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Ber [7]
3 years ago
9

On an air track, a 400.5 g glider moving to the right at 2.30 m/s collides elastically with a 500.0 g glider moving in the oppos

ite direction at 2.95 m/s . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Elastic collision on an air track. Part A Find the velocity of first glider after the collision.
Physics
1 answer:
Yanka [14]3 years ago
4 0

Answer:

3.53 m/s towards the left

Explanation:

m_1 = Mass of first glider = 400.5 g

m_2 = Mass of Second glider = 500 g

u_1 = Initial Velocity of first object = 2.3 m/s

u_2 = Initial Velocity of second object = -2.95 m/s

As momentum and Energy is conserved

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

{\tfrac {1}{2}}m_{1}u_{1}^{2}+{\tfrac {1}{2}}m_{2}u_{2}^{2}={\tfrac {1}{2}}m_{1}v_{1}^{2}+{\tfrac {1}{2}}m_{2}v_{2}^{2}

From the two equations we get

v_{1}=\frac{m_1-m_2}{m_1+m_2}u_{1}+\frac{2m_2}{m_1+m_2}u_2\\\Rightarrow v_1=\frac{0.4005-0.5}{0.4005+0.5}\times 2.3+\frac{2\times 0.5}{0.4005+0.5}\times -2.95\\\Rightarrow v_1=-3.53\ m/s

The velocity of first glider after collision is 3.53 m/s towards the left

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TEA [102]

Answer:

Reverse Osmosis and UV Treatment

Explanation:

Reverse Osmosis, commonly known as RO, process is very effective in removing salt content from water. This process is known as desalination. IN RO, a partially permeable membrane is used to remove unwanted molecules, ions, and large particles. RO will take care of both desalination and purification of water.

Further UV treatment can be done to kill any remaining microorganisms as these microorganisms can not withstand UV radiation.

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3 years ago
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A microwave oven operates at 2.50 GHzGHz . What is the wavelength of the radiation produced by this appliance? Express the wavel
sattari [20]

Answer:

The wavelength is \lambda  =  1.2  * 10^8 nm

Explanation:

From the question we are told that

   The frequency of operation of the microwave is  f =  2.50 GHz  =  2.50 *10^{9} \ Hz

     Generally the wavelength is mathematically represented as

          \lambda  =  \frac{c}{f}

Here c is the speed of light with value c =  3.0 *10^{8} \  m/s

So  

         \lambda  =  \frac{3.0 *10^{8}}{  2.50 *10^{9}}

=>       \lambda  =  0.12 \  m

converting to nanometer

           \lambda  =  1.2  * 10^8 nm

6 0
4 years ago
Hedea made a study chart about nuclear energy. Which best describes the error in Hedea’s chart? The nuclei of atoms are split ap
Lerok [7]

Option (d) is correct.Fission and fusion convert nuclear energy to both radiant and thermal energy.

fission is a process in which bigger nucleus breaks into two or more smaller nuclei with the liberation of a large amount of energy.

Fusion is a process in which two small nuclei fuse together to from an intermediate size nucleus with the liberation of tremendous amount of energy

During fission and fusion, energy is released in the form of both light (radiation) and heat.we can then convert that energy into useful form.

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3 years ago
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True or False. A projectile is an object that once set in motion, continues in motion by its own inertia.
bazaltina [42]
The answer is true.
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3 years ago
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At what net rate does heat radiate from a 300-m^2 black roof on a night when the roof's temperature is 33.0°C and the surroundin
Fynjy0 [20]

Answer:

24445.85 J/s

Explanation:

Area, A = 300 m^2

T = 33° C = 33 + 273 = 306 k

To = 18° C = 18 + 273 = 291 k

emissivity, e = 0.9

Use the Stefan's Boltzman law

E = \sigma  \times e \times A\times\left ( T^4 -T_{0}^{4}\right )

Where, e be the energy radiated per unit time, σ be the Stefan's constant, e be the emissivity, T be the temperature of the body and To be the absolute temperature of surroundings.

The value of Stefan's constant, σ = 5.67 x 10^-8 W/m^2k^4

By substituting the values

E = 5.64 \times 10^{-8}\times 0.9 \times 300 \times  (306^{4}-291^{4})

E = 24445.85 J/s

7 0
4 years ago
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