1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
mario62 [17]
4 years ago
12

An element's atomic number is 7. How many electrons would an atom of this element have?

Physics
1 answer:
almond37 [142]4 years ago
5 0

Hey There!

The atomic number represents the electrons and protons, therefore, if the atomic number of the element is 7 then there are 7 electrons and protons in the element.

In order to determine the amount of neutrons, you would subtract the atomic number from the atomic mass.

Hope this helped!

You might be interested in
A student wearing frictionless in-line skates on a horizontal surface is pushed by a friend with a constant force of 45 N. How f
kiruha [24]
Work is obtained by multiplying the force and the object's displacement. The force and displacement and force should be in the same direction in order to have work. 
                                      W = F x d
                                     d = W / F
Substituting the known values,
                                     d = 352 J / 45 N = 7.82 m
Thus, the displacement of the student is 7.82 m. 
8 0
4 years ago
A ball is thrown horizontally from the top of a building 0.10 km high. The ball strikes the ground at a point 65 m horizontally
Alex787 [66]

Answer:

47 ms

well the time it takes to fall 100m is the same time it takes to travel 65m horizontally.

The time to fall vertically, t is sqrt(2d/g). [this comes from d = 1/2at^2]

so t = sqrt(2*100/9.8) = 4.52s.

The vertical speed at that time is g*t = 9.8*4.52 = 44.3m/s

The horizontal speed is the horizontal distance over the same 4.52s, = 65/4.52 = 14.4m/s.

so the final velocity is = sqrt(44.3^2 + 14.4^2) = 47.ms

4 0
3 years ago
A thin, uniformly charged insulating rod has a linear charge density λ = 3 nC/m and lies along the x axis from x = 1m to x = 3m.
Contact [7]

Answer:

A) V_A = 11.93~V

B) The vector definition of E-field is

\vec{E} = -1.13\^x + 2.41\^y

where magnitude is E = 2.66 N/m.

Explanation:

The potential of a uniformly charged rod can be found by the method of integration. We will first choose an infinitesimal part on the rod. We will compute the potential of this part at point A. Then we will integrate this potential over the entire rod.

We will use the following formula for electric potential:

V = \frac{1}{4\pi \epsilon_0}\frac{Q}{r}

Let us choose the infinitesimal part a distance 'x' from the origin. Then the distance between this point and point A is

r = \sqrt{x^2+4^2}

The infinitesimal length is 'dx', and the potential of this length is dV. Let's apply the formula:

dV = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{\sqrt{x^2 + 4^2}}

Here, the charge Q is equal to the charge density multiplied by the length. Q = λdx

Now we have to integrate this infinitesimal potential over the rod:

V = \int\limits^3_1 {dV} \, dx = \frac{1}{4\pi \epsilon_0}\int\limits^3_1 {\frac{\lambda}{\sqrt{x^2 + 16}} \, dx

By using an integral table, this can be calculated:

V = \frac{3\times 10^{-9}}{4\pi\epsilon_0}\ln(|\sqrt{x^2+16}+x|)\left \{ {{x=3} \atop {x=1}} \right. \\V = 11.93~V

B) The electric field can be found by a similar approach, but a different formula:

\vec{E} = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}\^r

Let's apply this formula to the infinitesimal part we have chosen.

dE_x = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{x^2 + 4^2}\cos(\theta)\\dE_y = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{x^2 + 4^2}\sin(\theta)

By the geometry sine and cosine terms can be found:

\sin(\theta) = \frac{4}{\sqrt{x^2+16}}\\\cos(\theta) = \frac{x}{\sqrt{x^2 + 16}}

The x- and y-components of the E-field can be found separately by integrating the infinitesimal parts over the entire rod.

E_x = \int\limits^3_1 {dE_x} \, dx = \frac{\lambda}{4\pi\epsilon_0}\int\limits^3_1 {\frac{x}{(x^2+16)^{3/2}}} \, dx  = 1.13(-\^x)\\E_y = \int\limits^3_1 {dE_y} \, dx = \frac{4\lambda}{4\pi\epsilon_0}\int\limits^3_1 {\frac{1}{(x^2+16)^{3/2}}} \, dx  = 2.41(\^y)

So, the final E-field is

\vec{E} = -1.13\^x + 2.41\^y

The magnitude of the E-field is

E = 2.66 N/m

6 0
3 years ago
As an electron approaches a proton, the electron's force of attraction...
Tems11 [23]

Answer: B. Increases

Explanation:

An electron is a negatively charged particle, while a proton is positively charged.

Opposites attract so as it approaches the proton the force of attraction will increase.

7 0
3 years ago
Write down the SI unit of length and mass​
Vanyuwa [196]

Answer:

The SI unit for length is meters(m), for mass is kilograms(kg)

Explanation:

hope it helps

6 0
3 years ago
Other questions:
  • Usually, when the particle size is decreased, what will happen to the rate of dissolving?
    10·2 answers
  • How to convert units of volume in physics?
    6·1 answer
  • How does science help society?
    14·2 answers
  • What is echo?.....................<br>​
    6·2 answers
  • Ben starts walking along a path at 3 3 mi/h. One and a half hours after Ben leaves, his sister Amanda begins jogging along the s
    10·1 answer
  • Someone help
    7·1 answer
  • A charge of 6.65 mC is placed at each corner of a square 0.500 m on a side.
    11·1 answer
  • When atoms are split, they release energy. This concept applies to (2 points)
    5·1 answer
  • What is Focal Point<br>help me please​
    12·2 answers
  • a mountain goat starts a rock slide and the rocks crash down the slope 100 m. If the rocks reach the bottom in 5 s, what is thei
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!