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3 years ago

What's the pressure on solids

1 answer:

Studentka2010 [4]3 years ago

5 0

<span>As in all the solids, their atoms are arrange in certain intermolecular distances. When the pressure in the solid, it changes this intermolecular distances which causes selectively changes all those effects which are associated with the interactions between molecules. Pressure changes the phonon frequencies, changes in phonon line shape in the solid. This is also responsible for change in melting and boiling points. The pressure may also cause for phase transition in many solids.</span>

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The x-acis of a trajectory represents its C

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3 years ago

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2. A 56N force is exerted on an 8kg object. What is the object's acceleration?

White raven [17]

acceleration = 7 m/s²

Explanation:

To determine the object's acceleration we use the following formula:

force (N) = mass (kg) × acceleration (m/s²)

acceleration (m/s²) = force (N) / mass (kg)

acceleration = 56 N / 8 kg

acceleration = 7 m/s²

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3 years ago

A 50.0-g object connected to a spring with a force constant of 35.0 n/m oscillates with an amplitude of 4.00 cm on a frictionles

Dimas [21]

A) The total energy of the system is sum of kinetic energy and elastic potential energy:
$E=K+U= \frac{1}{2}mv^2 + \frac{1}{2}kx^2$
where
m is the mass
v is the speed
k is the spring constant
x is the elongation/compression of the spring

The total energy is conserved, so we can calculate its value at any point of the motion. If we take the point of maximum displacement:
$x=A=4.00 cm = 0.04 m$
then the velocity of the system is zero, so the total energy is just potential energy, and it is equal to
$E=U= \frac{1}{2}kA^2 = \frac{1}{2}(35.0 N/m)(0.04 m)^2=0.028 J$

b) When the position of the object is
$x=1.00 cm = 0.01 m$
the potential energy of the system is
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.01 m)^2 = 1.75 \cdot 10^{-3} J$
and so the kinetic energy is
$K=E-U=0.028 J - 1.75 \cdot 10^{-3}J =0.026 J$
since the mass is $m=50.0 g=0.05 kg$ , and the kinetic energy is given by
$K= \frac{1}{2}mv^2$
we can re-arrange the formula to find the speed of the object:
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.026 J}{0.05 kg} }=1.02 m/s$

c) The potential energy when the object is at
$x=3.00 cm=0.03 m$
is
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J$
Therefore the kinetic energy is
$K=E-U=0.028 J-0.016 J = 0.012 J$

d) We already found the potential energy at point c, and it is given by
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J$

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3 years ago

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Arturiano [62]

Answer:

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Explanation:

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