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zhenek [66]
2 years ago
7

What unit is used for electromagnetic fields.

Physics
1 answer:
eimsori [14]2 years ago
7 0
I believe the answer is hertz.
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A differnece in electric potential is required for an electric charge to flow through a wire.
cupoosta [38]

Answer:

TRUE

Explanation:

6 0
3 years ago
Read 2 more answers
Convertir:<br> A. 3Km a m<br> B. 250 ma Km<br> C. 1000Cm a m<br> D. 10000 mm a Cm
Katen [24]

Answer:

A. 3,000,000 m

B. 0.25 km

C. 10 m

D. 1,000 cm

Explanation:

no hablo español, así que solo ingrese esto en el traductor de G*ogle

A. One kilometer equals 1000 meters, so

3,000*1,000 = 3,000,000 m

B. One meter equals 0.001 kilometer, so

250*0.001 = 0.25 km

C. One centimeter equals 0.01 meter

1,000*0.01 = 10 m

D. One milimeter equals 0.1 centimer, so

10,000*0.1 = 1,000

4 0
3 years ago
Two stationary point charges of 3.00 nC and 2.00 nC are separated by a distance of 50.0 cm. An electron is released from rest at
andrew-mc [135]

Answer:

1. the electric potential energy of the electron when it is  at the midpoint is - 2.9 x 10^{-17} J

2. the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge is - 5.04 x  10^{-17} J

Explanation:

given information:

q_{1} =  3 nC = 3 x 10^{-9} C

q_{2} =  2 nC = 2 x 10^{-9} C

r = 50 cm = 0.5 m

the electric potential energy of the electron when it is  at the midpoint

potential energy of the charge, F

F = k \frac{q_{e}q}{r}

where

k = constant (8.99 x 10^{9} Nm^{2} /C^{2})

electron charge, q_{e} = - 1.6 x 10^{-19} C

since it is measured at the midpoint,

r = \frac{0.5}{2}

  = 0.25 m

thus,

F = F_{1}+ F_{2}

  = k\frac{q_{e} q_{1} }{r} + k\frac{q_{e} q_{2} }{r}

  = \frac{kq_{e} }{r} (q_{1} +q_{2})

  = (8.99 x 10^{9})( - 1.6 x 10^{-19} )(3 x 10^{-9} +2 x 10^{-9})/0.25

  = - 2.9 x 10^{-17} J

the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge

r_{1} = 10 cm = 0.1 m

r_{2} = 0.5 - 0.1 = 0.4 m

F = k\frac{q_{e} q_{1} }{r} + k\frac{q_{e} q_{2} }{r}

  = kq_{e}(\frac{q_{1} }{r_{1} }+\frac{q_{2} }{r_{2} })

  = (8.99 x 10^{9})( - 1.6 x 10^{-19} )(3 x 10^{-9} /0.1+2 x 10^{-9}/0.4)

  = - 5.04 x  10^{-17} J

3 0
3 years ago
Explain how an alpine glacier can change the topography of a mountainous area
igor_vitrenko [27]
<span>An alpine glacier can change the topography of a mountainous area through Glacial Erosion and Glacial Deposition. Glaciers are agents of erosion, it can pick up and carry large rocks and sediments. In the process, a deep cavity or hole can form when the glacier plucks a big rock from where it passed. Glaciers have shaped many Mountain Ranges and have created distinct landforms by its erosion process.  In Glacial Deposition, as glaciers melt, it deposits all that it carried and a landform is developed.</span>
7 0
3 years ago
An electric fan is running on HIGH. After fan has been running for 1.3 minutes, the LOW button is pushed. The fan slows down to
mafiozo [28]

Answer:

    wo = 18.75 rev / s

Explanation:

This is an exercise in endowment kinematics, it indicates that the final angular velocity is w_f = 109 rad / s, the time to reach this velocity is t = 1.87 s and the deceleration a = 4.7 rad / s²

         w_f = w₀ - a t

         w₀ = w_f + a t

         w₀ = 109 + 4.7 1.87

         w₀ = 117.8 rad / s

let's reduce to revolutions / s

         w₀ = 117.8 rad / s (1 rev / 2pi rad)

         w₀ = 18.75 rev / s

8 0
3 years ago
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