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garri49 [273]
3 years ago
5

Select True or False for the follwing statements about electric field lines.

Physics
1 answer:
aalyn [17]3 years ago
5 0

Answer:

Please see below as the answer is self explanatory.

Explanation:

1) A positive point charge released from rest will initially accelerate along an E-field line : TRUE.

The electric field lines have the same direction that will have a positive test charge, when acted on by the force exerted by the charge distribution that creates the field.

2) E-Field lines make circles around positive charges: FALSE.

E-Field lines start in positive charges and finish in negative charges, except at infinity.

3) E-field lines point inward toward negative charges: TRUE

As the E-field lines follow the trajectory that would have a positive test charge, as negative charges attract the positive charges, the positive test charge will travel towards the negative charges.

4) E-Field line may cross: FALSE

As the E-Field is the gradient of the electric potential (so they have the direction of the maximum change of the potential), if  they may cross, this would mean that the potential could be changing differently at the same point, which it is not possible.

5) E-Field lines point outward from positive charges: TRUE

As explained above, the E-field lines follow the trajectory that it would have a positive test charge, so if the field is created by a positive charge, the test charge would go away from it, so the field lines point outward from positive charges.

6) Where the E-field lines are dense the E-field must be weak : FALSE

By convention, the E-field lines are proportional to the charge that creates the field, so if the E-field lines are dense, this means that the field is strong.

7) E-field lines do not begin or end in a charge-free region except at infinity: TRUE.

Except at infinity, E-field lines originate at positive charges and finish at negative charges, so they can't originate in a charge-free region.

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<span>Inertia keeps us orbiting because any object with mass has the tendency to resist changes to their direction and speed of movement. Combine that with the interaction of the gravitational attraction of the sun, and that is what keeps Earth in orbit. The sun’s gravitational force is one that is proportional to Earth’s mass, and it acts in a way that is almost exactly perpendicular to Earth’s motion. This keeps Earth from spinning into the sun or far away from it.</span>
6 0
3 years ago
Acceleration is the magnitude of average velocity.​
lina2011 [118]

Answer:

false

Explanation:

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4 0
3 years ago
A commuter backs her car out of her garage with an acceleration of 1.40 m/s^2. (a) How long does it take her to reach a speed of
jonny [76]

Answer:

a) It takes her 1.43 s to reach a speed of 2.00 m/s.

b) Her deceleration is - 2.50 m/s²

Explanation:

The equation of velocity for an object that moves in a straight line with constant acceleration is as follows:

v = v0 + a · t

Where:

v = velocty.

v0 = initial velocity.

a = acceleration.

t = time.

a) Using the equation of velocity, let´s consider that the car moves in the positive direction. Then:

v = v0 + a · t

2.00 m/s = 0 m/s + 1.40 m/s² · t

t = 2.00 m/s / 1.40 m/s²

t = 1.43 s

It takes her 1.43 s to reach a speed of 2.00 m/s

b) Let´s use again the equation of velocity, knowing that at t = 0.800 s the velocity is 0 m/s:

v = v0 + a · t

0 = 2.00 m/s + a · 0.800 s

-2.00 m/s / 0.800 s = a

a = -2.50 m/s²

Her deceleration is - 2.50 m/s²

7 0
3 years ago
At t=0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When 3s. bullet B is fired upward with a muzzl
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Answer:

At time 10.28 s after A is fired bullet B passes A.

Passing of B occurs at 4108.31 height.

Explanation:

Let h be the height at which this occurs and t be the time after second bullet fires.

Distance traveled by first bullet can be calculated using equation of motion

s=ut+0.5at^2 \\

Here s = h,u = 450m/s a = -g and t = t+3

Substituting

h=450(t+3)-0.5\times 9.81\times (t+3)^2=450t+1350-4.9t^2-29.4t-44.1\\\\h=420.6t-4.9t^2+1305.9

Distance traveled by second bullet

Here s = h,u = 600m/s a = -g and t = t

Substituting

h=600t-0.5\times 9.81\times t^2=600t-4.9t^2\\\\h=600t-4.9t^2 \\

Solving both equations

600t-4.9t^2=420.6t-4.9t^2+1305.9\\\\179.4t=1305.9\\\\t=7.28s \\

So at time 10.28 s after A is fired bullet B passes A.

Height at t = 7.28 s

h=600\times 7.28-4.9\times 7.28^2\\\\h=4108.31m \\

Passing of B occurs at 4108.31 height.

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Which prediction of weather is the most accurate for the next two days if there is an occluded front over the area?
alisha [4.7K]

The most accurate weather for the next two days would be dry warm weather or severe thunderstorms if there is an occluded front over the area.

<u>Explanation:</u>

The weather front that is created during the cyclogenesis process is an occluded front in meteorology. Cyclogenesis process is the development of extra-tropical cyclone and its intensification.

During the occurrence of this, the warm air is occluded (separated) from the center of cyclone at the surface of the earth.

The cold front rotates the storm as it intensifies and holds up the warm front. This develops an occluded front, that is the boundary which separates the new cold air mass and the older cool air mass that is already in warm front's north.

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