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3 years ago

Xamine the graph. Select the statement that best describes the energy change in the particles of a substance during melting.

Physics

1 answer:

lesya692 [45]3 years ago

6 0

Answer:

Option B) Absorbed energy results in the change in potential energy.

Explanation:

Please, find attached the graph that accompanies this question.

The melting proces is the change from solid phase to liquid phase. It is represented with the lower flat line with the symbol ΔHfus over it.

The line is flat because the temperature remains constant during this process. Thus, you know the option "C) As the temperature increases during melting, the kinetic energy also increases" is FALSE.

What happens during this process is:

Most of the energy received by the particles from heating, during the melting process, goes to overcome the intermolecular bonds between the particles. This results in increasing the distance between the particles, so the internal potential energy increases. This is what the option "B) Absorbed energy results in the change in potential energy" correctly describes. Hence, option B) is TRUE.

Althoug most of the heat energy received is transformed into potential energy, yet a small part of the heat energy increases a bit the kinetic energy of the particles, because the particles will vibrate faster around their relatively fixed positions. Hence, the option "A) The kinetic energy of the particles remains unchanged" is FALSE.

As for option D) it is not reasonable at all: none chemical or physical priciple can be used to state that the kinetic energy decreases as the particles move farther apart. Thus, this is FALSE.

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A small steel wire of diameter 1.0mm is connected to an oscillator and is under a tension of 5.7N . The frequency of the oscilla

Fiesta28 [93]

Answer:

a) $P = 2 \pi^2 (57 Hz)^2 (0.0054 m)^2 \sqrt{\pi(0.0005m)^2 (7800 \frac{kg}{m^3}) (5.7N)}= 0.349 W$

b) For this case we se that $P \propto A^2 f^2$

Since the power is constant but the frequency is doubled we will see that $A^2 \propto \frac{P}{f^2}$

So the original amplitude is $A_i \propto \sqrt{\frac{P}{f^2}}$

And if the frequency is doubled we have that:

$A^2_f \propto \frac{P}{(2f)^2}= \frac{P}{4f^2}$

$A_f \propto \sqrt{\frac{P}{4f^2}}= \frac{1}{2} \sqrt{\frac{P}{f^2}}$

So then we will see that the amplitude would be reduced the half and for this case:

$A_f = \frac{0.54cm}{2}= 0.27 cm = 0.0027 m$

Explanation:

For this case we have the following data given:

$D= 1mm = 0.001 m$ represent the diameter

$r = D/2= 0.0005m$ represent the radius

$T= 5.7 N$ represent the tension

$f = 57 Hz$ represent the frequency of the oscillator

$A= 0.54 cm = 0.0054 m$ represent the amplitude of the wave

Part a

For this case we can assume that the power transmitted to the wave is the same power of the oscillator. and we have the following formula for the power:

$P= 2 \pi^2 \rho S v f^3 A^2$

This expression can be written in different ways:

$P= 2 \pi^2 \rho S \sqrt{\frac{T}{\mu}} f^2 A^2$

$P= 2\pi^2 \rho S \sqrt{\frac{T}{\rho S}} f^2 A^2$

$P= 2 \pi^2 f^2 A^2 \sqrt{S \rho T}$

Where f is the frequency , T the tension $rho= 7800 \frac{kg}{m^3}$ the density of the steel, A the amplitude and $S= \pi r^2$ the area, so then we have everuthing in order to replace and we got:

$P = 2 \pi^2 (57 Hz)^2 (0.0054 m)^2 \sqrt{\pi(0.0005m)^2 (7800 \frac{kg}{m^3}) (5.7N)}= 0.349 W$

Part b

For this case we se that $P \propto A^2 f^2$

Since the power is constant but the frequency is doubled we will see that $A^2 \propto \frac{P}{f^2}$

So the original amplitude is $A_i \propto \sqrt{\frac{P}{f^2}}$

And if the frequency is doubled we have that:

$A^2_f \propto \frac{P}{(2f)^2}= \frac{P}{4f^2}$

$A_f \propto \sqrt{\frac{P}{4f^2}}= \frac{1}{2} \sqrt{\frac{P}{f^2}}$

So then we will see that the amplitude would be reduced the half and for this case:

$A_f = \frac{0.54cm}{2}= 0.27 cm = 0.0027 m$

8 0

4 years ago

A 6.00-kg box sits on a ramp that is inclined at 37.0° above the horizontal. the coefficient of kinetic friction between the box

forsale [732]

We need to see what forces act on the box:

In the x direction:

Fh-Ff-Gsinα=ma, where Fh is the horizontal force that is pulling the box up the incline, Ff is the force of friction, Gsinα is the horizontal component of the gravitational force, m is mass of the box and a is the acceleration of the box.

