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katen-ka-za [31]
2 years ago
11

A car is being driven along a road at 25m/s when the driver suddenly notices that there is a fallen tree blocking the road 65m a

head. The driver immediately applies the brakes giving the car a constant velocity of 5m/s2. How far in front of the tree does the car come to rest?
Physics
1 answer:
Anuta_ua [19.1K]2 years ago
3 0

Hi there!

We can use the following equation to solve for displacement:

vf² = vi² + 2ad, where:

vf = final velocity

vi = initial velocity

a = acceleration

d = displacement

Since the car will come to rest, vf = 0, so:

0 = vi² + 2ad

Plug in the given values:

0 = 25² + 2(-5)d (Breaking is a negative acceleration in this instance)

Solve:

0 = 25² - 10d

10d = 625

d = 62.5m

Subtract from the total distance from the car to the tree:

65 - 62.5 = 2.5m

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Answer:

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Explanation:

Given:

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The force exerted by the power lines on each other is given by the relation:

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F = \frac{4\pi\times10^{-7}\times220\times130\times130 }{2\pi\times0.4}

F = 1.86 N

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Driving along a boring stretch of interstate in Illinois, you start experimenting using the average speed equation you learned i
astra-53 [7]
The average speed would be 33.29m/s.
The average speed equation is:

Average speed =  \frac{total distance}{total time}

First you will need to solve for the distance you traveled in each scenario. So we can solve this by getting the product of speed and the time traveled. 

Scenario 1:
Speed = 29m/s
Time = 120s
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Distance = (29m/s)(120s)
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Scenario 2
Speed = 35m/s
Time = 300s
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Distance = (35m/s)(300s)
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Now that you have the distance of both, you can solve for your average speed. 

Average speed = \frac{total distance}{total time}
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1. On each of your equipotential maps, draw some electric field lines with arrow heads indicating the direction of the field. (H
JulijaS [17]

Answer:

The angle between the electric field lines and the equipotential surface is 90 degree.

Explanation:

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The electric field lines are always perpendicular to the equipotential surface.

As

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For equipotential surface, dV = 0 so

0 = \overrightarrow{E} . d\overrightarrow{r}\\\\

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2 years ago
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