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3 years ago

6

The older, geocentric model of the solar system placed the____ at the center. Today's heliocentric model places the_

_ at the center of the solar system.

2 answers:

DedPeter [7]3 years ago

3 0

the answer is earth for the first one and sun for the second one

Allushta [10]3 years ago

3 0

Answer:

Earth

Sun

Explanation:

The heliocentric model opposes the geocentric model by placing the sun at the center of the solar system as opposed to the earth as it is with the geocentric model

You might be interested in

Consider a transformer. used to recharge rechargeable flashlight batteries, that has 500 turns in its primary coil, 3 turns in i

Rashid [163]

Answer:

<em>a) 0.72 V</em>

<em>b) 19.2 mA</em>

<em>c) 2.304 Watts</em>

Explanation:

A transformer is used to step-up or step-down voltage and current. It uses the principle of electromagnetic induction. When the primary coil is greater than the secondary coil, the it is a step-down transformer, and when the primary coil is less than the secondary coil, the it is a step-up transformer.

number of primary turns = $N_{p}$ = 500 turns

input voltage = $V_{p}$ = 120 V

number of secondary turns = $N_{s}$ = 3 turns

output voltage = $V_{s}$ = ?

using the equation for a transformer

$\frac{V_{s} }{V_{p} } = \frac{N_{s} }{N_{p} }$

substituting values, we have

$\frac{V_{s} }{120 } = \frac{3 }{500} }$

$500V_{p} = 120*3\\500V_{p} = 360$

$V_{p}$ = 360/500 =<em> 0.72 V</em>

<em></em>

b) by law of energy conservation,

$I_{P}V_{p} = I_{s}V_{s}$

where

$I_{p}$ = input current = ?

$I_{s}$ = output voltage = 3.2 A

$V_{s}$ = output voltage = 0.72 V

$V_{p}$ = input voltage = 120 V

substituting values, we have

120 $I_{p}$ = 3.2 x 0.72

120 $I_{p}$ = 2.304

$I_{p}$ = 2.304/120 = 0.0192 A

= <em>19.2 mA</em>

<em></em>

c) power input = $I_{p} V_{p}$

==> 0.0192 x 120 = <em>2.304 Watts</em>

7 0

3 years ago

A baseball player friend of yours wants to determine his pitching speed. You have him stand on a ledge and throw the ball horizo

Alla [95]

Answer:

$v_{ox}= 19.6\ m/s$

Explanation:

Data provided in the question:

Height above the ground, H= 5.0m

Range of the ball, R= 20 m

Initial horizontal velocity = $v_{ox}$

Initial vertical velocity= $v_{oy}$ (Since ball was thrown horizontally only)

Acceleration acting horizontally, $a_x$ = 0 m/s² [ Since no acceleration acts horizontally) ]

Vertical Acceleration, $a_y$ = 9.8 m/s² (Since only gravity acts on it)

Let 't' be the time taken to reach ground

Therefore, using equations of motion, we have

$H= v_{oy}t+\frac{1}{2}a_yt^2$

$5= (0)t+\frac{1}{2}(9.8)t^2$

$t= \frac{10}{9.8}=1.02 s$

Then using Equations of motion for horizontal motion,

$R= v_{ox}t+\frac{1}{2}a_xt^2$

$20= v_{ox}(1.02)+\frac{1}{2}(0)(1.02)^2$

$v_{ox}= 19.6\ m/s$

4 0

3 years ago

A runner starts from rest and in 3 s reaches a speed of 8 m/s. If we assume that the speed changed at a constant rate (constant

Stells [14]

Answer:

The average speed of the runner is 4 m/s.

Explanation:

Hi there!

