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Elena-2011 [213]
3 years ago
12

Equipotential surfaces a) make an angle of 45 degrees with the electric field. b) are parallel to the electric field. c) are per

pendicular to the electric field. d) have no relationship to the electric field
Physics
1 answer:
Schach [20]3 years ago
7 0

Answer:

Option c) are perpendicular to the electric field

Explanation:

Equipotential surfaces are perpendicular to the electric field. the electric field lines are projected outwards from the equipotential surface, i.e., the lines of the electric field are at 90^{\circ} to the equipotential surface.

Equipotential surface are those surfaces that have the same potential at any point on the surface. Thus the potential difference at any point on the surface is zero due to same potential.

Any charge particle on this surface will move in a perpendicular direction to the Coulombian force. No work is done by the force on a particle moving on an equipotential surface.

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Convert the following dB to decimal a. -12 b. 3 c. 10 d. 0
stiks02 [169]

Answer:

for -12db

 \frac{V1}{V2}=0.251\\\\\frac{P1}{P2}=0.0631

for 3db

\frac{V1}{V2}=\sqrt{2}\\\\\frac{P1}{P2}=2

for 10db

\frac{V1}{V2}=3.16\\\\\frac{P1}{P2}=10

for 0db

\frac{V1}{V2}=1\\\\\frac{P1}{P2}=1

Explanation:

The decibel is a logaritmic value given by:

db=10*log(\frac{P1}{P2})=20*log(\frac{V1}{V2})

we use 10 for power values and 20 for other values such voltages or currents.

\frac{V1}{V2}=10^{\frac{db}{10}}\\\\\frac{P1}{P2}=10^{\frac{db}{20}}

for -12db

\frac{V1}{V2}=10^{\frac{-12}{10}}=0.251\\\\\frac{P1}{P2}=10^{\frac{-12}{20}}=0.0631

for 3db

\frac{V1}{V2}=10^{\frac{3}{10}}=\sqrt{2}\\\\\frac{P1}{P2}=10^{\frac{3}{20}}=2

for 10db

\frac{V1}{V2}=10^{\frac{10}{10}}=3.16\\\\\frac{P1}{P2}=10^{\frac{10}{20}}=10

for 0db

\frac{V1}{V2}=1\\\\\frac{P1}{P2}=1

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A 0.25 kg softball has a velocity of 13 m/s at an angle of 46° below the horizontal just before making contact with the bat. Wha
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Answer:

a. 2.959kgm/s

b. 7.63kgm/s

Explanation:

We were given Mass= 0.25kg

Velocity = 13m/s

Angle = 46°

a) Along the x axis we have =

0- 0.25(13)cos(46°) = (- 2.26)i kgm/s

Along the y axis we have =

-0.25(20) +0.25(13)sin(46°) = (-1.91)j kgm/s

Magnitude of change in momentum

=√ i²+ j²

= √ (-2.26)²+ ( -1.91)²

= √8.7557

= 2.959 kgm/s

b.

Along the x axis

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Along the y axis

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Magnitude of change in momentum

= √ i²+ j²

= √( -7.26)² + (-2.34)²

= √58.1832

= 7.63kgm/s

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