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3 years ago

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1 answer:

aev [14]3 years ago

4 0

OD because Boyle’s law specifically states

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If a block of wood dropped from a tall building has attained a velocity of 78.4 m/s how long has it been falling

ryzh [129]

Gravity adds 9.8 m/s to the speed of a falling object every second.

An object dropped from 'rest' (v = 0) reaches the speed of 78.4 m/s after falling for (78.4 / 9.8) = <em>8.0 seconds</em> .

<u>Note:</u>

In order to test this, you'd have to drop the object from a really high cell- tower, building, or helicopter. After falling for 8 seconds and reaching a speed of 78.4 m/s, it has fallen 313.6 meters (1,029 feet) straight down.

The flat roof of the Aon Center . . . the 3rd highest building in Chicago, where I used to work when it was the Amoco Corporation Building . . . is 1,076 feet above the street.

5 0

3 years ago

Consider a 7 m stretched string that is clamped at both ends. What is the longest wavelength standing wave that it can support (

user100 [1]

A vibrating stretched string has nodes or fixed points at each end. The string will vibrate in its fundamental frequency with just one anti node in the middle - this gives half a wave.

$l=\frac{\lambda }{2}$

Rearranging for the wavelength

$\lambda=2l$

$\lambda =2(7)$

$\lambda = 14m$

Therefore the longest wavelength standing wave that it can support is 14m

7 0

4 years ago

Community psychologists are not<br> concerned with worker stress factors.<br> True<br> False

fgiga [73]

It would be false as your answer

8 0

2 years ago

Describe the path taken by electricity through an incandescent light bulb.

Nana76 [90]

Answer:

take it out

Explanation:

4 0

3 years ago

Astronomers observe the motion of four planets that orbit a star similar to the Sun. Each planet follows an elliptical orbit aro

Sergeu [11.5K]

Answer:

planet that is farthest away is planet X

kepler's third law

Explanation:

For this exercise we can use Kepler's third law which is an application of Newton's second law to the case of the orbits of the planets

T² = ( $\frac{4\pi ^2}{ G M_s}$ a³ = K_s a³

Let's apply this equation to our case

a = $\sqrt[3]{ \frac{T^2}{K_s} }$

for this particular exercise it is not necessary to reduce the period to seconds

Plant W

10² = K_s $a_{w}^3$

a_w = $\sqrt[3]{ \frac{100}{ K_s} }$

a_w = $\frac{1}{ \sqrt[3]{K_s} }$ 4.64

Planet X

a_x = $\sqrt[3]{ \frac{640^3}{K_s} }$

a_x = \frac{1}{ \sqrt[3]{K_s} } 74.3

Planet Y

a_y = $\sqrt[3]{ \frac{80^2}{K_s} }$

a_y = \frac{1}{ \sqrt[3]{K_s} } 18.6

Planet z

a_z = $\sqrt[3]{ \frac{270^2}{K_s} }$

a_z = \frac{1}{ \sqrt[3]{K_s} } 41.8

From the previous results we see that planet that is farthest away is planet X

where we have used kepler's third law

3 0

3 years ago