Sound is repeated near hills.why

How can you prove that acceleration is a derived unit

Aliun [14]

a = (dx / dt)²

Explanation: Unit of distance is m (metres) and unit of time is s (seconds) speed v is first derivative of distance x versus time:

v = dx / dt, unit is m/s. Acceleration is second derivative of

speed versus time a = (dx / dt)² = (dv/dt) , unit is m/s²

7 0

3 years ago

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A quantum system has three states, with energies 0, 1.6 × 10-21, and 1.6 × 10-21, in Joules. It is coupled to an environment wit

xenn [34]

To develop the problem it is necessary to apply two concepts, the first is related to the calculation of average data and the second is the Boltzmann distribution.

Boltzmann distribution is a probability distribution or probability measure that gives the probability that a system will be in a certain state as a function of that state's energy and the temperature of the system. It is given by

$z = \sum\limit_i e^{-\frac{\epsilon_i}{K_0T}}$

Where,

$\epsilon_i =$ energy of that state

k = Boltzmann's constant

T = Temperature

With our values we have that

T= 250K

$k = 1.381*10^{23} m^2 kg s^{-2} K^{-1}$

$\epsilon_1=0J$

$\epsilon_2=1.6*10^{-21}J$

$\epsilon_3=1.6*10^{-21}J$

To make the calculations easier we can assume that the temperature and Boltzmann constant can be summarized as

$\beta = \frac{1}{kT}$

$\beta = \frac{1}{(1.381*10^{23} m^2)(250)}$

$\beta = 2.9*10^{20}J$

Therefore the average energy would be,

$\bar{\epsilon} =\frac{\sum \epsilon_i e^{-\beta \epsilon_i}}{\sum e^{-\beta \epsilon_i}}$

Replacing with our values we have

$\bar{\epsilon} = \frac{0e^{-0}+1.6*10^{-21}*e^{-\Beta(1.6*10^{-21})}+1.6*10^{-2-1}*e^{-(2.9*10^{20})(1.6*10^{-21})}}{1+2e^{-2.9*10^{20}*1.6*10^{-21}}}$

$\bar{\epsilon} = 0.9*10^{-22}J$

Therefore the average internal energy is $\bar{\epsilon} = 0.9*10^{-22}J$

3 0

3 years ago

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Bingel [31]

C because it has 6 on the outer shell. 6+2=8 8 is a full shell

8 0

3 years ago

A charge of 7.00 mC is placed at opposite corners corner of a square 0.100 m on a side and a charge of -7.00 mC is placed at oth

andrew-mc [135]

Answer:

4.03\times10^{7}N[/tex], 135°

Explanation:

charge, q = 7 mC = 0.007 C

charge, - q = - 7 mC = - 0.007 C

d = 0.1 m

Let the force on charge placed at C due to charge placed at D is FD.

$F_{D}=\frac{kq^{2}}{DC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FD is along C to D.

Let the force on charge placed at C due to charge placed at B is FB.

$F_{B}=\frac{kq^{2}}{BC^{2}}$

$F_{B}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FB is along C to B.

Let the force on charge placed at C due to charge placed at A is FA.

$F_{A}=\frac{kq^{2}}{AC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1 \times\sqrt{2} \times 0.1 \times\sqrt{2}}=2.205 \times 10^{7}N$

The direction of FA is along A to C.

The net force along +X axis

$F_{x}=F_{A}Cos45-F_{D}$

$F_{x}=2.205\times10^{7}Cos45-4.41\times10^{7}=-2.85\times10^{7}N$

The net force along +Y axis

$F_{y}=F_{B}-F_{A}Sin45$

$F_{y}=4.41\times10^{7}-2.205\times10^{7}Sin45=2.85\times10^{7}N$

The resultant force is given by

$F=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{(-2.85\times10^{7})^{2}+(2.85\times10^{7})^{2}}$

$F = 4.03\times10^{7}N$

The angle from x axis is Ф

tan Ф = - 1

Ф = -45°

Angle from + X axis is 180° - 45° = 135°

5 0

3 years ago

a box is at rest on a ground. An unbalance force acts on the box causes it to start. The box moves 10 meters in 2 seconds and th

nordsb [41]

The box stopped moving because there was a negative acceleration in the x direction caused by friction.
To find the average speed, simply divide the displacement by time: 10 / 2 = 5 m/s

5 0

2 years ago

Sound is repeated near hills.​why

Sound is repeated near hills.why