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mr Goodwill [35]

3 years ago

6

What waves are passing through your house or classroom daily?(Tell what device is using them, think of the movie Tour of the EM

spectrum.)

1 answer:

aalyn [17]3 years ago

6 0

Radiation waves / EM spectrum - infrared light<span>, </span>ultraviolet light<span>, </span>Xrays<span> and </span>gamma.

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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

A car accelerates while trying to merge onto the freeway. Its speed goes from 0 km/h to 70km/h in 10 seconds. What is its accele

atroni [7]

Answer:

1.944m_s²

Explanation:

First convert speed from km/h to m/s the use the formula a=v-u

t

8 0

3 years ago

A 13.5 μF capacitor is connected to a power supply that keeps a constant potential difference of 22.0 V across the plates. A pie

-BARSIC- [3]

a) $3.27\cdot 10^{-3} J$

b) $11.60\cdot 10^{-3} J$

c) $8.33\cdot 10^{-3} J$

Explanation:

a)

The energy stored in a capacitor is given by

$U=\frac{1}{2}CV^2$

where

C is the capacitance of the capacitor

V is the potential difference across the plates of the capacitor

For the capacitor in this problem, before insering the dielectric, we have:

$C=13.5 \mu F = 13.5\cdot 10^{-6}F$ is its capacitance

V = 22.0 V is the potential difference across it

Therefore, the initial energy stored in the capacitor is:

$U=\frac{1}{2}(13.5\cdot 10^{-6})(22.0)^2=3.27\cdot 10^{-3} J$

b)

After the dielectric is inserted into the plates, the capacitance of the capacitor changes according to:

$C'=kC$

where

k = 3.55 is the dielectric constant of the material

C is the initial capacitance of the capacitor

Therefore, the energy stored now in the capacitor is:

$U'=\frac{1}{2}C'V^2=\frac{1}{2}kCV^2$

where:

$C=13.5\cdot 10^{-6}F$ is the initial capacitance

V = 22.0 V is the potential difference across the plate

Substituting, we find:

$U'=\frac{1}{2}(3.55)(13.5\cdot 10^{-6})(22.0)^2=11.60\cdot 10^{-3} J$

C)

The initial energy stored in the capacitor, before the dielectric is inserted, is

$U=3.27\cdot 10^{-3} J$

The final energy stored in the capacitor, after the dielectric is inserted, is

$U'=11.60\cdot 10^{-3} J$

Therefore, the change in energy of the capacitor during the insertion is:

$\Delta U=11.60\cdot 10^{-3}-3.27\cdot 10^{-3}=8.33\cdot 10^{-3} J$

So, the energy of the capacitor has increased by $8.33\cdot 10^{-3} J$

8 0

3 years ago

Will give Brainliest!! Question attached.

ss7ja [257]

Answer:

im pretty sure it is 3.0 K

Explanation:

8 0

3 years ago

Read 2 more answers

While following the directions on a treasure map a pirate walks 37.0 m north and then turns and walks 8.5 m east what is the mag

kogti [31]

Answer: $38\ m$

Explanation:

For this exercise you can use the Pythagorean theorem to find the magnitude of the resultant displacement.

Then:

$d^2=\triangle x^2+\triangle y^2$

You can observe that the square of the displacement is equal to the sum of the square of the horizontal displacement and the square of the vertical displacement.

Since the pirate walks 37.0 meters north and then turns and walks 8.5 meters east:

$\triangle x=37.0\ m\\\triangle y=8.5\ m$

Substituting values and solving for "d", you get:

$d=\sqrt{(37.0\ m)^2+(8.5\ m)^2}\\\\d=38\ m$

8 0

3 years ago