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babunello [35]
3 years ago
11

The charge imbalance that results from this movement of charge will generate an additional electric field in the region within t

he rod In what direction will this field point? A. It will point to the right and enhance the initial applied field B. It will point to the left and oppose the initial applied field
Physics
1 answer:
deff fn [24]3 years ago
8 0

Answer:

B. It will point to the left and oppose the initial applied field

Explanation:

Let the charge be positive and  and electric field be towards right . Charge will move towards right in the field due to which positive charge will accumulate towards right and excess of positive charge will lie there . This new charge distribution will create a field towards the left which is opposite to external electric field.

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A car is stopped at a traffic light. When the light turns green at t=0, a truck with a constant speed passes the car with a 20m/
s344n2d4d5 [400]

Answer:

At t = (70 / 3) \; {\rm s} (approximately 23.3 \; {\rm s}.)

Explanation:

Note that the acceleration of the car between t = 0\; {\rm s} and t = 20\; {\rm s} (\Delta t = 20\; {\rm s}) is constant. Initial velocity of the car was v_{0} = 0\; {\rm m\cdot s^{-1}}, whereas v_{1} = 35\; {\rm m\cdot s^{-1}} at t = 20\; {\rm s}\!. Hence, at t = 20\; {\rm s}\!\!, this car would have travelled a distance of:

\begin{aligned}x &= \frac{(v_{1} - v_{0})\, \Delta t}{2} \\ &= \frac{(35\; {\rm m\cdot s^{-1}} - 0\; {\rm m\cdot s^{-1}}) \times (20\; {\rm s})}{2} \\ &= 350\; {\rm m}\end{aligned}.

At t = 20\; {\rm s}, the truck would have travelled a distance of x = v\, t = 20\; {\rm m\cdot s^{-1}} \times 20\; {\rm s} = 400\; {\rm m}.

In other words, at t = 20\; {\rm s}, the truck was 400\; {\rm m} - 350\; {\rm m} = 50\; {\rm m} ahead of the car. The velocity of the car is greater than that of the truck by 35\; {\rm m\cdot s^{-1}} - 20\; {\rm m\cdot s^{-1}} = 15 \; {\rm m\cdot s^{-1}}. It would take another (50\; {\rm m}) / (15\; {\rm m\cdot s^{-1}}) = (10/3)\; {\rm s} before the car catches up with the truck.

Hence, the car would catch up with the truck at t = (20 + (10/3))\; {\rm s} = (70 / 3)\; {\rm s}.

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