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zalisa [80]
3 years ago
12

The electric field in a particular thundercloud is 3.8 x 105 N/C. What is the acceleration (in m/s2) of an electron in this fiel

d? (Enter the magnitude.) m/s2
Physics
1 answer:
Svetradugi [14.3K]3 years ago
8 0

Answer:

Answer:

6.68 x 10^16 m/s^2

Explanation:

Electric field, E = 3.8 x 10^5 N/C

charge of electron, q = 1.6 x 10^-19 C

mass of electron, m = 9.1 x 10^-31 kg

Let a be the acceleration of the electron.

The force due to electric field on electron is

F = q E

where q be the charge of electron and E be the electric field

F = 1.6 x 10^-19 x 3.8 x 10^5

F = 6.08 x 10^-14 N

According to Newton's second law

Force  = mass x acceleration

6.08 x 10^-14 = 9.1 x 10^-31 x a

a = 6.68 x 10^16 m/s^2

Explanation:

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Answer:

opinion a

Explanation:

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A spring is 6.0cm long when it is not stretched, and 10cm long when a 7.0N force is applied. What force is needed to make it 20c
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Answer:

Approximately 25\; {\rm N} (assuming that this spring is ideal.)

Explanation:

The displacement of a spring is the new length of the spring relative to the original length.

For example:

  • When the 6.0\; {\rm cm}-spring in this question is stretched to 10\; {\rm cm}, the displacement is x = (10\; {\rm cm} - 6.0\; {\rm cm}).
  • Likewise, if this spring is stretched to 20\; {\rm cm}, the displacement would be (20\; {\rm cm} - 6\; {\rm cm}).

If this spring is ideal, the force on the spring would be proportional to the displacement of the spring. In other words, if a force of F_{\text{a}} displaces this spring by x_{\text{a}}, while a force of F_{\text{b}} displaces this spring by x_{\text{b}}, then:

\displaystyle \frac{F_{\text{a}}}{x_{\text{a}}} = \frac{F_{\text{b}}}{x_{\text{b}}}.

In this question, it is given that a force of F_{\text{a}} = 7.0 \; {\rm N} would stretch this spring by x_{\text{a}} = (10\; {\rm cm} - 6.0\; {\rm cm}). Thus, the force F_{\text{b}} required to stretch this spring by x_{\text{a}} = (20\; {\rm cm} - 6.0\; {\rm cm}) would satisfy:

\displaystyle \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}}= \frac{F_{\text{b}}}{20\; {\rm cm} - 6.0\; {\rm cm}}.

Rearrange and solve for F_{\text{b}}:

\begin{aligned} F_{\text{b}} &= \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}} \, (20\; {\rm cm} - 6.0\; {\rm cm}) \\ &\approx 25\; {\rm N}\end{aligned}.

7 0
2 years ago
A car travelling at a constant speed of 70km/h passes a stationary police car. The police car immediately goes on the chase acce
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Answer:

18.24 seconds

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You know it took 10 seconds for the police to reach 85 km/h. Calculate the distance that the car is ahead of the police (175/9)*10=1750/9m. Then by divide 1750/9 with 425/18, you will get the value 8.24. Add the 10 seconds with the 8.24 you will get 18.24 sec which is the total time.

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I would believe that it would be the last option

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