the tangential velocity is $v= 0.868 m/s$

The magnitude of the resultant acceleration at that point is $a = 4.057 m/s^2$

Explanation:

From the question we are told that

The mass of the uniform disk is $m_d = 40.0kg$

The radius of the uniform disk is $R_d = 0.200m$

The force applied on the disk is $F_d = 30.0N$

Generally the angular speed i mathematically represented as

$w = \sqrt{2 \alpha \theta}$

Where $\theta$ is the angular displacement given from the question as

$\theta = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}$

$=1.257\ rad$

$\alpha$ is the angular acceleration which is mathematically represented as

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

The moment of inertial is mathematically represented as

$I = \frac{1}{2} m_dR^2_d$

Substituting values

$I = 0.5 * 40 * 0.200^2$

$= 0.8kg \cdot m^2$

Considering the equation for angular acceleration

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

Substituting values

$\alph$ $\alpha = \frac{(30.0)(0.200)}{0.8}$

$= 7.5 rad/s^2$

Considering the equation for angular velocity

$w = \sqrt{2 \alpha \theta}$

Substituting values

$w =\sqrt{2 * (7.5) * 1.257}$

$= 4.34 \ rad/s$

The tangential velocity of a given point on the rim is mathematically represented as

$v = R_d w$

Substituting values

$= (0.200)(4.34)$

$v= 0.868 m/s$

The radial acceleration at hat point is mathematically represented as

$\alpha_r = \frac{v^2}{R}$

$= \frac{0.868^2}{0.200^2}$

$= 3.7699 \ m/s^2$

The tangential acceleration at that point is mathematically represented as

$\alpha _t = R \alpha$

Substituting values

$\alpha _t = (0.200) (7.5)$

$= 1.5 m/s^2$

The magnitude of resultant acceleration at that point is

$a = \sqrt{\alpha_r ^2+ \alpha_t^2 }$

Substituting values

$a = \sqrt{(3.7699)^2 + (1.5)^2}$

$a = 4.057 m/s^2$

7 0

3 years ago

A 5-cm-high peg is placed in front of a concave mirror with a radius of curvature of 20 cm.

Andrei [34K]

Answer:

Explanation:

Using the magnification formula.

Magnification = Image distance(v)/object distance(u) = Image Height(H1)/Object Height(H2)

M = v/u = H1/H2

v/u = H1/H2...1

3) Given the radius of curvature of the concave lens R = 20cm

Focal length F = R/2

f = 20/2

f = 10cm

Object distance u = 5cm

Object height H2= 5cm

To get the image distance v, we will use the mirror formula

1/f = 1/u+1/v

1/v = 1/10-1/5

1/v = (1-2)/10

1/v =-1/10

v = -10cm

Using the magnification formula

(10)/5 = H1/5

10 = H1

H1 = 10cm

Image height of the peg is 10cm

4) If u = 15cm

1/v = 1/f-1/u

1/v = 1/10-1/15

1/v = 3-2/30

1/v = 1/30

v = 30cm

30/15 = H1/5

15H1 = 150

H1/= 10cm

5) if u = 20cm

1/v = 1/f-1/u

1/v = 1/10-1/20

1/v = 2-1/20

1/v = 1/20

v = 20cm

20/20 = H1/5

20H1 = 100

H1 = 5cm

6) If u = 30cm

1/v = 1/f-1/u

1/v = 1/10-1/30

1/v = 3-1/30

1/v = 2/30

v = 30/2 cm

v =>15cm

15/30 = Hi/5

30H1 = 75

H1 = 75/30

H1 = 2.5cm

4 0

2 years ago

A particle was moving in a straight line at 172.8 km/hr. If it decelerated over 120 meters to come to rest, find the time taken

Bezzdna [24]

Answer:

v=s/t

s=vt

t=s/v

t=(120×10‐³)/172.8

(the distance meters has been changed to kilometres)

t=1/1440 hrs

Given ,

8 0

2 years ago

Suppose that the height of the incline is h = 14.7 m. Find the speed at the bottom for each of the following objects. 1.solid sp

tensa zangetsu [6.8K]

Answer:

1. 14.4 m/s 2. 13.2 m/s 3. 12.0 m/s 4. 13.9 m/s

Explanation:

Assuming no friction present, the different objects roll without slipping, so there is a constant relationship between linear and angular velocity, as follows:

ω= v/r

If no friction exists, the change in total kinetic energy must be equal in magnitude to the change in the gravitational potential energy:

∆K = -∆U

½ *m*v² + ½* I* ω² = m*g*h

Simplifying and replacing the value of the angular velocity:

½ * v² + ½ I *(v/r)² = g*h (1)

In order to answer the question, we just need to replace h by the value given, and I (moment of inertia) for the value for each different object, as follows:

Solid Sphere I = 2/5* m *r²

Replacing in (1):

½ * v² + ½ (2/5 *m*r²) *(v/r)² = g*h

Replacing by the value given for h, and solving for v:

v = √(10/7*9.8 m/s2*14.7 m) = 14. 4 m/s

Spherical shell I=2/3*m*r²

Replacing in (1):

½ * v² + ½ (2/3 *m*r²) *(v/r)² = g*h

Replacing by the value given for h, and solving for v:

v = √(6/5*9.8 m/s2*14.7 m) = 13.2 m/s

Hoop I= m*r²

Replacing in (1):

½ * v² + ½ (m*r²) *(v/r)² = g*h

Replacing by the value given for h, and solving for v:

v = √(9.8 m/s2*14.7 m) = 12.0 m/s

Cylinder I = 1/2 * m* r²

Replacing in (1):

½ * v² + ½ (1/2 *m*r²) *(v/r)²= g*h

Replacing by the value given for h, and solving for v:

v = 2*√(1/3*9.8 m/s2*14.7 m) = 13.9 m/s

5 0

3 years ago

Every chemical reaction demonstrate what important scientific principle

Korolek [52]

The answer to this question would be the Law of conservation of energy.

In every chemical reaction, 10 atoms of reacted will make 10 atoms as result. The number of atoms produced will be same as the number of ingredients. This will prove that the energy is <span>can neither be created nor destroyed; The energy is just changed into a different shape.</span>

4 0

2 years ago