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3 years ago

4. The chart shows melting and boiling points of

Physics

1 answer:

kondor19780726 [428]3 years ago

6 0

Answer:

The correct option is A

Explanation:

Firstly, it should be noted that the freezing point of a substance can be assumed to be melting point of that substance because a substance will normally change from liquid to solid (freezes) at the same point it changes from solid to liquid (melts). For example, water freezes at 0°C and also starts melting at 0°C.

Thus, the substance with the lowest melting point among the substances mentioned in the question is alcohol (ethanol) with the melting point of -114°C. Hence, <u>ethanol also has the lowest freezing point thereby freezing at the lowest temperature.</u>

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Calculate the de Broglie wavelength of: a) A person running across the room (assume 180 kg at 1 m/s) b) A 5.0 MeV proton

solmaris [256]

Answer:

$\lambda = 3.68 *10^{-36} \ m$

$\lambda_p = 1.28*10^{-14} \ m$

Explanation:

From the question we are told that

The mass of the person is $m = 180 \ kg$

The speed of the person is $v = 1 \ m/s$

The energy of the proton is $E_ p = 5 MeV = 5 *10^{6} eV = 5.0 *10^6 * 1.60 *10^{-19} = 8.0 *10^{-13} \ J$

Generally the de Broglie wavelength is mathematically represented as

$\lambda = \frac{h}{m * v }$

Here h is the Planck constant with the value

$h = 6.62607015 * 10^{-34} J \cdot s$

$\lambda = \frac{6.62607015 * 10^{-34}}{ 180 * 1 }$

=> $\lambda = 3.68 *10^{-36} \ m$

Generally the energy of the proton is mathematically represented as

$E_p = \frac{1}{2} * m_p * v^2_p$

Here $m_p$ is the mass of proton with value $m_p = 1.67 *10^{-27} \ kg$

=> $8.0*10^{-13} = \frac{1}{2} * 1.67 *10^{-27} * v^2$

=> $v _p= \sqrt{\frac{8.0 *10^{-13}}{ 0.5 * 1.67 *10^{-27}} }$

=> $v = 3.09529 *10^{7} \ m/s$

$\lambda_p = \frac{h}{m_p * v_p }$

so $\lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }$

=> $\lambda_p = 1.28*10^{-14} \ m$

5 0

3 years ago

Two college teams, A and B, are playing a game of tug of war. Team A applies a force of 103 newtons, but the rope does not move

Kipish [7]

Given that the rope is not moving (acceleration is zero), by the second Law of Newton (F=m*a), the net force acting on the rope is zero.

Then, the force applied by the team B equals the force applied by the tema A: 103 N.

7 0

3 years ago

Read 2 more answers

How much energy is required to heat 70 g of water at 20°C to boiling

choli [55]

Answer:

$Q=23,430J$