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3 years ago

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You are walking around your neighborhood and you see a child on top of a roof of a building kick a soccer ball. The soccer ball

is kicked at 31° from the edge of the building with an initial velocity of 15 m/s and lands 63 meters away from the wall. How tall, in meters, is the building that the child is standing on?

1 answer:

irinina [24]3 years ago

3 0

Answer:

69.58 m tall

Explanation:

Pls see attached file

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A team of ecologists are studying four different ecosystems with varying levels of biodiversity. The ecologists categorize the d

Ann [662]

Answer:

B. Ecosystem B, because its high species diversity could have resulted from increased competition among its members.

Explanation:

This is because, in the ecosystem with varying level of biodiversity, Ecosystem B has medium level of species diversity found in them with High medium level of habitat diversity which causes increasing competitions among them.

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3 years ago

A concave mirror has a focal length of 13.5 cm. This mirror forms an image located 37.5 cm in front of the mirror. Find the magn

77julia77 [94]

Explanation:

It is given that,

Focal length of the concave mirror, f = -13.5 cm

Image distance, v = -37.5 cm (in front of mirror)

Let u is the object distance. It can be calculated using the mirror's formula as :

$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$

$\dfrac{1}{u}=\dfrac{1}{f}-\dfrac{1}{v}$

$\dfrac{1}{u}=\dfrac{1}{(-13.5)}-\dfrac{1}{(-37.5)}$

u = -21.09 cm

The magnification of the mirror is given by :

$m=\dfrac{-v}{u}$

$m=\dfrac{-(-37.5)}{(-21.09)}$

m = -1.77

So, the magnification produced by the mirror is (-1.77). Hence, this is the required solution.

7 0

3 years ago

A sound source is moving at 80 m/s toward a stationary listener that is standing in still air (a) Find the wavelength of the sou

Setler [38]

Answer:

a. wavelength of the sound, $\vartheta = 1.315\vartheta_{o}$

b. observed frequecy, $\lambda = 0.7604\lambda_{o}$

Given:

speed of sound source, $v_{s}$ = 80 m/s

speed of sound in air or vacuum, $v_{a}$ = 343 m/s

speed of sound observed, $v_{o}$ = 0 m/s

Solution:

From the relation:

v = $\vartheta \lambda$ (1)

where

v = velocity of sound

$\vartheta$ = observed frequency of sound

$\lambda$ = wavelength

(a) The wavelength of the sound between source and the listener is given by:

$\lambda = \frac{v_{a}}{\vartheta }$ (2)

(b) The observed frequency is given by:

$\vartheta = \frac{v_{a}}{v_{a} - v_{s}}\vartheta_{o}$

$\vartheta = \frac{334}{334 - 80}\vartheta_{o}$

$\vartheta = 1.315\vartheta_{o}$ (3)

Using eqn (2) and (3):

$\lambda = \frac{334}{1.315} = \frac{1}{1.315}\frac{v_{a}}{\vartheta_{o}}$

$\lambda = 0.7604\lambda_{o}$

4 0

3 years ago

How does the search for exoplanets help us describe how science requires creativity?

Aliun [14]

Answer:

As the planets are very small and dark in comparison with stars, it makes them very hard to be found from earth.

Explanation:

Astronomy, of course, has a solution for this. As astronomers can't observe planets directly, they decided to observe the stars and search for the effects that planets have on them.

There are many ways of observing the exoplanets: Radial Velocity, Transit Photometry, Microlensing, Astrometry, Direct Imaging, etc.

Before all of this, scientist had to find ways to prove their theories. Most of their time they have spent in giving the creative answers.

Science and creativity are very much connected when we speak about the development of science. Rationality and creativity always go together.

In order to create an idea that other people will consider useful, it is important to use creativity. As no one has the exact answer when it comes to science, the adventure is to research the unknown.

5 0

3 years ago

You attach a meter stick to an oak tree, such that the top of the meter stick is 1.87 meters above the ground. Later, an acorn f

Verdich [7]

To solve this problem we will apply the concepts related to the kinematic equations of linear motion. We will calculate the initial velocity of the object, and from it, we will calculate the final position. With the considerations made in the statement we will obtain the total height. Initial velocity of the acorn,

$u = 0m/s$

Also, it is given that the acorn takes 0.201s to pass the length of the meter stick.

$s = ut+\frac{1}{2} at^2$

Replacing,

$1 = u(0.141)+ \frac{1}{2} (9.8)(0.141)^2$

$u =6.4013m/s$

The height of the acorn above the meter stick can be calculated as,

$v^2 = u^2 +2gh$

$h = \frac{v^2-u^2}{2g}$

$h = \frac{6.4013^2-0^2}{2(9.8)}$

$h = 2.0906m$

Also the top of the meter stick is 1.87m above the ground hence the height of the acorn above the ground is

$h = 2.0906+1.87$

$h = 3.9606m$

4 0

3 years ago