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Molodets [167]
3 years ago
15

"A jack-in-the-box is a toy in which a figure in an open box is pushed down, compressing a spring. The lid of the box is then cl

osed. When the box is opened, the figure is pushed up by the spring. The spring in the toy is compressed 0.070 meter by using a downward force of 12.0 newtons.
Calculate the total amount of elastic potential energy stored in the spring when it is compressed. [Show all work, including the equation and substitution with units.] "

Physics
2 answers:
klio [65]3 years ago
8 0

Answer:

0.42 J

Explanation:

The elastic potential energy of a spring is given by:

U=\frac{1}{2}kx^2 (1)

where

k is the spring constant

x is the compression/stretching of the spring

In order to find the elastic potential energy, we must find the spring constant first. We know by Hook's law that the force compressing the spring is:

F=kx

Re-arranging the equation and substituting F=12.0 N and x=0.070 m, we find:

k=\frac{F}{x}=\frac{12.0 N}{0.070 m}=171.4 N/m

And now we can use eq.(1) to calculate the elastic potential energy:

U=\frac{1}{2}(171.4 N/m)(0.070 m)^2=0.42 J

Svetlanka [38]3 years ago
7 0
We know, Applied force(f) = Spring constant (K) * Extension of material (x)
Here, f = 12 N
x = 0.070 m

Substitute their values,   
12 = k * 0.070
k = 12 / 0.070
k = 171.43 N/m

In short, Your Answer would be 171.43 N/m

Hope this helps!
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Where have you heard the word calorie before what do you think a calorie is
Ainat [17]

Answer:

Explanation:

Calorie is a common term used to describe the amount of energy that can be derived from food products.

We quantify foods based on the calories of energy they possess. A high calorific food will yield more energy to the body and is often desired for intense physical activities.

Calorie is defined as the amount of heat energy needed to raise the temperature of 1g of a substance by 1°C.

Food calories slightly differs in definition and they imply the amount of heat needed to raise the temperature of 1kg of water by 1°C.

6 0
3 years ago
What is the frequency of an X-ray with wavelength 0.13 nm ? Assume that the wave travels in free space. Express your answer to t
dem82 [27]

Answer:

Frequency, f=2.30\times 10^{18}\ Hz

Explanation:

Given that,

The wavelength of the x-rays, \lambda=0.13\ nm=0.13\times 10^{-9}\ m

We need to find the frequency of an x-ray. All electromagnetic wave travel with a speed of light. It is given by the formula as :

c=f\lambda

f is the frequency

f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{0.13\times 10^{-9}}\\\\f=2.30\times 10^{18}\ Hz

So, the frequency of an x-ray is 2.30\times 10^{18}\ Hz. Hence, this is the required solution.

5 0
3 years ago
A car drives past a pole at 40km/hr. Describe the motion from the point of view of a) the car, and b) the pole. Thanks in advanc
ki77a [65]
I was going to beg off until tomorrow, but this one is nothing like those others.
Why, at only 40km/hr, we can ignore any relativistic correction, and just go with Newton.

To put a finer point on it, let's give the car a direction.  Say it's driving North.

a).  From the point of view of the car, its driver, and passengers if any,
the pole moves past them, heading south, at 40 km/hour .

b).  From the point of view of the pole, and any bugs or birds that may be
sitting on it at the moment, the car and its contents whiz past them, heading
north, at 40 km/hour.

c).  A train, steaming North at 80 km/hour on a track that exactly parallels
the road, overtakes and passes the car at just about the same time as
the drama in (a) and (b) above is unfolding.

The rail motorman, fireman, and conductor all agree on what they have
seen. From their point of view, they see the car moving south at 40 km/hr,
and the pole moving south at 80 km/hr.

Now follow me here . . .

The car and the pole are both seen to be moving south.  BUT ... Since the
pole is moving south faster than the car is, it easily overtakes the car, and
passes it . . . going south.

That's what everybody on the train sees.

==============================================

Finally ... since you posed this question as having something to do with your
fixation on Relativity, there's one more question that needs to be considered
before we can put this whole thing away:

You glibly stated in the question that the car is driving along at 40 km/hour ...
AS IF we didn't need to know with respect to what, or in whose reference frame.
Now I ask you ... was that sloppy or what ? ! ? 

Of course, I came along later and did the same thing with the train, but I am
not here to make fun of myself !  Only of others.

The point is . . . the whole purpose of this question, obviously, is to get the student accustomed to the concept that speed has no meaning in and of itself, only relative to something else.  And if the given speed of the car ...40 km/hour ... was measured relative to anything else but the ground on which it drove, as we assumed it was, then all of the answers in (a) and (b) could have been different.

And now I believe that I have adequately milked this one for 50 points worth.


7 0
3 years ago
Help me quick!!! please!!
Rudik [331]

Answer:

39.240 W

Explanation:

Let's start by calculating the work done by the engine. We can assume that it is the same work done by the weight of the object to bring it from 40m to the surface: as much energy it takes to bring it up, the same ammount it takes to bring it down. Said work is w= \vec F\cdot \vec{h} = mg h = 1000 \times 9.81\times 40 = 392.400 J

At this point we can simply apply the definition of power, that is P = \frac wt, to get the power of the engine is 39.240 W

4 0
3 years ago
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Yuliya22 [10]

Answer:

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3 0
3 years ago
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