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3 years ago

The model of the universe that suggests that the sun is the center of the universe was first brought by

2 answers:

5 0

I think the correct answer from the choices listed above is option D. The model of the universe that suggests that the sun is the center of the universe was first brought by Copernicus. His model is known as the "Sun centered model".

arlik [135]3 years ago

4 0

Copernicus is the right answer. On page 125 of your study material explains that Copernicus put a rotating earth in a sun-centered model. The rotation of earth was able to account for the rising and setting of stars. The orbital motion of the earth and moon also accounted for the motion of the sun and moon with respect to the stars. This was easier to understand but encountered scrutiny due to its differences from religious teachings.

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A spring is 6.0cm long when it is not stretched, and 10cm long when a 7.0N force is applied. What force is needed to make it 20c

Artist 52 [7]

Answer:

Approximately $25\; {\rm N}$ (assuming that this spring is ideal.)

Explanation:

The displacement of a spring is the new length of the spring relative to the original length.

For example:

When the $6.0\; {\rm cm}$ -spring in this question is stretched to $10\; {\rm cm}$ , the displacement is $x = (10\; {\rm cm} - 6.0\; {\rm cm})$ .
Likewise, if this spring is stretched to $20\; {\rm cm}$ , the displacement would be $(20\; {\rm cm} - 6\; {\rm cm})$ .

If this spring is ideal, the force on the spring would be proportional to the displacement of the spring. In other words, if a force of $F_{\text{a}}$ displaces this spring by $x_{\text{a}}$ , while a force of $F_{\text{b}}$ displaces this spring by $x_{\text{b}}$ , then:

$\displaystyle \frac{F_{\text{a}}}{x_{\text{a}}} = \frac{F_{\text{b}}}{x_{\text{b}}}$ .

In this question, it is given that a force of $F_{\text{a}} = 7.0 \; {\rm N}$ would stretch this spring by $x_{\text{a}} = (10\; {\rm cm} - 6.0\; {\rm cm})$ . Thus, the force $F_{\text{b}}$ required to stretch this spring by $x_{\text{a}} = (20\; {\rm cm} - 6.0\; {\rm cm})$ would satisfy:

$\displaystyle \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}}= \frac{F_{\text{b}}}{20\; {\rm cm} - 6.0\; {\rm cm}}$ .

Rearrange and solve for $F_{\text{b}}$ :

$\begin{aligned} F_{\text{b}} &= \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}} \, (20\; {\rm cm} - 6.0\; {\rm cm}) \\ &\approx 25\; {\rm N}\end{aligned}$ .

7 0

1 year ago

Two loudspeakers are placed side by side and driven by the same source at 500 Hz. A listener is positioned in front of the two s

Oliga [24]

Answer:

0.68 m

Explanation:

We know that the speed of sound in air is a product of frequency and wavelength. Taking speed of sound in air as 340 m/s

V=frequency*wavelength

Then wavelength is given by 350/500=0.68 m

Therefore, to repeat constructive interference at the listener's ear, a distance of 0.68 m is needed

4 0

2 years ago

a runner covers the last straigjt stretch of a race in 4 s. during that time, he speeds up from 5m/s to 9m/s.

PtichkaEL [24]

During that final period of time,
his acceleration is
(9 m/s - 5 m/s) / (4 sec) = 1 m/s² .

Did you have a question to ask ?

8 0

3 years ago

A plane mirror of circular shape with radius r=20cm is fixed to the ceiling. A bulb is to be placed on the axis of the mirror. A

KIM [24]

Answer:

0.75 m

Explanation:

Let's call the distance between the bulb and the mirror x.

The bulb and the length of the mirror form a triangle. The mirror and the illuminated area on the floor form a trapezoid. If we extend the lines from the mirror edge to the reflected image of the bulb, we turn that trapezoid into a large triangle. This triangle and the small triangle are similar. So we can say:

x / 0.4 = (3 + x) / 2

Solving for x:

2x = 0.4 (3 + x)

2x = 1.2 + 0.4 x

1.6 x = 1.2

x = 0.75

So the bulb should located no more than 0.75 m from the mirror.