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3 years ago

PLEASE HELP ME I AM TIMED!

Physics

1 answer:

Arte-miy333 [17]3 years ago

6 0

The sequence of rock layers can be disturbed by erosion and earthquakes

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I think to see further away

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3 years ago

Read 2 more answers

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: x-component and y-component. Being their main equations as follows for both snowballs:

<h3>Snowball 1:</h3>

x-component:

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

y-component:

$y=y_{o}+V_{o}sin\theta_{1} t_{1}+\frac{gt_{1}^{2}}{2}$ (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>Snowball 2:</h3>

x-component:

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

y-component:

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

WILL GIVE BRAINIEST TO CORRECT ANSWER

Illusion [34]

Answer:

Hey mate, here's ur answer

-------------------------------------------------------------

Loudness refers to how a sound seems to a listener, whether it's loud or soft.

___________________________

Intensity  is the sound power per unit area. It is independent of the sensitivity of the human ears.

___________________________

The loudness of a sound relates the intensity of any given sound to the intensity at the threshold of hearing. It is measured in decibels (dB).

___________________________

Hope this helps

#stayhomestaysafemate

:D

5 0

4 years ago

How much work does a 65 kg person climbing a 2000 m high cliff do?

qaws [65]

The answer for the following question is explained below.

Therefore the work done is 130 kilo Joules.

Explanation:

Work:

A force causing the movement or displacement of an object.

Given:

mass of the person (m) = 65 kg

height of the cliff (h) = 2000 m

To calculate:

work done (W)

We know;

According to the formula:

W = m × g × h

Where;

m represents mass of the person

g represents the acceleration due to gravity

where the value of g is;

g = 10 m/ s²

h represents the height of the cliff

From the above formula;

W = 65 × 10 × 2000

W = 130,000 J

W = 130 Kilo Joules

Therefore the work done is 130 kilo Joules.

3 0

3 years ago

A flutist assembles her flute in a room where the speed of sound is 342 m/sm/s . When she plays the note A, it is in perfect tun

dezoksy [38]

Answer:

A) beats per second she will hear if she now plays the note A as the tuning fork is sounded = 5.13 beats per second

B) length she needs to extend the "tuning joint" of her flute to be in tune with the tuning fork = 0.0045m

Explanation:

A) First of all, wavelength = v/f

Where v is speed of wave and f is frequency.

Thus, wavelength of the sound wave of Note A is;

f2 = 440 Hz and v = 342m/s

λ = 342/440 = 0.7773m

Now, since the air inside the note was warmed after a while, the wave will will have a new frequency which we'll call (f1) and and new speed (v'), thus;

f2 = v'/λ = 346/0.7773 = 445.13 Hz

Now let's calculate beat frequency(fbeat).

fbeat = (f1 - f2)

So fbeat = 445.13 - 440 = 5.13Hz or 5.13 beats per second

B) Now, frequency of standing wave models (fm) = n(v/2L)

Where n is a positive integer and L is the open tube length

Making L the subject of the formula, we have; L = nv/2fm

Now from earlier derivation, we see that v = fλ and in this case, v=fλ

Thus, let's replace v with fλ to het;

L = nλ/2

If we take, n=1, L = (1 x 0.7773)/2 = 0.3887m

Now, when the air inside the tube has warmed, it will have a new length to eliminate beats and give same frequency of 440Hz.

So let's call this new length L1;

So L1 = v'/2(f2) = 346/(2x440) = 346/880 = 0.3932m

So the length she needs to extend the "tuning joint" of her flute to be in tune with the tuning fork will be;

ΔL = L1 - L = 0.3932 - 0.3887 = 0.0045m

5 0

4 years ago