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3 years ago

Please help will make brainliest

Physics

2 answers:

natka813 [3]3 years ago

4 0

<h2>Answer:</h2>

<u>The answer is </u><u>D) In the middle of the ocean</u>

<h2>Explanation:</h2>

Offshore drilling, the process of extracting oil and gas resources from underwater locations such as ocean, seas or even lakes. If we look into the history, mostly offshore drilling began by extending known coastal oil- and gas-producing trends out into the ocean. For this reason, most US offshore drilling has taken place offshore Louisiana, Texas, California, and Alaska, areas with coastal onshore oil and gas fields.

kumpel [21]3 years ago

3 0

The Gulf of Mexico
OR
near trenches, in the middle of the ocean, the continental shelf, and lastly, in the United States

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A) One Strategy in a snowball fight the snowball at a hangover level ground. While your opponent is watching this first snowfall

Alexandra [31]

Answers:

a) $\theta_{2}=38\°$

b) $t=0.495 s$

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which the travel of the snowball has two components: <u>x-component</u> and <u>y-component</u>. Being their main equations as follows for both snowballs:

<h3><u>Snowball 1:</u></h3>

<u>x-component: </u>

$x=V_{o}cos\theta_{1} t_{1}$ (1)

Where:

$V_{o}=14.1 m/s$ is the initial speed of snowball 1 (and snowball 2, as well)

$\theta_{1}=52\°$ is the angle for snowball 1

$t_{1}$ is the time since the snowball 1 is thrown until it hits the opponent

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta_{1} t_{1}+\frac{gt_{1}^{2}}{2}$ (2)

Where:

$y_{o}=0$ is the initial height of the snowball 1 (assuming that both people are only on the x axis of the frame of reference, therefore the value of the position in the y-component is zero.)

$y=0$ is the final height of the snowball 1

$g=-9.8m/s^{2}$ is the acceleration due gravity (always directed downwards)

<h3><u>Snowball 2:</u></h3>

<u>x-component: </u>

$x=V_{o}cos\theta_{2} t_{2}$ (3)

Where:

$\theta_{2}$ is the angle for snowball 2

$t_{2}$ is the time since the snowball 2 is thrown until it hits the opponent

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta_{2} t_{2}+\frac{gt_{2}^{2}}{2}$ (4)

Having this clear, let's begin with the answers:

<h2>a) Angle for snowball 2</h2>

Firstly, we have to isolate $t_{1}$ from (2):

$0=0+V_{o}sin\theta_{1} t_{1}+\frac{gt_{1}^{2}}{2}$ (5)

$t_{1}=-\frac{2V_{o}sin\theta_{1}}{g}$ (6)

Substituting (6) in (1):

$x=V_{o}cos\theta_{1}(-\frac{2V_{o}sin\theta_{1}}{g})$ (7)

Rewritting (7) and knowing $sin(2\theta)=sen\theta cos\theta$ :

$x=-\frac{V_{o}^{2}}{g} sin(2\theta_{1})$ (8)

$x=-\frac{(14.1 m/s)^{2}}{-9.8 m/s^{2}} sin(2(52\°))$ (9)

$x=19.684 m$ (10) This is the point at which snowball 1 hits and snowball 2 should hit, too.

With this in mind, we have to isolate $t_{2}$ from (4) and substitute it on (3):

$t_{2}=-\frac{2V_{o}sin\theta_{2}}{g}$ (11)

$x=V_{o}cos\theta_{2} (-\frac{2V_{o}sin\theta_{2}}{g})$ (12)

Rewritting (12):

$x=-\frac{V_{o}^{2}}{g} sin(2\theta_{2})$ (13)

Finding $\theta_{2}$ :

$2\theta_{2}=sin^{-1}(\frac{-xg}{V_{o}^{2}})$ (14)

$2\theta_{2}=75.99\°$

$\theta_{2}=37.99\° \approx 38\°$ (15) This is the second angle at which snowball 2 must be thrown. Note this angle is lower than the first angle $(\theta_{2} < \theta_{1})$ .

<h2>b) Time difference between both snowballs</h2>

Now we will find the value of $t_{1}$ and $t_{2}$ from (6) and (11), respectively:

$t_{1}=-\frac{2V_{o}sin\theta_{1}}{g}$

$t_{1}=-\frac{2(14.1 m/s)sin(52\°)}{-9.8m/s^{2}}$ (16)

$t_{1}=2.267 s$ (17)

$t_{2}=-\frac{2V_{o}sin\theta_{2}}{g}$

$t_{2}=-\frac{2(14.1 m/s)sin(38\°)}{-9.8m/s^{2}}$ (18)

$t_{2}=1.771 s$ (19)

Since snowball 1 was thrown before snowball 2, we have:

$t_{1}-t=t_{2}$ (20)

Finding the time difference $t$ between both:

$t=t_{1}-t_{2}$ (21)

$t=2.267 s - 1.771 s$

Finally:

$t=0.495 s$

4 0

4 years ago

Which change will always result in an increase in the gravitational force between two objects?

Jet001 [13]

It's C. increasing the masses of the objects and decreasing the distance between the objects

3 0

3 years ago

Read 2 more answers

You fire a projectile 35° above horizontal with an initial velocity of 200m/s. It lands in a valley 300m below the launch point.

Brilliant_brown [7]

Answer:

See below

Explanation:

Vertical component of initial velocity = 200 sin 35° = 114.72 m/s

then use position formula a = 9.81 m/s^2

0 = 300 + 114.72 t - 1/2 (9.81)(t^2)

use quadratic formula with a = - 4.905 b = 114.72 c = 300

to find t = <u>25.76 seconds </u>

<u />

To find the range

( horizontal distance the projectile lands from launch point)

Horizontal component of initial velocity 200 cos 35 = 163.83 m/s

( it flies horizontally at this speed for <u>25.76 seconds <==found above</u>)

163.83 m/s * 25.76 s = <u>4220.3 meters</u>

3 0

2 years ago

There are 92 naturally occurring elements. How does this relate to the number of compounds?

raketka [301]

C) Elements combine in different ways to create many more than 92 compounds.

Explanation:

The number of elements we have relates to the compound formation in that, elements combine in different ways to create many more than 92.

Elements are distinct substances that cannot be split up into simpler substances. Such substances are only made up of one kind of atom.

Compounds are substances that are made up of two or more kinds of atoms joined together in a definite grouping.

Several millions of compounds are known because elements can combine in different ways.
The number of elements do not increase because new elements are discovered, it is because elements have several ways by they combine.

Learn more:

Chemical reactions brainly.com/question/3953793

#learnwithBrainly

3 0

3 years ago

Vector A has a magnitude 5 m, vector B has a magnitude 7 m. Which of the following cannot be the magnitude of the vector A + B?

bulgar [2K]

I think it should be 7m

7 0

4 years ago