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vova2212 [387]
3 years ago
8

The atmosphere of Neptune and Uranus have a blue color because of which gas?

Physics
1 answer:
kirza4 [7]3 years ago
5 0
They are blue because of hydrogen helium and methane
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Why are vectors so important?
Zina [86]
Vectors represent relationships according to the position, velocity, and acceleration (of a moving object).
Because so many physical things are vectors, you have to be able to add and subtract them, helping you understand how the world around you behaves.

I had to do some research (I was never good with physics) so if you have a couple questions, I will try to explain.
6 0
3 years ago
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 disk ring of inner radius a and outer radius b lying on the y − z plane has a uniform surface electric charge density +σ(&gt;
solmaris [256]

Answer:

a)  V = k 2π σ (√(b² + x²) - √ (a² + x²)) ,  

b)  E = - k 2π σ x (1 /√(b² + x²) - 1 /√(a² + x²))

Explanation:

a) The expression for the electric potential is

        V = k ∫ dq / r

For this case, consider the disk formed by a series of concentric rings of radius r and width dr, the distance of each ring to point P

         R = √(x² + r²)

The charge on a ring is

        σ = dq / dA

The area of ​​a ring is

        A = π r

        dA = 2π r dr

So the charge is

        dq = σ  2π r dr

We substitute

       V = k σ 2pi ∫ r dr / √(r² + x²)

We integrate

       V = k 2π σ √(r² + x²)

We evaluate from the lower limit r = a to the upper limit r = b

      V = k 2π σ (√(b² + x²) - √ (a² + x²))

 

b) the electric field and the potential are related

        E = - dV / dx

        E = - k 2π σ (1/2 2x /√(b² + x²) - ½ 2x /√(a² + x²))

        E = - k 2π σ x (1 /√(b² + x²) - 1 /√(a² + x²))

7 0
3 years ago
A bear fells from 10m long grass <br> From where the bear is
Elina [12.6K]
He from the 10m long grass
3 0
3 years ago
A sound wave has a frequency of 300 Hz. If the wavelength is .50 m, then what is the speed of
kherson [118]

Answer:

15000 m/s

Explanation:

You just need to multiply the wavelength with the frequency.

7 0
3 years ago
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