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sdas [7]
3 years ago
10

Matt plays badminton. He serves the birdie with a velocity of 4.25 m/s [right] covering a displacement of 5.2 m [forward]. How m

uch time does this birdie’s trip take? show your work below
Physics
1 answer:
OLEGan [10]3 years ago
6 0

Answer:

Time taken by birdie’s trip = 1.2235 s (Approx)

Explanation:

Given:

Velocity = 4.25 m/s

Displacement = 5.2 m

Find:

Time taken by birdie’s trip = ?

Computation:

⇒ Time taken = Displacement / Velocity

⇒ Time taken by birdie’s trip = Displacement / Velocity

⇒ Time taken by birdie’s trip = 5.2 m / 4.25 m/s

⇒ Time taken by birdie’s trip = 1.2235 s (Approx)

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A mixture contains salt, sand, pebbles, and grass. Which part of this mixture is soluble in water?
melamori03 [73]

Answer:

hey!

the answer is

salt

Explanation:

this is due to sand, grass and pebbles are not soluble (they do not dissolve in water)

pls put brainliest

3 0
3 years ago
Name two everyday examples in which stored elastic potential energy is made use of. In each case state the energy transfer which
trasher [3.6K]

Answer:

A raised weight.

Water that is behind a dam.

Energy transfer takes place when energy moves from one place to another. Energy can move from one object to another, like when the energy from your moving foot is transferred to a soccer ball, or energy can change from one form to another.

Explanation:

7 0
3 years ago
Please help. All the information is in the image.
kondaur [170]

The tiger's position at time <em>t</em> is given by

<em>x</em> (horizontal) = (4.5 m/s) <em>t</em>

<em>y</em> (vertical) = 7.5 m - 1/2 <em>gt</em> ²

Solve <em>y</em> = 0 for <em>t</em> to find the time it takes for the tiger to reach the ground :

0 = 7.5 m - 1/2 (9.8 m/s²) <em>t</em> ²

===>   <em>t</em> = √(2 (7.5 m) / (9.8 m/s²)) ≈ 1.2 s

Evaluate <em>x</em> at this time :

<em>x</em> = (4.5 m/s) (1.2 s) ≈ 5.6 m

5 0
3 years ago
I
Rasek [7]

When conducting a search to identify a text's credibility and reliability, you have to check the following characteristics:

  • Sources: where the information is obtained and it is supposed to be true. It is said that you need at least three different sources that explain the same information for it to be validated.

  • Is the article is current?: articles could change all the time, specially scientific ones because new discoveries can change what was discovered before, so it is important to check if the text you are reading is current or not, and if it is not, you need to check if something has changed during all those years.

<h3>Article's credibility</h3>

In this exercise, you have to present an article and describe the purpose of the source and if the article is current or not.

For example, let's select an article called "How using social media affects teenagers".

The purpose of the sources in this article is to demonstrate how social media affects teenagers using different surveys.

Check more information about sources here brainly.com/question/24708478

6 0
3 years ago
Compare the gravitational acceleration on the following objects compared to the Sun using:
arsen [322]

The gravitational acceleration of White dwarf compared to Sun is 13,675.86.

The gravitational acceleration of Neutron star compared to Sun is 6.79 x 10⁻²⁴.

The gravitational acceleration of Star Betelgeuse compared to Sun is 8.5 x 10¹⁰.

<h3>Mass of the planets</h3>

Mass of sun = 2 x 10³⁰ kg

Mass of white dwarf = 2.765  x 10³⁰ kg

Mass of Neutron star = 5.5 x 10¹² kg

Mass of star Betelgeuse = 2.188 x 10³¹ kg

<h3>Radius of the planets</h3>

Radius of sun = 696,340 km

Radius of white dwarf = 7000 km

Radius of Neutron star = 11 km

Radius of star Betelgeuse = 617.1 x 10⁶ km

<h3>Gravitational acceleration of White dwarf compared to Sun</h3>

\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.765 \times 10^{30}}{2\times 10^{30}} \times [\frac{696,340,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 13,675.86

<h3>Gravitational acceleration of Neutron star compared to Sun</h3>

\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{5.5 \times 10^{12}}{2\times 10^{30}} \times [\frac{11,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 6.79\times 10^{-24}

<h3>Gravitational acceleration of Star Betelgeuse compared to Sun</h3>

\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.188 \times 10^{31}}{2\times 10^{30}} \times [\frac{617.1 \times 10^9}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 8.5\times 10 ^{10}

Learn more about acceleration due to gravity here: brainly.com/question/88039

3 0
3 years ago
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