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3 years ago

600 J of work is done on a system in a process that decreases the thermal energy of the system by 300 J.

Physics

1 answer:

Daniel [21]3 years ago

6 0

Answer:

900 J

Explanation:

$\Delta U$ = Change in entropy = -300 J (Decrease)

$W$ = Work done = 600 J

$Q$ = Heat transferred

Change in internal energy is given by

$\Delta U=W+Q\\\Rightarrow Q=\Delta U-W\\\Rightarrow Q=-300-600\\\Rightarrow Q=-900\ J$

The heat transferred to or from the system as heat is -900 J.

900 J is transferred. As heat is transferred from the system the sign is negative.

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A spring stretches 0.150 m when a 0.30 kg mass is hung from it. The spring is then stretched an additional 0.100 m from this equ

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Answer:

a) $k=19.6N/m$

b) $V_m=0.81m/s$

c) $a_m=6.561m/s^2$

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e) $T=0.78sec$ & $F=1.29sec$

f) $mx'' + kx' =0$

Explanation:

From the question we are told that:

Stretch Length $L=0.150m$

Mass $m=0.30kg$

Total stretch length $L_t=0.150+0.100=>0.25$

Generally the equation for Force F on the spring is mathematically given by

$F=-km\\\\k=F/m\\\\k=\frac{m*g}{x}\\\\k=\frac{0.30*9.8}{0.15}$

$k=19.6N/m$

b)Generally the equation for Max Velocity of Mass on the spring is mathematically given by

$V_m=A\omega$

Where

A=Amplitude

$A=0.100m$

And

$\omega=angulat Velocity\\\\\omega=\sqrt{\frac{k}{m}}\\\\\omega=\sqrt{\frac{19.6}{0.3}}\\\\\omega=8.1rad/s$

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$a_m=\omega^2A$

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$a_m=6.561m/s^2$

Generally the equation for Total mechanical energy of Mass on the spring is mathematically given by

$K.E=\frac{1}{2}mv^2$

$K.E=\frac{1}{2}*0.3*0.8^2$

$K.E=0.096J$

Generally the equation for the period T is mathematically given by

$\omega=\frac{2\pi}{T}$

$T=\frac{2*3.142}{8.1}$

$T=0.78sec$

Generally the equation for the Frequency is mathematically given by

$F=\frac{1}{T}$

$F=1.29sec$

Generally the Equation of time-dependent vertical position of the mass is mathematically given by

$mx'' + kx' =0$

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'= signify order of differentiation

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Students perform a set of experiments by placing a block of mass m against a spring, compressing the spring a distance x along a

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Increasing the angle of inclination of the plane decreases the velocity of the block as it leaves the spring.

The statement that indicates how the relationship between v and x changes is; As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of x.

Reasons:

The energy given to the block by the spring = $\mathbf{0.5 \cdot k \cdot x^2}$

According to the principle of conservation of energy, we have;

On a flat plane, energy given to the block = $0.5 \cdot k \cdot x^2$ = kinetic energy of

block = $0.5 \cdot m \cdot v^2$

Therefore;

0.5·k·x² = 0.5·m·v²

Which gives;

x² ∝ v²

x ∝ v

On a plane inclined at an angle θ, we have;

The energy of the spring = $\mathbf{0.5 \cdot k \cdot x^2}$

The force of the weight of the block on the string, $F = m \cdot g \cdot sin(\theta)$

The energy given to the block = $0.5 \cdot k \cdot x^2 - m \cdot g \cdot sin(\theta)$ = The kinetic energy of block as it leaves the spring = $\mathbf{0.5 \cdot m \cdot v^2}$

Which gives;

$0.5 \cdot k \cdot x^2 - m \cdot g \cdot sin(\theta) = 0.5 \cdot m \cdot v^2$

Which is of the form;

a·x² - b = c·v²

a·x² + c·v² = b

Where;

a, b, and c are constants

The graph of the equation a·x² + c·v² = b is an ellipse

Therefore;

As x increases, v increases, however, the value of v obtained will be lesser than the same value of x as when the block is on a flat plane.

Please find attached a drawing related to the question obtained from a similar question online

The possible question options are;

As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of x

The relationship is no longer linear and v will be more for the same value of x

The relationship is still linear, with lesser value of v

The relationship is still linear, with higher value of v

The relationship is still linear, but vary inversely, such that as x increases, v decreases

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