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Nimfa-mama [501]

3 years ago

15

An object of mass m attached to a spring of force constant k oscillates with simple harmonic motion. The maximum displacement fr

om equilibrium is a and the total mechanical energy of the system is e

1 answer:

Lady_Fox [76]3 years ago

6 0

An object of mass m attached to a spring of force constant k oscillates with simple harmonic motion. The system's potential energy when kinetic energy of (3/4) E is (1/8) k A².

<h3>What is mechanical energy?</h3>

Mechanical energy is the sum of potential energy and kinetic energy.

Total mechanical energy = P.E max = K.E max

Total mechanical energy = K.E +P.E

Given is the kinetic energy is (3/4)E.

E= (3/4)E + P.E

P.E = (1/4) E

Maximum potential energy =E = (1/2) k A²

Here. A is the maximum displacement and k is the spring constant.

The potential energy at kinetic energy of (3/4) E is

P.E = (1/4)E = (1/8) k A²

Therefore, the system's potential energy when kinetic energy of (3/4) E is (1/8) k A².

Learn more about mechanical energy.

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A block of mass 3.5 kg, sliding on a horizontal plane, is released with a velocity of 3.3 m/s. The blocks slides and stops at a

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Answer:

If the initial velocity were quadrupled the block would slide a distance of $\Delta x_{2} =25.6\ m$ .

Explanation:

We are told the mass of the block is $m=3.5\ kg$ , the initial velocity in the first case is $v_{i}=3.3\ \frac{m}{s}$ , the final velocity is $v_{f}=0\ \frac{m}{s}$ and the distance at which the block stops in the first case is $\Delta x_{1}=1.6\ m$ .

In the second case we are told the initial velocity is quadrupled.

Because the block doesn't move in the vertical direction the sum of forces is:

$F_{normal}-F_{weight}=0\ \Longrightarrow\ F_{normal}=F_{weight}=m.g$

$F_{normal}=m.g=3.5\ kg.\ 9.8\frac{m}{s^{2}}\ \Longrightarrow\ F_{normal}=34.3\ N$

In the horizontal direction the only force we have is the force of kinetic friction is $F_{friction}=\mu_{d}.\ F_{normal}$ . We don't know the value of the coefficient of kinetic friction $\mu_{d}$ but this force is the same in both cases.

We use that the change in kinetic energy is equal to the work done by $F_{friction}$ :

$W=\Delta K$

$-F_{friction}.\ \Delta x=\frac{1}{2}mv_{f} ^{2}-\frac{1}{2}mv_{i} ^{2}$

Because $v_{f}=0\ \frac{m}{s}$ and $F_{friction}$ is the same in both cases we have that:

$F_{friction}=\frac{1}{2}\frac{m}{\Delta x}\ v_{i} ^{2}$

$\Delta x_{2}$ is the distance the block will travel until stopping when the initial velocity is quadrupled. If we equal the above equation in the first and second case we have that:

$\frac{1}{2}\frac{m}{\Delta x_{1}}\ (v_{i})^{2}=\frac{1}{2}\frac{m}{\Delta x_{2}}\ (4v_{i})^{2}$

If we manipulate the equation we get that:

$\Delta x_{2}=16\ \Delta x_{1}$

$\Delta x_{2}=16\ . \ 1.6\ m$

$\Delta x_{2}=25.6\ m$

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Tungsten, W-181, is a radioactive isotope with a half life of 121 days. If a medical lab purchases 24 kg of W-181, how much will

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A balloon filled with helium gas has an average density of rhob = 0.27 kg/m3. The density of the air is about rhoa = 1.23 kg/m3.

8_murik_8 [283]

Answer: $34.84 m/s^2$

Explanation:

Given

density of balloon $\rho _{He}=0.27 kg/m^3$

density of air $\rho _a=1.23 kg/m^3$

volume of balloon $V=0.084 m^3$

If balloon is rising then

$F_b-mg=ma$

where $F_b=buoyant\ Force=\rho _a\times V\times g$

$mg=\rho _{He}\times V\times g=weight of gas$

$a=acceleration$

$\rho _a\times V\times g-\rho _{He}\times V\times g=\rho _{He}\times V\times a$

$\rho _a\times g-\rho _{He}\times g=\rho _{He}\times a$

divide by $\rho _{He}$

$\frac{\rho _a}{\rho _{He}}\times g-g=a$

$a=g(\frac{\rho _a}{\rho _{He}}-1)$

$a=g(\frac{1.23}{0.27}-1)$

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Determine the pressure P of a 450K Oxygen gas in a gas chamber when its initial pressure is 175Pa has a temp 300K

FrozenT [24]

Answer:

262.5Pa

Explanation:

Given parameters:

P1 = 175Pa

T1 = 300K

T2 = 450K

Unknown:

P2 = ?

Solution:

To solve this problem, we are going to use an adaptation of the combined gas law.

$\frac{P1}{T1}$ = $\frac{P2}{T2}$

Insert the parameters and solve;

$\frac{175}{300}$ = $\frac{P2}{450}$

P2 = 262.5Pa

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