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Setler [38]
2 years ago
5

With a thunderstorm brewing, an electric field of magnitude 2.0 × 102 newtons/coulomb exists at a certain point in the earth’s a

tmosphere. If an electron at that point has a charge of -1.6 × 10-19 coulombs, what is the magnitude of the force on the electron as a result of the electric field?
A.
2.0 × 10-6 newtons
B.
3.2 × 10-17 newtons
C.
5.8 × 105 newtons
D.
6.4 × 10-3 newtons
E.
1.7 × 104 newtons
Physics
1 answer:
Aleks04 [339]2 years ago
5 0

Answer:

The correct option is (b).

Explanation:

Given that,

Electric field, E=2\times 10^2\ N/C

We need to find the magnitude of the force on the electron as a result of the electric field.

We know that, the electric force is given by :

F=qE\\\\=1.6\times 10^{-19}\times 2\times 10^2\\\\F=3.2\times 10^{-17}\ N

So, the required force on the electron is equal to 3.2\times 10^{-17}\ N.

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Moving a neutral wire in a(n) ____ field will induce a(n) _____.
frutty [35]

the answer is b) magnetic current

7 0
3 years ago
Two square air-filled parallel plates that are initially uncharged are separated by 1.2 mm, and each of them has an area of 190
julia-pushkina [17]

Answer:

5.5\cdot 10^{-11} C

Explanation:

The capacitance of the parallel-plate capacitor is given by:

C=\epsilon_0 \frac{A}{d}

where

\epsilon_0 = 8.85\cdot 10^{-12} F/m is the vacuum permittivity

A=190 mm^2 = 190 \cdot 10^{-6} m^2 is the area of the plates

d=1.2 mm = 0.0012 m is the separation between the plates

Substituting,

C=(8.85\cdot 10^{-12}) \frac{190 \cdot 10^{-6}}{0.0012}=1.4\cdot 10^{-12}F

The energy stored in the capacitor is given by

U=\frac{Q^2}{2C}

Since we know the energy

U=1.1 nJ = 1.1 \cdot 10^{-9} J

we can re-arrange the formula to find the charge, Q:

Q=\sqrt{2UC}=\sqrt{2(1.1\cdot 10^{-9} J)(1.4\cdot 10^{-12}F )}=5.5\cdot 10^{-11} C

8 0
2 years ago
Please help me with this physics prooblem
zaharov [31]

Take the missile's starting position to be the origin. Assuming the angles given are taken to be counterclockwise from the positive horizontal axis, the missile has position vector with components

x=v_0\cos20.0^\circ t+\dfrac12a_xt^2

y=v_0\sin20.0^\circ t+\dfrac12a_yt^2

The missile's final position after 9.20 s has to be a vector whose distance from the origin is 19,500 m and situated 32.0 deg relative the positive horizontal axis. This means the final position should have components

x_{9.20\,\mathrm s}=(19,500\,\mathrm m)\cos32.0^\circ

y_{9.20\,\mathrm s}=(19,500\,\mathrm m)\sin32.0^\circ

So we have enough information to solve for the components of the acceleration vector, a_x and a_y:

x_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\cos20.0^\circ(9.20\,\mathrm s)+\dfrac12a_x(9.20\,\mathrm s)^2\implies a_x=21.0\,\dfrac{\mathrm m}{\mathrm s^2}

y_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\sin20.0^\circ(9.20\,\mathrm s)+\dfrac12a_y(9.20\,\mathrm s)^2\implies a_y=110\,\dfrac{\mathrm m}{\mathrm s^2}

The acceleration vector then has direction \theta where

\tan\theta=\dfrac{a_y}{a_x}\implies\theta=79.2^\circ

5 0
2 years ago
HELP ME PLSSS I WILL GIVE YOU ANYTHING HELPPPPPPPP MEEEEEEEE
Minchanka [31]

Answer:

B)

Explanation:

I hope this help's :)

5 0
3 years ago
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When an object is in orbit, it is falling at the same rate at which the Earth is curving. Please select the best answer from the
olga2289 [7]

The statement “When an object is in orbit, it is falling at the same rate at which the Earth is curving” is true. The speed of a satellite orbiting the earth depends only on the mass of the earth and the mass of the satellite.

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