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wlad13 [49]
3 years ago
7

The ____________________ of your voice refers to its intensity or loudness.

Physics
1 answer:
JulijaS [17]3 years ago
7 0
That's the amplitude.
You might be interested in
One strategy in a snowball fight is to throw
faltersainse [42]

Answers:

a) \theta_{2}=23\°

b) t=1.199 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}=11.1 m/s is the initial speed  of snowball 1 (and snowball 2, as well)

\theta_{1}=67\° 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{(11.1 m/s)^{2}}{-9.8 m/s^{2}} sin(2(67\°))   (9)

x=9.043 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}=45.99\°  

\theta_{2}=22.99\° \approx 23\°  (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(11.1 m/s)sin(67\°)}{-9.8m/s^{2}}   (16)

t_{1}=2.085 s   (17)

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

t_{2}=-\frac{2(11.1 m/s)sin(23\°)}{-9.8m/s^{2}}   (18)

t_{2}=0.885 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=2.085 s - 0.885 s  

Finally:

t=1.199 s  

7 0
3 years ago
A 0.5-kilogram apple falls from a height of 2 meters to 1.50 meters. Ignoring frictional effects, what is the kinetic energy of
spin [16.1K]

The  final kinetic energy of the ball is 2.45 J

Explanation:

We can solve this problem by using the law of conservation of energy.

In absence of frictional effect, the mechanical energy of the apple must be conserved during the fall. So we can write:

U_i +K_i = U_f + K_f

where :

U_i is the initial potential energy, at the top

K_i is the initial kinetic energy, at the top

U_f is the final potential energy, at the bottom

K_f is the final kinetic energy, at the bottom

By explicing the potential energy, we can rewrite the equation as:

mgh_i + K_i = mgh_f + K_f

where:

m = 0.5 kg is the mass of the apple

g=9.8 m/s^2 is the acceleration of gravity

h_i = 2 m is the initial height

h_f=1.50 m is the final height

The initial kinetic energy is zero, since the ball starts from rest:

K_i = 0

Therefore we can solve the equation for K_f, the final kinetic energy of the ball:

K_f = mg(h_i-h_f)=(0.5)(9.8)(2-1.50)=2.45 J

Learn more about kinetic energy and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647  

brainly.com/question/10770261  

#LearnwithBrainly

3 0
3 years ago
Read 2 more answers
How do i solve?<br><br>a = bc<br>(b)​
stira [4]
I don’t even know I’m so dumb.
7 0
3 years ago
Object A attracts object B with a gravitational force of 5 newtons from a given distance. If the distance between the two object
enyata [817]

Answer:

10 newtons, because the gravitaional force willbe stronger the closer it gets.

7 0
3 years ago
Which formula can be used to find the magnitude of the resultant vector? R2 = Rx2 + Ry2 R = Rx + Ry R = Rx(cosθ) R = Rx(sinθ)
12345 [234]
R^2 = rx^2 + ry^2 !!!!!!!!!!
5 0
3 years ago
Read 2 more answers
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