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3 years ago

Metric conversions

Physics

1 answer:

AfilCa [17]3 years ago

5 0

1m=10dm

$\\ \sf\longmapsto 15m$

$\\ \sf\longmapsto 15(10)$

$\\ \sf\longmapsto 150dm$

1L=1000cm^3

$\\ \sf\longmapsto 0.023cm^3$

$\\ \sf\longmapsto 0.023\times 0.001$

$\\ \sf\longmapsto 0.00023L$

1km=1000m

$\\ \sf\longmapsto 0.00049km$

$\\ \sf\longmapsto 0.00049(1000)$

$\\ \sf\longmapsto 0.49m$

4:-

1kg=1000g

$\\ \sf\longmapsto 0.25kg$

$\\ \sf\longmapsto 0.25(1000)$

$\\ \sf\longmapsto 250g$

5:-

1g=0.0001hg

$\\ \sf\longmapsto 15g$

$\\ \sf\longmapsto 15(0.0001)$

$\\ \sf\longmapsto 0.0015hg$

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3 years ago

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PLEASE HELP !!!!!!! Fastttt !!!!!!!!! You are using a flashlight to find your phone in the dark but the light isn’t bright enoug

My name is Ann [436]

Answer:

The part of the wave is the Amplitude.

Reason: The amplitude of light waves is responsible for our perception of brightness, with larger amplitudes appearing brighter than lower amplitudes.

Explanation:

The amplitude of light waves is responsible for our perception of brightness, with larger amplitudes appearing brighter than lower amplitudes.

Thus, when you turn up the brightness of the light so that you can better see, you are increasing the amplitude of the light wave, which in turn increases the intensity of the light, seen from the increased brightness of the flash light.

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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:

$t=0.495 s$

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4 years ago

During most of its lifetime, s star maintains an equilibrium size in which the inward force of gravity on each atom is balanced

frez [133]

Answer:

$r_f=137493m$

$v=8638940m/s$

Explanation:

During this process the mass $M=2\times10^{30}Kg$ will be considered constant. We start from a radius $r_i=7\times10^8m$ and a period $T_i=30\ days=(30)(24)(60)(60)s=2592000s$ . The final period is $T_f=0.1s$ .

Angular momentum <em>L</em> is conserved in this process. We can use the formula $L=I\omega$ , where I is the momentum of inertia (which for a solid sphere is $I=\frac{2mr^2}{5}$ ) and $\omega=\frac{2\pi }{T}$ is the angular velocity, so we can write the star's angular momentum as: