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An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr

Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

$F= mg$

Where, f = restoring force

$kx=mg$

$k=\dfrac{mg}{x}$

Put the value into the formula

$k=\dfrac{2.15\times9.8}{0.0895}$

$k=235.41\ N/m$

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

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

Here, $v = A\omega$

$K.E=\dfrac{1}{2}m\times(A\omega)^2$

Here, $\omega=\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}kA^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2$

$K.E=0.06500\ J$

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

8 0

2 years ago

liquid helium has a very low boiling point, 4.2 k, as well as a very low latent heat of vaporization, 2.00 104 j/kg. if energy i

aksik [14]

$4.80 \times 10^3 \text { seconds }$ long does it take to boil away 2.40 kg of the liquid.

Boiling point of He is $T=4.2 \mathrm{k}$

Latent heat of vapourization $L=2.00 \times 10^4 \mathrm{~J} / \mathrm{kg}$

Power of electrical heater $P=30 \mathrm{w}$

mass of liquid is $m=2.40 \mathrm{~kg}$

amount of heat required to boil

$\begin{aligned}&Q=m L \\&Q=2.40 \times 2 \times 10^4 \mathrm{~J} \\&Q=4.80 \times 10^4 \mathrm{~J}\end{aligned}$

Power $p=\frac{\text { work }}{\text { time }}=\frac{\text { Energy }}{\text { Time }}$

$\begin{aligned}P &=\frac{Q}{t} \\\text { tine } t &=\frac{Q}{P}=\frac{4.80 \times 10^4 \mathrm{~J}}{10} \\t &=4.80 \times 10^3 \text { seconds }\end{aligned}$

The heat or energy that is absorbed or released during a substance's phase shift is known as latent heat. It could go from a solid to a liquid or from a liquid to a gas, or vice versa. Enthalpy, a characteristic of heat, is connected to latent heat.

The heat that is used or lost as matter melts and transitions from a solid to a fluid form at a constant temperature is known as the latent heat of fusion.

Due to the fact that during softening the heat energy anticipated to transform the substance from solid to fluid at air pressure is the latent heat of fusion and that the temperature remains constant during the process, the "enthalpy" of fusion is a latent heat. The enthalpy change of any quantity of material during dissolution is known as the latent heat of fusion.

For learn more about Latent heat of vaporization, visit: brainly.com/question/14980744

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3 0

8 months ago

I need help calculating Velocity,

velikii [3]

Average velocity = (800+1600)/(4+10)
= 171.42m/s

5 0

2 years ago

Read 2 more answers

A circular window of 30 cm diameter in a submarine can withstand a maximum force of 5.20 × 105 N. The maximum depth in a lake to

Svetach [21]

Answer: Option (b) is the correct answer.

Explanation:

The given data is as follows.

F = $5.20 \times 10^{5}$ N

g = 9.8 m/s

radius = $\frac{diameter}{2}$

= $\frac{30 cm}{2}$ = 15 cm = 0.15 m (as 1 m = 100 cm)

Formula to calculate depth is as follows.

F = $\rho \times g \times h \times A$

or, h = $\frac{F}{\rho \times g \times A}$

h = $\frac{5.2 \times 10^{5}}{1000 \times 9.8 \times (3.1416 \times (0.15 m^{2})}$

= 751 m

Thus, we can conclude that the maximum depth in a lake to which the submarine can go without damaging the window is closest 750 m.

3 0

3 years ago

5. An acrobat, starting from rest, swings freely on a trapeze of

34kurt

The energy conservation and trigonometry we can find the results for the questions about the movement of the acrobat are;

a) The maximum speed is v = 4.89 m / s

b) The maximum height is h = 1.22 m

The energy conservation is one of the most fundamental principles of physics, stable that if there are no friction forces the mechanistic energy remains constant. Mechanical energy is the sum of the kinetic energy plus the potential energies.

Em = K + U

Let's write the energy in two points.

Starting point. Highest part of the oscillation

Em₀ = U = m g h

Final point. Lower part of the movement

$Em_f$ = K = ½ m v²

Energy is conserved.

Emo = $Em_f$

m g h = ½ m v²

v² = 2 gh

Let's use trigonometry to find the height, see attached.

h = L - L cos θ

h = L (1- cos θ)

They indicate that the initial angle is tea = 48º and the length is L = 3.7 m, let's calculate.

h = 3.7 (1- cos 48)

h = 1.22 m

this is the maximum height of the movement.

Let's calculate the velocity.

$v= \sqrt{2 \ 9.8 \ 1.22}$

v = 4.89 m / s

In conclusion using the conservation of energy and trigonometry we can find the results for the questions about the movement of the acrobat are;

a) The maximum speed is v = 4.89 m / s

b) The maximum height is h = 1.22 m

Learn more here: brainly.com/question/13010190

5 0

2 years ago