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3 years ago

Which of the following electromagnetic waves is not used for communication?

1 answer:

8 0

Microwaves and radio waves are employed in radio and satellite communications while infrared waves are used in remote controls and infrared features of new phones and other electrons. However, gamma rays have far too much energy and cause damage to the body. They are not used in communication.
The answer is A.

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A car is traveling 80km/h is 130m behind a truck traveling at 75km/h. How long will it take the car to reach the truck?

rjkz [21]

The car approaches the truck at a speed of 80-75=5km/h. To travel the 130m, it takes
$\frac{0.130km}{5 \frac{km}{h} } =0.026h=93.6s$

8 0

3 years ago

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

A 90.0-kg man runs at a constant speed of 11.5 m/s. what is his kinetic energy?

harina [27]

The kinetic energy possessed by the man is 517.5 Joules.

<u>Given the following data:</u>

Mass = 90.0 kg

Velocity = 11.5 m/s.

To find the kinetic energy of the man:

Kinetic energy is an energy that is possessed by a physical object or body due to its motion.

Mathematically, kinetic energy is calculated by using the formula;

$K.E = \frac{1}{2} MV^2$

<u>Where:</u>

K.E is the kinetic energy.

M is the mass of an object.

V is the velocity of an object.

Substituting the given values, we have;

$K.E = \frac{1}{2}$ × $90$ × $11.5$

$K.E = 45$ × $11.5$

<em>Kinetic energy = 517.5 Joules.</em>

Therefore, the kinetic energy possessed by the man is 517.5 Joules.

8 0

3 years ago

The demand for your hand-made skateboards, in weekly sales, is q = −3p + 700 if the selling price is $p. You are prepared to sup

Y_Kistochka [10]

Answer:

you should sell your skateboards at $240

Explanation:

The price p to sell your skateboards for so that there is neither a shortage nor a surplus is the price that makes equal the quantity of sales and the quantity of supply, so p is equal to:

q (sales) = q (supply)

-3p + 700 = 2p - 500

700 + 500 = 2p + 3p

1200 = 5p

1200/5 = p

$240 = p

3 0

3 years ago

Please help! I'll give brainliest.

Mamont248 [21]

Answer:

Below

Explanation:

Surface waves cannot pass through the Earth's mantle but travel along the Earth's crust. They are more destructive than body waves.

Have a good night ((:

7 0

3 years ago