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4vir4ik [10]
3 years ago
8

Wish if these Are not examples of matter?

Physics
2 answers:
neonofarm [45]3 years ago
6 0

Answer:

I'd love to help, but there isn't anything to choose from.

lidiya [134]3 years ago
4 0

Answer: Lots of answers.

Explanation:

Love, reflection, Memory, Time, sunlight, and thoughts.

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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
2 years ago
A plane leaves the airport in Galisteo and flies 170 km at 68.0° east of north; then it changes direction to fly 230 km at 36.0°
luda_lava [24]

Answer:

The direction will be 84.86^\circ and the distance 250.75km.

Explanation:

Let's say A is the displacement vector which represents the first 170km and B the one for the next 230km. Then the components of these vector will be:

A_x=170cos(68^{\circ})\\ A_y=170sin(68^\circ)\\\\B_x=230cos(-36^\circ)\\B_y=230sin(-36^\circ)

The vector which point from the origin to the final position of the plane will be R=A+B. We sum components on <em>x </em>and <em>y </em>independetly (vector property):

R_x=A_x+B_x=170cos(68^{\circ})+230cos(-36^\circ)=63.68km+186.07km=249.75km

R_y=A_y+B_y=170sin(68^\circ)+230sin(-36^\circ)=157.62km-135.19km=22.43km

If \theta is the direction of R then:

tan(\theta )=\frac{R_x}{R_y} ⇒ \theta = arctan(\frac{R_x}{R_y}) ⇒ \theta = 84.86^\circ.

The distance will be given by the magnitud of the vector R:

R=\sqrt{R_x^2 + R_y^2} ⇒ R=\sqrt{R_x^2 + R_y^2} = 250.75.

4 0
3 years ago
What kind of energy does an unlit match have?
Rudiy27
It’s is none of the above answers.
It is d. Potential energy
6 0
3 years ago
What is the frequency of this wave?
Anastasy [175]
In telecommunication systems, Carrier frequency is a technical term used to indicate: ... The frequency of the unmodulated electromagnetic wave at the output of a conventional amplitude-modulated (AM-unsupressed carrier), or frequency-modulated (FM), or phase-modulated (PM) radio transmitter.
6 0
3 years ago
Read 2 more answers
Which of the following is a subsurface event takes place during the rock cycle
alukav5142 [94]

Answer:

The answer is A. Cementing...

Explanation:

hope this helps

4 0
3 years ago
Read 2 more answers
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