When a car's acceleration is negative, what is happening to the car's motion?

Are positively charged ions form when atoms lose electrons

Sedbober [7]

Answer: Atoms that lose electrons acquire a positive charge as a result because they are left with fewer negatively charged electrons to balance the positive charges of the protons in the nucleus. Positively charged ions are called cations. Most metals become cations when they make ionic compounds.

Explanation: hope this helps

7 0

3 years ago

If two waves are traveling at the same speed, which statement will be true about their wavelengths? 1.The wave with higher frequ

Margarita [4]

I beleive that the wave with higher frequency will have shorter wavelength

4 0

4 years ago

A motorcyclist drives from a to b with the uniform speed of 30 km/h-1 and returns back with the speed of 20 km/h-1.find the aver

maxonik [38]

30+20 =50
For average 50/2=25

4 0

3 years ago

Read 2 more answers

For a proton in the ground state of a 1-dimensional infinite square well, what is the probability of finding the proton in the c

Butoxors [25]

Answer:

The probability of finding the proton at the central 2% of the well is almost exactly 4%

Explanation:

If we solve Schrödinger's equation for the infinite square well, we find that its eigenfunctions are sinusoidal functions, in particular, the ground state is a sinusoidal function for which only half a cycle fits inside the well.

let $L$ be the well's length, the boundary conditions for the wavefunction are:

$\psi(0) = \psi(L) =0$

And Schrödinger's equation is:

$- \frac{\hbar^2}{2m} \frac{d^2\psi}{dx^2} = E\psi$

The solution to this equation are sines and cosines, but the boundary conditions only allow for sine waves. As we pointed out, the ground state is the sine wave with the largest wavelength possible (that is, with the smallest energy).

$\psi_0(x)=\sqrt[]{\frac{2}{L} }\, \sin(\frac{\pi x}{L} )\\$

here the leading constant is just there to normalise the wavefunction.

Now, if we know the wavefunction, we can know what the probability density function is, it is:

$f_X(x) = |\psi|^2$

So in our case:

$f_X(x) = \frac{2}{L} \sin^2(\frac{\pi x}{L})$

And to find the probability of finding the particle in a strip at the centre of the well of width 2% of L we only have to integrate:

$P(X \in [0.49 L, 0.51L ])= \int\limits^{0.49L}_{0.51L } {\frac{2}{L} \sin^2(\frac{\pi x}{L})} \, dx$

If we do a substitution:

$x = u \, L$

We get the integral:

$\int\limits^{0.49}_{0.51 } 2\, \sin^2(\pi u)} \, du$

This integral can be computed analytically, and it's numerical value is .0399868, that is, almost a 4% probability.

5 0

3 years ago

A cylindrically shaped piece of collagen (a substance found in the body in connective tissue) is being stretched by a force that

Marat540 [252]

Answer:

Part(a): The value of the spring constant is $3.11 \times 10^{2}~Kg~s^{-2}$ .

Part(b): The work done by the variable force that stretches the collagen is $1.5 \times 10^{-6}~J$ .

Explanation:

Part(a):

If ' $k$ ' be the force constant and if due the application of a force ' $F$ ' on the collagen ' $\Delta l$ ' be it's increase in length, then from Hook's law

$F = k~\Delta l....................................................................(I)$

Also, Young's modulus of a material is given by

$Y = \dfrac{F/A}{\Delta l/l}...............................................................(II)$

where ' $A$ ' is the area of the material and ' $l$ ' is the length.

Comparing equation ( $I$ ) and ( $II$ ) we can write

$&& Y = \dfrac{l~k}{A}\\&or,& k = \dfrac{Y~A}{l}\\&or,& k = \dfrac{Y~(\pi~r^{2})}{l}$

Here, we have to consider only the circular surface of the collagen as force is applied only perpendicular to this surface.

Substituting the given values in equation ( $III$ ), we have

$k = \dfrac{3.10 \times 10^{6}~N~m^{-2} \times \pi \times (0.00093)^{2}~m^{2}}{.027~m} = 3.11 \times 10^{2}~Kg~s^{-2}$

Part(b):

We know the amount of work done ( $W$ ) on the collagen is stored as a potential energy ( $U$ ) within it. Now, the amount of work done by the variable force that stretches the collagen can be written as

$W = \dfrac{1}{2}k~x^{2} = \dfrac{(\dfrac{F}{k})^{2}k}{2} = \dfrac{F^{2}}{2~k}...................................(IV)$

Substituting all the values, we can write

$W = \dfrac{(3.06 \times 10^{-2})^{2}~N^{2}}{2 \times 3.11 \times 10^{2}~Kg~s^{-2}} = 1.5 \times 10^{-6}~J$

3 0

3 years ago