The change of state from a gas to a liquid?

You are out running and the first mile takes you 10 minutes. The second mile takes you 20 minutes. This is an example of

cupoosta [38]

If Equal distance is traveled in equal interval of time then it is known as uniform motion in which velocity of object will remain same.

Then if distance covered will be same and the time taken to cover same distance is decreasing then it shows that speed is increasing with time due to which it took less time to cover same distance. This is also known as positive acceleration.

Now if the distance covered will be same and time taken to cover same distance is increasing then it shows that speed is decreasing with time due to which it took more time to cover the same distance. This is also known as negative acceleration.

Now in the above case it is given that the first mile takes you 10 minutes. The second mile takes you 20 minutes. So the time taken is increasing while we cover same distance so this is an example of <u>Negative Acceleration</u>

5 0

3 years ago

A Truck with a mass of 1500 kg is decelerated At a rate of 5m/s2. how much force did this require

Marianna [84]

(1500 kg)*(5 m/s^2) = 7500 N

6 0

3 years ago

A photon of wavelength 7.33 pm scatters at an angle of 157° from an initially stationary, unbound electron. What is the de Brogl

Ann [662]

Answer:

4.63 p.m.

Explanation:

The problem given here can be solved by the Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=157^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42(1-cos157^\circ ) p.m.$

Therfore,

$\lambda^{'}-\lambda=4.64 p.m.$

Here, the photon's incident wavelength is $\lamda=7.33pm$

So,

$\lambda^{'}=7.33+4.64=11.97 p.m$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

here, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda$ and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therfore,

$\lambda_{e}=\frac{7.33\times 11.97}{\sqrt{7.33^{2}+11.97^{2}-2\times 7.33\times 11.97\times cos157^\circ }} p.m.\\\lambda_{e}=\frac{87.7401}{18.935} = 4.63 p.m.$

This is the de Broglie wavelength of the electron after scattering.

8 0

3 years ago

What is the peacocks displacement from 2 to 3 seconds

mash [69]

1.5 second difference

7 0

3 years ago

A parallel plate capacitor with air between the plates has a potential difference of 71.0 V. Determine the potential difference

Gwar [14]

Answer:

15.8 V

Explanation:

The relationship between capacitance and potential difference across a capacitor is:

$q=CV$

where

q is the charge stored on the capacitor

C is the capacitance

V is the potential difference

Here we call C and V the initial capacitance and potential difference across the capacitor, so that the initial charge stored is q.

Later, a dielectric material is inserted between the two plates, so the capacitance changes according to

$C'=kC$

where k is the dielectric constant of the material. As a result, the potential difference will change (V'). Since the charge stored by the capacitor remains constant,

$q=C'V'$

So we can combine the two equations:

$CV=CV'\\CV=(kC)V'\\V'=\frac{V}{k}$

and since we have

V = 71.0 V

k = 4.50

We find the new potential difference:

$V'=\frac{71.0}{4.50}=15.8 V$

6 0

3 years ago