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cupoosta [38]
3 years ago
8

Which formula can be used to find the tangential speed of an orbiting object?

Physics
2 answers:
Naily [24]3 years ago
8 0

Answer:

I think the answer is C. V= G  m central/r

Explanation:

Luden [163]3 years ago
6 0
The correct formula for calculating the tangential speed of an orbiting object is V(t)=wr.
V(t)= Tangential Speed
w= Angular Velocity
r= Radius of the Path

Hope this helps.
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A circuit is constructed with six resistors and two batteries as shown. the battery voltages are v1 = 18 v and v2 = 12 v. the po
VladimirAG [237]

Answer:

V4=9.197v

Explanation:

Given:

V1= 18v ,V2= 12v ,r1=r5=58ohms ,r2=r6=124ohms , r3=47ohms ,r4= 125ohms

V4= I4R4 = V2/(R4 + R5)×R4

V4= 12×125 /(125 + 58)

V4=1500/183 =9.197v

5 0
3 years ago
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 capacitor's dielectric material is a vacuum. The capacitor's dielectric constant, or K, will be equal to
dlinn [17]

If  a capacitor's dielectric constant is vacuum its dielectric constant is k will be equal to 1.

<u>Explanation:</u>

Relative permittivity of  a dielectric substance is referred to as its dielectric constant. Relative permittivity/dielectric constant k is a dimensionless quantity that is the ratio of absolute permittivity and vacuum permittivity.

It is given by the expression

k=k=ε /ε0

where ε denotes absolute permittivity and  ε0 denotes permittivity of vacuuum.

Absolute permittivity  ε of vacuum= ε0

therefore k= ε0/ ε0=1

dielectric constant of vacuum is 1 .

4 0
3 years ago
A uniform solid disk and a uniform hoop are placed side by side at the top of an incline of height h. If they are released from
svlad2 [7]

Answer:

) the uniform disk has a lower moment of inertia and arrives first.

Explanation:

(a) the uniform disk has a lower moment of inertia and arrives first.

(b) Let's say the disk has mass m and radius r, and

the hoop has mass M and radius R.

disk: initial E = PE = mgh

I = ½mr², so KE = ½mv² + ½Iω² = ½mv² + ½(½mr²)(v/r)² = (3/4)mv² = mgh

m cancels, leaving v² = 4gh / 3

hoop: initial E = Mgh

I = MR², so KE = ½MV² + ½(MR²)(V/R)² = MV² = Mgh

M cancels, leaving V² = gh

Vdisk = √(4gh/3) > Vhoop = √(gh)

3 0
3 years ago
Which of the following has a mass of approximately one gram?
Vilka [71]
A paper clip is a example of a thing that has a mass of one gram.
3 0
3 years ago
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