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Simora [160]
3 years ago
9

A ball rolls horizontally off an incline, at velocity of 8.00m/s l. If the incline is 3.50m above the ground below determine:

Physics
1 answer:
Nesterboy [21]3 years ago
8 0

Answer:

sadsadasadsdassda

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3 years ago
Light shines through a single slit whose width is 5.7 10-4 m. A diffraction pattern is formed on a flat screen located 4.0 m awa
polet [3.4K]

Answer: 557nm

Explanation:

Visible light has been known to have its wavelengths in the range between 400 to 700 nanometres (nm), or 4*10^-7 to 7*10^-7 m. This wavelength is between infrared and ultraviolet. Infrared has longer wavelengths and while on the other hand ultraviolet has shorter wavelengths. The wavelength of visible light has a frequency of approximately 430 to 750 terahertz (THz).

tanΦ= 0.0039/4 = 9.75*10^-4

Φ= tan^-1 (9.75*10^-4)

Φ= 0.056

λ= (5.7*10^-4) * sin 0.056

λ= 5.57*10^-7m

λ= 557nm

5 0
4 years ago
What is the mass of 9.11 moles of<br> ozone, 03?
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72) What is the freezing point (°C) of a solution prepared by dissolving 11.3 g of Ca(NO3)2 (formula weight = 164 g/mol) in 115
Rina8888 [55]

<u>Answer:</u> The freezing point of solution is -3.34°C

<u>Explanation:</u>

Vant hoff factor for ionic solute is the number of ions that are present in a solution. The equation for the ionization of calcium nitrate follows:

Ca(NO_3)_2(aq.)\rightarrow Ca^{2+}(aq.)+2NO_3^-(aq.)

The total number of ions present in the solution are 3.

  • To calculate the molality of solution, we use the equation:

Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

Where,

m_{solute} = Given mass of solute (Ca(NO_3)_2) = 11.3 g

M_{solute} = Molar mass of solute (Ca(NO_3)_2) = 164  g/mol

W_{solvent} = Mass of solvent (water) = 115 g

Putting values in above equation, we get:

\text{Molality of }Ca(NO_3)_2=\frac{11.3\times 1000}{164\times 115}\\\\\text{Molality of }Ca(NO_3)_2=0.599m

  • To calculate the depression in freezing point, we use the equation:

\Delta T=iK_fm

where,

i = Vant hoff factor = 3

K_f = molal freezing point depression constant = 1.86°C/m.g

m = molality of solution = 0.599 m

Putting values in above equation, we get:

\Delta T=3\times 1.86^oC/m.g\times 0.599m\\\\\Delta T=3.34^oC

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

\Delta T=\text{freezing point of water}-\text{freezing point of solution}

\Delta T = 3.34 °C

Freezing point of water = 0°C

Freezing point of solution = ?

Putting values in above equation, we get:

3.34^oC=0^oC-\text{Freezing point of solution}\\\\\text{Freezing point of solution}=-3.34^oC

Hence, the freezing point of solution is -3.34°C

4 0
4 years ago
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