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aleksklad [387]
3 years ago
7

When a certain rubber band is stretched a distance x, it exerts a restoring force of magnitude f = ax, where a is a constant. th

e work done by a person in stretching this rubber band from x = 0 to x = l is:?
Physics
1 answer:
Veseljchak [2.6K]3 years ago
4 0
Given:
F = ax
where
x = distance by which the rubber band is stretched
a =  constant

The work done in stretching the rubber band from x = 0 to x = L is
W=\int_{0}^{L} Fdx = \int_{0}^{L}ax \, dx = \frac{a}{2}  [x^{2} ]_{0}^{L} =  \frac{aL^{2}}{2}

Answer:  \frac{aL^{2}}{2}

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You are driving down the highway late one night at 20 m/s when a deer steps onto the road 35m in front of you. Your reaction tim
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Answer:

- Distance between car and the deer when the car stopped = 20 m

- The time required for you to stop once you press the brakes = less than 5 s in order not to hit the deer.

Explanation:

Using the equations of motion,

In the 0.5 s reaction time, we need to first calculate how far he has travelled in that time.

a = 0 m/s² (Since the car is travelling at constant velocity)

x = ?

Initial velocity = u = 20 m/s

x = ut + at²/2

x = 20×0.5 + 0 = 10 m

From that moment,

a = - 10 m/s²

u = initial velocity at the start of the deceleration = 10 m/s

v = final velocity = 0 m/s

x = ?

v² = u² + 2ax

0² = 10² + 2(-10)(x)

20x = 100

x = 5 m

Total distance travelled from when the deer stepped onto the road = 10 + 5 = 15 m

Distance between car and the deer when the car stopped = 35 - 15 = 20 m

b) To determine the time required to stop once you step on the brakes

u = 10 m/s

t = ?

v = 0 m/s²

x = distance from when the brake was stepped on to the deer = 35 - 10 = 25 m

x = (u + v)t/2

25 = (10 + 0)t/2

10t = 50

t = 5 s

Meaning the time required to stop once you step on the brakes is less than 5s.

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