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3 years ago

What is the best approximate value for the elastic potential energy (EPE) of the spring elongated by 3.0 meters?

Physics

1 answer:

DIA [1.3K]3 years ago

3 0

The elastic potential energy of the spring is 6.8 J

Explanation:

The elastic potential energy of a compressed/stretched spring is given by the equation:

$E=\frac{1}{2}kx^2$

where

k is the spring constant

x is the elongation of the spring

The spring constant of the spring in this problem can be found by keeping in mind the relationship between force (F) and elongation (x) (Hooke's law):

$F=kx$

By looking at the graph and comparing it with the formula, we realize that the slope of the force-elongation graph corresponds to the spring constant. Therefore in this case,

$k=\frac{15.0-0}{10.0-0}=1.5 N/m$

Therefore when the spring has a elongation of $x=3.0 m$ , its potential energy is

$E=\frac{1}{2}(1.5)(3.0)^2=6.8 J$

Learn more about potential energy:

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Answer:

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b) The ice reaches first 80°C ,

Explanation:

Since the heat Q that should be provided to ice

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since the rate of heat addition q = constant and the time t taken to reach the final temperature is the same , then the heat absorbed Q=q*t is the same for both, therefore

m water * c water * ( T final - T equil ) = m ice* [c ice *( T equil -T initial ) + L + c water * ( T final - T equil)]

m water/ m ice = [c ice * ( T equil -T initial ) + L + c water * ( T final - T equil)]/ [ c water * ( T final - T equil)]

m water/ m ice = [c ice * ( T equil -T initial ) + L ]/[c water * ( T final - T equil) ] + 1

since [c ice * ( T equil -T initial ) + L ]/[c water * ( T final - T equil) ] >0 , then

m water/ m ice > 1

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so the mass of ice is smaller that the mass of water

b) Since the heat Q that should be provided to the ice, starting from 55°C mass would be

Q ice= m ice * c water * ( T final2 - T final1 )

and for the water mass

Q water = m water * c water * ( T final2 - T final1 )

dividing both equations

Q water / Q ice = m water / m ice >1

thus

Q water > Q ice

since the heat addition rate is constant

Q water = q* t water and Q ice=q* t ice

therefore

q* t water > q* t ice

t water > t ice

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3 years ago

Find the energy in Joules required to lift a 55.0 Megagram object a distance of 500 cm.

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Energy to lift something =

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This simple formula only works if you use the right units.

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Height . . . meters

For this question . . .

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So we have ...

Energy = (5.5 x 10⁴ kilogram) x (9.8 m/s²) x (5 m)

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Answer:

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