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Rufina [12.5K]
3 years ago
6

A spring-loaded toy gun is used to launch a 12.0-g plastic ball. The spring which has a spring constant of 15.0 N/m is compresse

d 8.00 cm as the ball is pushed into the barrell. When the trigger is pulled the swing is released and shoots the ball back out horizontally. What is the elastic potential energy stored in the spring when it is compressed 8.00 cm?
Physics
1 answer:
ziro4ka [17]3 years ago
4 0

Answer:

the elastic potencial energy stored when the spring is compressed x=8 cm is K(x)= 0.048 J

Explanation:

since the work is related with the force through

W=∫F dx

for a spring of constant k :

F=k*x , where F= compression force, x= compression length

then for a compression from 0 until x

W=∫F dx = ∫ k*x dx =  k ∫x dx =1/2*k*x² - 1/2*k*0² = V(x) - V(0)

since the work depends only on the final value of compression and not on the process 1/2*k*x² represents the elastic potential energy V stored in the spring, then

V(0) = 1/2*k*0² = 0  

V(x) = 1/2*k*x²

when the spring is compressed x= 8 cm = 0.08m , the elastic potencial energy is

V (x) = 1/2*k*x² = 1/2 * 15 N/m* (0.08m)² = 0.048 J

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Answer:

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6 0
3 years ago
SP1b.
nata0808 [166]

Answer:

2 m/s^2, west

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Vf=final velcoity

Vi=initial velocity

t=timw

a =  \frac{vf - vi}{t}

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\frac{15 - 25}{5}

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The - changes direction and makes it opposite

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3 years ago
List the types of electromagnetic radiation in order from lowest energy photons to highest energy photons.
frutty [35]

radio waves,X-rays,

Explanation:

In order from highest to lowest energy, the sections of the EM spectrum are named: gamma rays, X-rays, ultraviolet radiation, visible light, infrared radiation, and radio waves. Microwaves (like the ones used in microwave ovens) are a subsection of the radio wave segment of the EM spectrum.

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Which type of reaction is this?<br> CH3COOH + H20 2 H+ + CH3C00
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dissociation of acetic acid in water

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Starting from rest, a disk rotates about its central axis with constant angular acceleration. in 6.00 s, it rotates 44.5 rad. du
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\theta=\dfrac\alpha2t^2

It rotates 44.5 rad in this time, so we have

44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}

b. Since acceleration is constant, the average angular velocity is

\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2

where \omega_f is the angular velocity achieved after 6.00 s. The velocity of the disk at time t is

\omega=\alpha t

so we have

\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}

making the average velocity

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c. We already found this using the first method in part (b),

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so this is just a matter of plugging in t=12.0\,\mathrm s. We get

\theta=179\,\mathrm{rad}

Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2

Then for t=6.00\,\rm s we would get the same \theta=179\,\rm rad.

7 0
3 years ago
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