True or False? There are equal numbers of protons and electrons in neutral atoms.

Physics

2 answers:

Alecsey [184]3 years ago

8 0

Answer: True
Hope this helped

storchak [24]3 years ago

8 0

The answer is true since it is a neutral atom

You might be interested in

What two variables are multiplied together to calculate mass?

Aleksandr [31]

Volume x Density = Mass

Weight / Gravity = Mass

8 0

3 years ago

your friend sit in a sked in the snow. if you apply a force of 120 N to them, they have an acceleration of 1.3 m/s2. what is the

levacccp [35]

Need to know the equation for force
F=MA
F is force
M is mass- we need to know the mass
A is acceleration
use "x" for mass
120 N= 1.3x
divide 1.3 in both side
kg unit for mass
X=92.31 kg
or
mass = 92.31 kg
Hope this helps

7 0

3 years ago

A student analyzes data of the motion of a planet as it orbits a star that is in deep space. The orbit of the planet is consider

Valentin [98]

Explanation:

The orbit of the planet is considered to be stable and does not change over time. This means that the force of gravity and inertia are perfectly balanced. Leading to forward motion of Planet and moons under force of gravity of some other large body(Most probably stars).This is in turns leads to formation of orbit.

8 0

2 years ago

Read 2 more answers

Will students do better in school if teachers are

marissa [1.9K]

Higher pay for teachers means students do better When teachers get paid more, students do better. In one study, a 10% increase in teacher pay was estimated to produce a 5 to 10% increase in student performance. Teacher pay also has long-term benefits for students.

3 0

2 years ago

What is the potential difference between a point 0.48 mm from a charge of 2.9 nc and a point at infinity?

Nuetrik [128]

The potential at a distance r from a charge Q is given by
$V(r) = k_e \frac{Q}{r}$
where ke is the Coulomb's constant.

The charge in our problem is $Q=2.9 nC=2.9 \cdot 10^{-9} C$ ; for the point at $r=0.48 mm=0.48 \cdot 10^{-3} m$ , the potential is
$V_1 = k_e \frac{Q}{r}= (8.99 \cdot 10^9 Nm^2 C^{-2}) \frac{2.9 \cdot 10^{-9} C}{0.48 \cdot 10^{-3} m}= 5.43 \cdot 10^4 V$

For the point at infinity, we immediately see that the potential is zero, because $r= \infty$ and so $V_2 = 0$ .

Therefore, the potential difference between the two points is
$\Delta V = V_1 - V_2 = V_1 = 5.43 \cdot 10^4 V$

6 0

2 years ago