The average speed (a.s) is calculated by dividing the traveled distance (d) over the time needed to travel that distance (t):

a.s = d / t

So, let´s find the distance traveled in those 3 s. For that, we can use the equation of position of an object moving in a straight line with constant acceleration:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the object at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

If we place the origin of the frame of reference at the point where the runner starts, then, x0 = 0. Since the runner starts from rest, v0 = 0. So, the equation gets reduced to this:

x = 1/2 · a · t²

We have the time (3 s), so let´s find the acceleration. For that, we can use the equation of velocity of an object moving in a straight line with constant acceleration:

v = v0 + a · t

Where "v" is the velocity at a time "t". Since v0 = 0, then:

v = a · t

At t = 3 s, v = 8 m/s

8 m/s = a · 3 s

8/3 m/s² = a

So let´s find the position of the runner at t = 3 s (In this case, the position of the runner will be equal to the traveled distance):

x = 1/2 · a · t²

x = 1/2 · 8/3 m/s² · (3 s)²

x = 12 m

Now, we can calcualte the average speed:

a.s = d/t

a.s = 12 m / 3 s

a.s = 4 m/s

The average speed of the runner is 4 m/s.

4 0

3 years ago

nekit [7.7K]

Answer:

The image is formed at a ‘distance of 16.66 cm’ away from the lens as a diminished image of height 3.332 cm. The image formed is a real image.

Solution:

The given quantities are

Height of the object h = 5 cm

Object distance u = -25 cm

Focal length f = 10 cm

The object distance is the distance between the object position and the lens position. In order to find the position, size and nature of the image formed, we need to find the ‘image distance’ and ‘image height’.

The image distance is the distance between the position of convex lens and the position where the image is formed.

We know that the ‘focal length’ of a convex lens can be found using the below formula

1f=1v−1u\frac{1}{f}=\frac{1}{v}-\frac{1}{u}

f

1

=

v

1

−

u

1

Here f is the focal length, v is the image distance which is known to us and u is the object distance.

The image height can be derived from the magnification equation, we know that

Magnification=h′h=vu\text {Magnification}=\frac{h^{\prime}}{h}=\frac{v}{u}Magnification=

h

h

′

=

u

v

Thus,

h′h=vu\frac{h^{\prime}}{h}=\frac{v}{u}

h

h

′

=

u

v

First consider the focal length equation to find the image distance and then we can find the image height from magnification relation. So,

1f=1v−1(−25)\frac{1}{f}=\frac{1}{v}-\frac{1}{(-25)}

f

1

=

v

1

−

(−25)

1

1v=1f+1(−25)=110−125\frac{1}{v}=\frac{1}{f}+\frac{1}{(-25)}=\frac{1}{10}-\frac{1}{25}

v

1

=

f

1

+

(−25)

1

=

10

1

−

25

1

1v=25−10250=15250\frac{1}{v}=\frac{25-10}{250}=\frac{15}{250}

v

1

=

250

25−10

=

250

15

v=25015=503=16.66 cmv=\frac{250}{15}=\frac{50}{3}=16.66\ \mathrm{cm}v=

15

250

=

3

50

=16.66 cm

Then using the magnification relation, we can get the image height as follows

h′5=−16.6625\frac{h^{\prime}}{5}=-\frac{16.66}{25}

5

h

′

=−

25

16.66

So, the image height will be

h′=−5×16.6625=−3.332 cmh^{\prime}=-5 \times \frac{16.66}{25}=-3.332\ \mathrm{cm}h

′

=−5×

25

16.66

=−3.332 cm

Thus the image is formed at a distance of 16.66 cm away from the lens as a diminished image of height 3.332 cm. The image formed is a ‘real image’.

5 0

3 years ago

How to find acceleration in non uniform motion. Please give two ways

Olenka [21]

<u>The two ways to find acceleration in non uniform motion are as follows:</u>

By Graphs
By Calculus

<u>Explanation:</u>

Non-uniform acceleration comprises the most common description of motion. Acceleration refers to the rate of changes of velocity per unit time. Basically, it implies that acceleration changes during motion. This variety can be communicated either as far as position (x) or time (t).

Accordingly, non-uniform acceleration motion can be carried out in 2 ways:

By Calculus
By graphs

Calculus analysis is general and accurate, but limited to the availability of speed and acceleration expressions. It is not always possible to get the expression of motion attributes in the form "x" or "t". On the other hand, the graphic method is not accurate enough, but it can be used accurately if the graphic has the correct shapes.

The use of calculations involves differentiation and integration. Integration enables evaluation of the expression of acceleration of speed and expression of movement at a distance. Similarly, differentiation allows us to evaluate expression of speed position and expression speed to acceleration.

6 0

3 years ago