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MariettaO [177]
3 years ago
12

URGENT HELP IT IS DUE BY TONIGHT PLEASE HELP

Physics
1 answer:
Law Incorporation [45]3 years ago
5 0

Answer:

45.8

Explanation:

becuse 9.8+36=45.8 simple

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Mercury's surface has many craters because _____.
Alexeev081 [22]
I think it would be that it has no atmosphere
5 0
3 years ago
For each of the following decays or reactions, name at least. one conservation law that prevents it from occurring.(c) p → π⁺ +
Amanda [17]

(c)p→π⁺₊π⁺₊π

Baryon number is +1 on the left side of the equation, 0 on the

right side. Baryon number is not conserved.

<h3>How do you determine whether a baryon number is conserved?</h3>

  • According to the law of conservation of baryon number, the sum of the baryon numbers of all incoming particles equals the sum of the baryon numbers of all particles produced by the reaction. Energy, and so on, are conserved even if the incoming proton has sufficient energy and charge.
<h3>What is Baryon Number</h3>
  • In particle physics, the baryon number denotes which particles are baryons and which particles are not. Each baryon has a baryon number of 1, and each antibaryon has a baryon number of -1. Other non-baryonic particles have a baryon number of 0. Since there are exotic hadrons like pentaquarks and tetraquarks, there is a general definition of baryon number as:
  • B=1/3(n_{q} -n_q^{-} })
  • wheren_{q} represents the number of quarks and nq represents the number of antiquarks.

To learn more about Baryon Number refer to

brainly.com/question/10358797

#SPJ4

4 0
2 years ago
As the mass and distance between two objects decrease, so does the gravitational force exerted between them. true or false
marta [7]
I believe it's true,..hope this helps!

7 0
3 years ago
Find the range of a projectile launched at an angle of 30° with an initial velocity of 20m/s.​
Tems11 [23]

Answer:

<em>The range is 35.35 m</em>

Explanation:

<u>Projectile Motion</u>

It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.

Being vo the initial speed of the object, θ the initial launch angle, and g=9.8m/s^2 the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:

\displaystyle d={\frac  {v_o^{2}\sin(2\theta )}{g}}

The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:

\displaystyle d={\frac  {20^{2}\sin(2\cdot 30^\circ )}{9.8}}

\displaystyle d={\frac  {400\sin(60^\circ )}{9.8}}

d=35.35\ m

The range is 35.35 m

7 0
3 years ago
Given a double slit apparatus with slit distance 2 mm, what is the theoretical maximum number of bright spots that I would see w
katen-ka-za [31]

Answer:

The values is  m_{max} = 8001 \  bright \ spots

Explanation:

From the question we are told that

    The slit distance is  d  =  2 \ mm  =  2*10^{-3} \ m

    The  wavelength is  \lambda =  500 \ nm =  500 *10^{-9} \ m

At the first half of the screen from the central maxima

   The number of bright spot according to the condition for constructive interference is  

          n  =  \frac{d *  sin (\theta )}{\lambda}

For maximum number of spot \theta =  90^o

So  

       n  =  \frac{2*10^{-3} *  sin (90 )}{500 *10^{-9}}

        n  =4000

Now for the both sides plus the central maxima  we have

      m_{max} = 2 * n  + 1

substituting values

       m_{max} = 2 *  4000 + 1

       m_{max} = 8001 \  bright \ spots

   

6 0
4 years ago
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