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Mars2501 [29]
3 years ago
6

Where is each subatomic particle is found in an atom

Physics
1 answer:
Ronch [10]3 years ago
7 0
Protons and neutrons are located in the nucleus, a dense central core in the middle of the atom, while the electrons are located outside the nucleus.
You might be interested in
When an atom releases gamma radiation____________.
wel

Answer:

d. the atomic number remains the same.

Explanation:

First at all it's important to know how to read nuclear information

_{Z}^{A}X

X is the atomic symbol, A the mass number and Z the atomic number of the element.

Gamma rays emitted on gamma decay are characterized as _{0}^{0}\gamma

If we write the nuclear equation for the decay, we have that:

_{Z}^{A}X\,\rightarrow{}_{0}^{0}\gamma\,+\,{}_{Z}^{A}Y

The sum of the mass numbers and atomic numbers on the right side has to be equal to the left side numbers of the equation, that means the mass number and the atomic number remains the same for the resulting atom to preserve the equality.

3 0
3 years ago
Saturn moves in an orbit around the Sun with radius 10 AU. How many degrees does it move on the Celestial in one year? (Hint: Ca
Lana71 [14]

Answer:

B. About 12 degrees

Explanation:

The orbital period is calculated using the following expression:

T = 2π*(\sqrt{\frac{r^3}{Gm}})

Where r is the distance of the planet to the sun, G is the gravitational constant and m is the mass of the sun.

Now, we don't actually need to solve the values of the constants, since we now that the distance from the sun to Saturn is 10 times the distance from the sun to the earth. We now this because 1 AU is the distance from the earth to the sun.

Now, we divide the expression used to calculate the orbital period of Saturn by the expression used to calculate the orbital period of the earth. Notice that the constants will cancel and we will get the rate of orbital periods in terms of the distances to the sun:

\frac{Tsaturn}{Tearth} = \sqrt{\frac{rSaturn^3}{rEarth^3} } = \sqrt{10^3}}

Knowing that the orbital period of the earth is 1 year, the orbital period of Saturn will be \sqrt{10^3}} years, or 31.62 years.

We find the amount of degrees it moves in 1 year:

1year * \frac{360degrees}{31.62years} = 11.38 degrees

or about 12 degrees.

6 0
3 years ago
Alicia can row 6 miles downstream in the same time it takes her to row 4 miles upstream. She rows downstream 3 miles/hour faster
m_a_m_a [10]
Let us assume the upstream rowing rate of Alicia = x
Let us assume the downstream rowing rate of Alicia = y
We already know that
Travelling time = Distance traveled/rowing rate
Then
6/(x + 3) = 4/x
6x = 4x + 12
6x - 4x = 12
2x = 12
x = 6
Then
Rowing rate of Alicia going upstream = 6 miles per hour
Rowing rate of Alicia going downstream = 9 miles per hour.
4 0
3 years ago
Read 2 more answers
The fastest pitched baseball was clocked at 46 m/s. assume that the pitcher exerted his force (assumed to be horizontal and cons
Zielflug [23.3K]
Using the Equation:
                                 v² = vi² + 2 · a · s    → Eq.1
where,
v = final velocity 
vi = initial velocity 
a = acceleration 
s = distance 

<span><span>We know that vi = 0 because the ball was at rest initially.
</span><span>
Therefore,

Solving Eq.1 for acceleration,
 
</span></span> v² = vi² + 2 · a · s
 v² = 0 + 2 · a · s
 v² = 2 · a · s
Rearranging for a,
a = v ²/2·<span>s
Substituting the values,
a = 46</span>²/2×1<span> 
a = 1058 m/s</span>² 

<span>Now applying Newton's 2nd law of motion,
 </span>
<span>F = ma
   = 0.145</span>×<span>1058

F = 153.4 N</span>
8 0
3 years ago
Based on Kepler's work, which best describes the orbit If a planet around the Sun?
krek1111 [17]

Answer:

an ellipse with the Sun at one focus  or D

Explanation:

Answer for edgenuity

7 0
3 years ago
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