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Veronika [31]
3 years ago
12

A spring with a spring constant of 34 N/m is stretched 14 m. What is the energy stored in the

Physics
2 answers:
ehidna [41]3 years ago
8 0

Answer:

Elastic potential energy

Elastic potential energy is stored in the spring. Provided inelastic deformation has not happened, the work done is equal to the elastic potential energy stored.

Explanation:

becuse

iris [78.8K]3 years ago
3 0

Answer:10

Explanation:

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HELP!!!! WILL MARK BRAINLIEST!!!! IF YOU LEAVE AN ANSWER EXPLAIN! THANKS
Travka [436]

Answer:

the answer is C

Explanation:

we know this because if you compare the graphs and look at the direction. it isn't always in the explanation or the few sentences they gave you at the top. also, look at the waves, you can see in Davids drawing that it is directly straight up, A and B do not represent that. A isn't even a valid answer. Notice also in A that the arrow is going in the completely different direction than in Davids drawing. B is also going a different direction even though it is only turned a little bit although if it was straight up like Davids drawing then it would most likely be a correct answer. C does have one arrow going a different direction but look at how it has two, showing in which if the waves were to turn then the arrow is still valid

7 0
3 years ago
3. A museum curator moves artifacts into place on various different display
Alex

Explanation:

a. moving a 145 kg aluminum sculpture across a horizontal steel platform

b. pulling a 15 kg steel sword across a horizontal steel shield

c. pushing a 250 kg wood bed on a horizontal wood floor

d. sliding a 0.55 kg glass amulet on a horizontal glass display case

4 0
3 years ago
if an object is being acted on by two forces a push and a pull of 6N what is the net force of the object
Paul [167]
The concepts of force<span>, mass, and weight play critical roles. Newton's Laws of. Motion ... the person stops </span>pushing<span>? ... F </span>net<span> =10 N </span>2<span> N. = 8 N (to the right) a = F </span>net<span> m. = 8 N. 5 kg. =1.6 m s. </span>2<span> ... </span>Two equal forces<span> act on an </span>object<span> in the directions shown. </span>If<span> these ... </span>Two<span> connected carts </span>being accelerated by a force<span> F applied by.</span>
6 0
3 years ago
Considerable scientific work is currently under way to determine whether weak oscillating magnetic fields such as those found ne
VLD [36.1K]

Answer:

The the maximum emf is 2.52\times10^{-11}\ V

Explanation:

Given that,

Magnetic field B = 1.40\times10^{-3}\ T

Frequency = 60 Hz

Diameter = 7.8 μm

We need to calculate the maximum emf

Using formula of emf

\epsilon=NBA\omega

Where, N = number of turns

B= magnetic field

A = area

Put the value in to the formula

\epsilon=1\times1.40\times10^{-3}\times\pi\times(\dfrac{7.8\times10^{-6}}{2})^2\times2\times\pi\times60

\epsilon=2.52\times10^{-11}\ V

Hence, The the maximum emf is 2.52\times10^{-11}\ V

5 0
3 years ago
Two ice skaters, Daniel (mass 70.0 kg) and Rebecca (mass 45.0 kg), are practicing. Daniel stops to tie his shoelace and, while a
wel

Answer:

a) v=7.32m/s

b) \alpha =-35º

c) ΔK=-1094.62J

Explanation:

From the exercise we know that the collision between Daniel and Rebecca is elastic which means they do not stick together

So, If we analyze the collision we got

p_{1x}=p_{2x}

To simplify the problem, lets name D for Daniel and R for Rebecca

a) p_{D1x}+p_{R1x}=p_{D2x}+p_{R2x}

Since Daniel's initial velocity is 0

m_{R}v_{Rx}=m_{D}v_{D2x}+m_{R}v_{R2x}

v_{D2x}=\frac{m_{R}*v_{R1x}-m_{R}*v_{R2x}}{m_{D}}

v_{D2x}=\frac{(45kg)(14m/s)-(45kg)(8cos(55.1)m/s)}{(70kg)}=6m/s

Now, lets analyze the movement in the vertical direction

p_{1y}=p_{2y}

Since p_{1y}=0

0=m_{D}v_{D2y}+m_{R}v_{R2y}

v_{D2y}=-\frac{m_{R}v_{2Ry}}{m_{D}}=-\frac{(45kg)(8sin(55.1)m/s)}{(70kg)}=-4.21m/s

Now, we can find the magnitude of Daniel's velocity after de collision

v_{D}=\sqrt{(6m/s)^2+(-4.21m/s)^2}=7.32m/s

b) To know whats the direction of Daniel's velocity we need to solve the arctan of the angle

\alpha =tan^{-1}(\frac{v_{y}}{v_{x}})=tan^{-1}(\frac{-4.21}{6})=-35º

c) The change in the total kinetic energy is:

ΔK=K_{2}-K_{1}

ΔK=\frac{1}{2}[(45kg)(8m/s)^2+(70kg)(7.32m/s)^2-(45kg)(14m/s)^2]=-1094.62J

That means that the kinetic energy decreases

5 0
3 years ago
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