In the y direction:

N-Gcosα = 0, where N is the force of the ramp and Gcosα is the vertical component of the gravitational force.

From N-Gcosα=0 we get:

N=Gcosα, we will need this for the force of friction.

Now to solve for Fh:

Fh=ma + Ff + Gsinα,

Ff=μN=μGcosα, this is the friction force where μ is the coefficient of friction. We put that into the equation for Fh.
G=mg, where m is the mass of the box and g=9.81 m/s²

Fh=ma + μmgcosα+mgsinα

Now we plug in the numbers and get:

Fh=6*3.6 + 0.3*6*9.81*0.8 + 6*9.81*0.6 = 21.6 + 14.1 + 35.3 = 71 N

The horizontal force for pulling the body up the ramp needs to be Fh=71 N.

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3 years ago

The ability to react with air is a

seraphim [82]

I already said it but its reactivity

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3 years ago

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A string with a length of 4.00 m is held under a constant tension. The string has a linear mass density of μ = 0.006 kg/m. Two r

Maksim231197 [3]

Answer:

a) λn = 1.6m , λ(n+1) = 1.33

b) F(t) = 2457.6N

Explanation:

L = 4.0m

μ = 0.006kg/m

F1 =F(n) = 400Hz

F2 = F(n+1) = 480Hz

The natural frequencies of normal nodes for waves on a string is

F(n) = n(v / 2L)

F(n) = n / 2L √(F(t) / μ)

where F(t) = tension acting on the string

the speed (v) on a wave is dependent on the tension acting on the string F(t) and mass per unit length μ

v = √(F(t) / μ)

F(n) = frequency of the nth normal node

F(n+1) = frequency of the successive normal node.

frequency of thr nth normal node F(n) = n(v / 2L).......equation (i)

frequency of the (n+1)th normal node F(n+1) = (n+1) * (v / 2L).....equation (ii)

dividing equation (ii) by (i)

F(n+1) / F(n) = [(n+1) * (v / 2L)] / [n (v / 2L)]

F(n+1) / F(n) = (n + 1) / n

F(n +1) / Fn = 1 + (1/n)

1/n = [F(n+1) / Fn] - 1

1/n = (480 / 400) - 1

1/n = 1.2 - 1

1 / n = 0.2

n = 1 / 0.2

n = 5

the wavelength of the resonant nodes (5&6) nodes are

λ = 2L / n

λn = (2 * 4) / 5

λn = 8 / 5

λn = 1.6

λ(n+1) = (2 * 4) / (5 +1)

λ(n+1) = 8 / 6

λ(n+1) = 1.33m

The tension F(t) acting on the string is

v = √(F(t) / μ)

v² = F(t) / μ

F(t) = μv²

but Fn = n(v / 2L)

nv = 2F(n)L

v = 2F(n)L / n

v = (2 * 400 * 4) / 5

v = 3200 / 5

v = 640m/s

substituting v = 640m/s into F(t) = v²μ

F(t) =(640)² * 0.006

F(t) = 2457.6N

5 0

4 years ago

A square loop of wire is held in a uniform 0.24 T magnetic field directed perpendicular to the plane of the loop. The length of

NNADVOKAT [17]

Answer:

Explanation:

Given that,

Magnetic field of 0.24T

B = 0.24T

Field perpendicular to plane i.e 90°

Rate of decrease of length of side of square is 5.4cm/s

dL/dt = 5.4cm/s = 0.054m/s

Since it is decreasing

Then, dL/dt = -0.054m/s

When L is 14cm, what is the EMF induced?

L = 14cm = 0.14m

EMF is give as

ε = - dΦ/dt

Where flux is given as

Φ = BA

Where A is the area of the square

A = L²

Then, Φ = BL²

Substituting this into the EMF

ε = - dΦ/dt

ε = - d(BL²)/dt

B is constant

ε = - Bd(L²)/dt

ε = -2BL dL/dr

ε = -2 × 0.24 × 0.14 × -0.054

ε = 3.63 × 10^-3 V

ε = 3.63mV

8 0

4 years ago