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4 years ago

Whose geocentric model was accepted for 1400 years

1 answer:

8 0

IDK the answer but I think it was "Ptolemy" whose geocentric model of the solar system was accepted for 1,400 years, and was embraced by the the Church until Galileo called it into question.

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. In science, which of the following is NOT true about scientific theories a. They are generated by the testing of many hypothes

nydimaria [60]

Answer:

The correct option is b) By the time they get to be theories, they cannot be disproven. It's not necessary that the theories are always right. They can be proven wrong.

Explanation:

5 0

4 years ago

Ultrasound is the name given to frequencies above the human range of hearing True or false

gulaghasi [49]

Yes. That's a true statement. Ultrasound is indeed the name given to sounds with frequencies above the human range of hearing.

Ultrasound is not different from "normal" (audible) sound in its physical properties, except that humans can't hear it.

5 0

3 years ago

Explain the difference between displacement and distance

alisha [4.7K]

Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion. Displacement is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.

6 0

4 years ago

In the Milky Way Galaxy, where would you expect to find the bulge?

nalin [4]

Answer:

at the sun???

Explanation:

3 0

3 years ago

A 20-foot ladder is leaning against the wall. If the base of the ladder is sliding away from the wall at the rate of 3 feet per

tamaranim1 [39]

Answer:

6.87 ft/s is the rate at which the top of ladder slides down.

Explanation:

Given:

Length of the ladder is, $L=20\ ft$

Let the top of ladder be at height of 'h' and the bottom of the ladder be at a distance of 'b' from the wall.

Now, from triangle ABC,

AB² + BC² = AC²

$h^2+b^2=L^2\\h^2+b^2=20^2\\h^2+b^2=400----1$

Differentiating the above equation with respect to time, 't'. This gives,

$\frac{d}{dt}(h^2+b^2)=\frac{d}{dt}(400)\\\\\frac{d}{dt}(h^2)+\frac{d}{dt}(b^2)=0\\\\2h\frac{dh}{dt}+2b\frac{db}{dt}=0\\\\h\frac{dh}{dt}+b\frac{db}{dt}=0--------2$

In the above equation the term $\frac{dh}{dt}$ is the rate at which top of ladder slides down and $\frac{db}{dt}$ is the rate at which bottom of ladder slides away.

Now, as per question, $h=8\ ft, \frac{db}{dt}=3\ ft/s$

Plug in $h=8$ in equation (1) and solve for $b$ . This gives,

$8^2+b^2=400\\64+b^2=400\\b^2=400-64\\b^2=336\\b=\sqrt{336}=18.33\ ft$

Now, plug in all the given values in equation (2) and solve for $\frac{dh}{dt}$

$8\times \frac{dh}{dt}+18.33\times 3=0\\8\times \frac{dh}{dt}+54.99=0\\8\times \frac{dh}{dt}=-54.99\\ \frac{dh}{dt}=-\frac{54.99}{8}=-6.87\ ft/s$

Therefore, the rate at which the top of ladder slide down is 6.87 ft/s. The negative sign implies that the height is reducing with time which is true because it is sliding down.

8 0

4 years ago