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AVprozaik [17]
3 years ago
13

In which scenario is elastic potential energy present? A bird sits in its nest in a tree. A snake slithers through a leaf pile.

A dog sniffs at the base of a tree. A fly stretches a spider web.
Physics
1 answer:
aleksklad [387]3 years ago
8 0

Answer: A fly stretches a spider web.

Elastic potential energy is an energy which is required to distort the shape and volume of an elastic flexible object. For example energy required to stretch the spring, the spring will not come back to it's original shape after this stretching process. A fly stretches a spider web is also the example of elastic potential energy because fly is applying elastic potential energy to destroy the spider web.

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A force of 60 N is used to stretch two springs that are initially the same length. Spring A has a spring constant of 4 N/m, and
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We can calculate the length of each spring by using the relationship:
F=kx
where
F is the force applied to the spring
k is the spring constant
x is the length of the spring (measured with respect to its rest position)

Re-arranging the equation, we have
x= \frac{F}{k}

The force applied to both spring is F=60 N. Spring A has spring constant of k=4 N/m, therefore its length with respect to its rest position is
x_A= \frac{F}{k_A}= \frac{60 N}{4 N/m}=15 m
Spring B has spring constant of k=5 N/m, so its length with respect to its rest position is
x_B= \frac{F}{k}= \frac{60 N}{5 N/m}=12 m

Therefore, the correct answer is
<span>D.Spring A is 3 m longer than spring B because 15 – 12 = 3.</span>
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3 years ago
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Which is the best aerobic exercise plan?
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Answer:

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(a) Calculate the force needed to bring a 950-kg car to rest from a speed of 90.0 km/h in a distance of 120 m (a fairly typical
musickatia [10]

(a)

mass of the car: m=950 kg

Initial speed: u = 90.0 km/h=25 m/s

Final speed: v=0 (the car comes to rest)

distance: S=120 m

We can find the acceleration by using the following SUVAT equation:

v^2 -u^2 =2aS

Re-arranging it and replacing the numbers, we find the acceleration

a=\frac{v^2-u^2}{2S}=\frac{0-(25 m/s)^2}{2(120 m)}=-2.6 m/s^2

So now we can calculate the force using Newton's second law:

F=ma=(950 kg)(-2.6 m/s^2)=-2470 N

And the negative sign means the force is applied against the direction of motion.

(b)

In this case, the distance is different:

S=2.00 m

so, the acceleration in this case is

a=\frac{v^2-u^2}{2S}=\frac{0-(25 m/s)^2}{2(2 m)}=-156.3 m/s^2

And so, the force applied in this case is

F=ma=(950 kg)(-156.3 m/s^2)=-148 500 N

which is much larger than the force exerted in the previous exercise.


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Answer:

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