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sashaice [31]
3 years ago
11

A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.

Physics
1 answer:
MakcuM [25]3 years ago
4 0

Answer:

b. 9.5°C

Explanation:

m_i = Mass of ice = 50 g

T_i = Initial temperature of water and Aluminum = 30°C

L_f = Latent heat of fusion = 3.33\times 10^5\ J/kg^{\circ}C

m_w = Mass of water = 200 g

c_w = Specific heat of water = 4186 J/kg⋅°C

m_{Al} = Mass of Aluminum = 80 g

c_{Al} = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C

The final equilibrium temperature is 9.50022°C

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Explanation:

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3 years ago
Does the theory of relativity show that Newtonian mechanics is wrong?
valina [46]

Answer:

Einstein extended the rules of Newton for high speeds. For applications of mechanics at low speeds, Newtonian ideas are almost equal to reality. That is the reason we use Newtonian mechanics in practice at low speeds.

Explanation:

<em>But on a conceptual level, Einstein did prove Newtonian ideas quite wrong in some cases, e.g. the relativity of simultaneity. But again, in calculations, Newtonian ideas give pretty close to correct answer in low-speed regimes. So, the numerical validity of Newtonian laws in those regimes is something that no one can ever prove completely wrong - because they have been proven correct experimentally to a good approximation.</em>

4 0
3 years ago
Monochromatic light from a distant source is incident on a slit 0.750 mm wide. On a screen 2.00 m away, the distance from the ce
jasenka [17]

Answer:

\lambda= 506.25 nm

Explanation:

Diffraction is observed when a wave is distorted by an obstacle whose dimensions are comparable to the wavelength. The simplest case corresponds to the Fraunhofer diffraction, in which the obstacle is a long, narrow slit, so we can ignore the effects of extremes.

This is a simple case, in which we can use the Fraunhofer single slit diffraction equation:

y=\frac{m \lambda D}{a}

Where:

y=Displacement\hspace{3}from\hspace{3} the\hspace{3} centerline \hspace{3}for \hspace{3}minimum\hspace{3} intensity =1.35mm\\\lambda=Light\hspace{3} wavelength \\D=Distance\hspace{3}between\hspace{3}the\hspace{3}screen\hspace{3}and\hspace{3}the\hspace{3}slit=2m\\a=width\hspace{3}of\hspace{3}the\hspace{3}slit=0.750mm\\m=Order\hspace{3}number=1

Solving for λ:

\lambda=\frac{y*a}{mD}

Replacing the data provided by the problem:

\lambda=\frac{(1.35\times 10^{-3})*(0.750\times 10^{-3})}{1*2} =5.0625\times 10^{-7}m =506.25nm

7 0
3 years ago
2.10-kg box is moving to the right with speed 8.50 m&gt;s on a horizontal, frictionless surface. At t = 0 a horizontal force is
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Explanation:

Below is an attachment containing the solution.

8 0
3 years ago
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An MRI technician moves his hand from a region of very low (B=0) magnetic field strength into an MRI scanner’s 3.50 T field with
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Answer:

The change in the magnetic flux in the ring is 10.99 Wb.

Explanation:

Given that,

An MRI technician moves his hand from a region of very low (B=0) magnetic field strength into an MRI scanner’s 3.50 T field with his fingers pointing in the direction of the field.

Diameter of the ring, d = 2 cm

Radius, r = 1 cm

It takes 0.4 s to move it into the field. We need to find the change in magnetic flux in the ring. Magnetic flux is given by :

\phi=BA\\\\\phi=B\pi r^2\\\\\phi=3.5\pi \times (1)^2\\\\\phi=10.99\ Wb

So, the change in the magnetic flux in the ring is 10.99 Wb. Hence, this is the required solution.

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3 years ago
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