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Elenna [48]
2 years ago
15

A pan hangs from a 50 cm spring. When a 10 kg mass is placed in the pan, it stretches the spring 6 cm. What is a function rule l

(w) that models the length of the spring based on the mass of the object in the pan?
Physics
1 answer:
MrRissso [65]2 years ago
3 0

Answer:

Explanation:

A Spring stretches / compresses when force is applied on them and they are governed by the Hookes Law which states that the force required to stretch or compress a spring is directly proportional to the distance it is stretched.

F = -kx

F is the force applied and x is the elongation of the spring

k is the spring constant.

negative sign indicates the change in direction from equilibrium position.

In the given question, we dont have force but we know that the pan is hanging. We also know from the Newton's second law of motion that

F=mg

Inserting this into Hooke's Law

mg=-kx

computing it for x,

-x=mg/k

This is the model which will tell the length of the spring against change in the mass located in the pan.

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what is the energy (in j) of a photon required to excite an electron from n = 2 to n = 8 in a he⁺ ion? submit an answer to three
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Since the result needs to be accurate to three significant figures, keep at least four significant figures in the calculations.

Look up the Rydberg constant for hydrogen: R_{\text{H}} \approx 1.0968\times 10^{7}\; {\rm m^{-1}.

Look up the speed of light in vacuum: c \approx 2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}.

Look up Planck's constant: h \approx 6.6261 \times 10^{-34}\; {\rm J \cdot s}.

Apply the Rydberg formula to find the wavelength \lambda (in vacuum) of the photon in question:

\begin{aligned}\frac{1}{\lambda} &= R_{\text{H}} \, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\end{aligned}.

The frequency of that photon would be:

\begin{aligned}f &= \frac{c}{\lambda}\end{aligned}.

Combine this expression with the Rydberg formula to find the frequency of this photon:

\begin{aligned}f &= \frac{c}{\lambda} \\ &= c\, \left(\frac{1}{\lambda}\right) \\ &= c\, \left(R_{\text{H}}\, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\right) \\ &\approx (2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}) \\ &\quad \times (1.0968 \times 10^{7}\; {\rm m^{-1}}) \times \left(\frac{1}{2^{2}} - \frac{1}{8^{2}}\right)\\ &\approx 7.7065 \times 10^{14}\; {\rm s^{-1}} \end{aligned}.

Apply the Einstein-Planck equation to find the energy of this photon:

\begin{aligned}E &= h\, f \\ &\approx (6.6261 \times 10^{-34}\; {\rm J \cdot s}) \times (7.7065 \times 10^{14}\; {\rm s^{-1}) \\ &\approx 5.11 \times 10^{-19}\; {\rm J}\end{aligned}.

(Rounded to three significant figures.)

6 0
2 years ago
The freezer compartment in a conventional refrigerator can be modeled as a rectangular cavity 0.3 m high and 0.25 m wide with a
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Answer:

Thickness of Styrofoam insulation is 0.02741 m.

Explanation:

Given that,

Height = 0.25 m

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Power = 400 W

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Using formula of area

A=2(lb+bh+hl)

Put the value into the formula

A=2(0.3\times0.5+0.25\times0.5+0.5\times0.3)

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Outer surface temperature of freezer

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We need to calculate the thickness of Styrofoam insulation

Using Fourier law,

q=\dfrac{kA}{L}(T_{o}-T_{i})

L=\dfrac{kA}{q}(T_{o}-T_{i})

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Hence, Thickness of Styrofoam insulation is 0.02741 m.

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