Answer: Atoms are electrically neutral because they have equal numbers of protons (positively charged) and electrons (negatively charged). If an atom gains or loses one or more electrons, it becomes an ion.

8 0

3 years ago

*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude

kvv77 [185]

Answer:

Approximately $3.86\; {\rm N}$ (given that the magnitude of this charge is $-7.35\; {\rm \mu C}$ .)

Explanation:

If a charge of magnitude $q$ is placed in an electric field of magnitude $E$ , the magnitude of the electrostatic force on that charge would be $F = E\, q$ .

The magnitude of this charge is $q = 7.35\; {\rm \mu C}$ . Apply the unit conversion $1\; {\rm \mu C} = 10^{-6}\; {\rm C}$ :

$\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}$ .

An electric field of magnitude $E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}}$ would exert on this charge a force with a magnitude of:

$\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}$ .

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

4 0

2 years ago

How much force (in Newtons) acts on a 1700 kg car going around a

Daniel [21]

Answer:

b) 4781 N

Explanation:

Because there is a redius do this question is talking about the acceleration force which= mv^2/r

so a=15^2/80=2.8125 m^2/s

so the force will be = m.a

F =1700×2.8125=4781.25 N

5 0

3 years ago

True or Flase The fastest moving traffic on the expressway will be traveling in the right lane

Yakvenalex [24]

I think this answer is False because i believe the right lane is the slowest lane you could be in.

5 0

3 years ago